physics lectures

Upload: muhotasim-ahmed

Post on 16-Oct-2015

64 views

Category:

Documents


0 download

DESCRIPTION

A LECTURE ON MODERN PHYSICS

TRANSCRIPT

  • Lecture notes for the spring semester of 2005

    MODERN PHYSICS

    Fam Le Kien

    Department of Applied Physics and Chemistry,

    University of Electro-Communications, Chofu, Tokyo 182-8585, Japan

    Textbook: CONCEPTS OF MODERN PHYSICS (sixth edition, 2003)

    By Arthur Beiser

    McGraw-Hill

    i

  • Contents

    I. The special theory of relativity 1

    A. The Michelson-Morley experiment 1

    B. The special theory of relativity 4

    C. The Galilean transformation 5

    D. The Lorentz transformation 7

    E. Length contraction 9

    F. Time dilation 11

    G. Doppler effect 13

    1. Doppler effect in sound 13

    2. Doppler effect in light 14

    H. The relativity of mass 15

    I. Mass and energy 18

    J. Velocity addition 21

    II. Wave properties of particles 23

    A. De Broglie waves 23

    B. Wave function: de Broglie waves are waves of probability amplitude 24

    C. Describing a wave 25

    D. Phase and group velocities of de Broglie waves 27

    III. Particle diffraction 32

    IV. Uncertainty principle 37

    Average value and standard deviation 40

    Compatible observables 40

    Proof of the uncertainty principle 40

    Uncertainty principle from the particle approach 41

    Uncertainty principle for energy and time 42

    V. Atomic spectra 46

    A. Spectral series 46

    B. The Bohr atom 47

    ii

  • C. Energy levels and spectra 50

    D. Origin of line spectra 51

    VI. Correspondence principle 55

    VII. The lasers 57

    VIII. Quantum mechanics 61

    IX. Schrodinger equation 64

    X. Particle in a box 69

    XI. Finite potential well 75

    XII. Tunnel effect 79

    iii

  • I. THE SPECIAL THEORY OF RELATIVITY

    A. The Michelson-Morley experiment

    In the past, it was assumed that light is a wave propagating in an all-pervading elastic

    medium called the ether. In other words, it was assumed that there exists a universal frame

    of reference. Let us see what this idea means by considering a simple analogy.

    Consider a river of width D which flows with the speed v, see Fig. 1. Two boats start

    out from one bank of the river with the same speed V (with respect to the water). Boat

    A crosses the river to a point on the other bank directly opposite the starting point and

    then returns to the starting point. Boat B heads downstream for the distance D and then

    returns to the starting point. Lets calculate the time required for each round trip.

    We first consider boat A. In order to compensate the water current, boat A must head

    somewhat upstream, see Fig. 2. The upstream component of the velocity of boat A should

    be exactly v in order to cancel out the river current v. The perpendicular component V is the actual speed across the river. We have the relation

    V 2 = V 2 + v2. (1)

    Hence, the actual speed across the river is

    V =V 2 v2 = V

    1 v2/V 2. (2)

    The time for the initial crossing is D/V . The total round-trip time tA is twice D/V , that

    is,

    tA =2D/V1 v2/V 2 . (3)

    The case of boat B is somewhat different. As boat B heads downstream, its speed relative

    to the shore is V + v, and it travels the distance D in the time

    D

    V + v. (4)

    On the return trip, the speed of boat B relative to the shore is V v, and boat B travelsthe distance D in the time

    D

    V v . (5)

    1

  • FIG. 1: Boat A goes directly across the river and returns to its starting point, while boat B heads

    downstream for an identical distance and then returns.

    FIG. 2: Boat A must head upstream in order to compensate for the river current.

    The total round-trip time tB is the sum of these times, namely,

    tB =D

    V + v+

    D

    V v =2D/V

    1 v2/V 2 . (6)

    The ratio between the times tA and tB is

    tAtB

    =1 v2/V 2. (7)

    If we know the common speed V of the two boats and measure the ratio tA/tB, we can

    determine the speed v of the river current.

    2

  • ether current

    FIG. 3: The Michelson-Morley experiment.

    The reasoning used in this problem may be transferred to the analogous problem of the

    passage of light waves through the ether. If there is an ether pervading space, we move

    through it with at least the speed of the earths orbital motion about the sun. From the

    point of view of an observer on the earth, the ether is moving past the earth. To detect this

    motion, we can use a pair of light beams formed by a beamsplitter instead of a pair of boats,

    see Fig. 3. One of these light beams is directed to a mirror along a path perpendicular to the

    ether current. The other beam goes to a mirror along a path parallel to the ether current.

    The optical arrangement is such that both beams return to the same viewing screen. The

    path lengths of the two beams are chosen to be exactly the same.

    If there is no ether current, the two beams will arrive at the screen in phase and will

    interfere constructively to yield a bright field of view. The presence of an ether current,

    however, would cause the beams to have different transit times, so that they would no

    longer arrive at the screen in phase but would interfere destructively. This is the essence of

    the famous experiment performed by American physicists Michelson and Morley in 1887.

    In the Michelson-Morley experiment, although the sensitivity was enough to detect the

    expected ether current, no ether current was detected.

    The negative result of the Michelson-Morley experiment had two consequences. First, it

    3

  • said that the hypothesis of the ether is wrong. Second, it suggested that the speed of light

    in free space is the same everywhere, regardless of any motion of the source or the observer.

    B. The special theory of relativity

    When we speak of motion, we mean motion relative to a frame of reference. Without a

    frame of reference the concept of motion has no meaning. The frame of reference may be

    a road, the earths surface, the sun, the center of our galaxy; but in every case we must

    specify it. The absence of an ether means that there is no universal frame of reference.

    Therefore, all motion exists solely relative to the person or instrument observing it. Should

    we be isolated in the universe, there would be no way in which we could determine whether

    we are in motion or not.

    Inertial frame of reference: An inertial frame of reference is a frame in which Newtons

    first law of motion holds true. In such a frame, an object at rest remains at rest and an

    object in motion continues to move at constant velocity if no force acts on it. Any frame

    of reference that moves at constant velocity with respect to an inertial frame is itself an

    inertial frame.

    The special theory of relativity, developed by Einstein in 1905, treats problems involving

    the motion of inertial frames of references at constant velocity with respect to one another.

    It has a profound influence on all of physics.

    The special theory of relativity is based upon two postulates:

    1) The first postulate: The laws of physics may be expressed in equations having the

    same form in all inertial frames of reference moving at constant velocity with respect to one

    another. This postulate expresses the absence of a universal frame of reference.

    2) The second postulate: The speed of light in free space has the same value in all inertial

    frames of reference. This postulate follows directly from the result of the Michelson-Morley

    experiment.

    The above postulates subvert almost all of the intuitive concepts of time and space we

    form on the basis of our daily experience. A simple example will illustrate this statement.

    Example: We have two boats on a lake, with boat A stationary in the water and boat B

    drifts at the constant velocity v. At the instant that B is abreast of A, a flare is fired by

    somebody on one of the boats. The light from the flare travels uniformly in all directions,

    4

  • according to the second postulate of special relativity. An observer in either boat must see a

    sphere of light expanding with himself at its center, even though one of the boat is changing

    its position with respect to the point where the flare went off.

    The above situation is unusual. Why? Let us consider a more familiar analog. Instead

    of firing a flare, one of the observers drops a stone into water when the boats are abreast of

    each other. A circular pattern of ripples spreads out. The center of the circular pattern is

    the point where the stone was dropped. Therefore, the pattern appears different to observers

    on each boat.

    It is important to recognize that motion and waves in water are entirely different from

    motion and waves of light in space; water is in itself a frame of reference while space is not,

    and the wave speed in water varies with the observers motion while the wave speed of light

    does not.

    C. The Galilean transformation

    Suppose that we are in a frame of reference S and find that an event occurs at the time

    t and has the coordinates x, y, z. Consider a different frame of reference S moving with

    respect to S at the constant velocity v, see Fig. 4. An observer located in S will find

    that the same event occurs at the time t and has the coordinates x, y, z. How are the

    measurements x, y, z, t related to x, y, z, t.

    For simplicity, we assume that v is in the +x direction and that time in both systems is

    measured from the instant when the origins of S and S coincide. We intuitively expect that

    the measurement x will exceed the measurement x by the amount vt while the measurements

    y, z, t are the same as the measurements y, z, t:

    x = x vt,y = y,

    z = z.

    (8)

    We also intuitively expect that times are the same in both frames of reference:

    t = t. (9)

    The set of equations (8) and (9) is known as the Galilean transformation.

    5

  • FIG. 4: Frame S moves in the +x direction with the speed v relative to frame S.

    To convert velocity components measured in the frame S to their equivalents in the frame

    S , we simply differentiate Eqs. (8) with respect to time. The results are

    V x = Vx v,V y = Vy,

    V z = Vz.

    (10)

    The Galilean transformation and the corresponding velocity transformation are in accord

    with our intuitive expectations. However, they violate both of the postulates of special

    relativity. The first postulate calls for identical equations of physics in both the S and S

    frames of reference, but the fundamental equations of electricity and magnetism assume

    very different forms when Eqs. (8) and (9) are used. The second postulate calls for the

    same value of the speed of light whether determined in S or S . If the speed of light in

    the direction x in the S system is c, then in the system S we have c = c v, not correct.Clearly, a different transformation is required if the postulates of special relativity are to be

    satisfied.

    6

  • D. The Lorentz transformation

    We now develop a set of transformation equations directly from the postulates of special

    relativity. A reasonable guess is

    x = k(x vt), (11)

    where k is a factor of proportionality that does not depend on x and t but may be a function

    of v. The choice of Eq. (11) follows from several considerations: it is linear in x and x, so

    that a single event in the frame S corresponds to a single event in the frame S ; it is simple;

    and it has the possibility of reducing to the equation x = x vt, which is valid in ordinarymechanics. Because the equations of physics must have the same form in both S and S , we

    need only change the sign of v to write the corresponding equation for x in terms of x and

    t:

    x = k(x + vt). (12)

    As in the case of the Galilean transformation, there is nothing to indicate that there

    might be differences between y and y and between z and z. Hence we again take

    y = y

    z = z.(13)

    To get the time transformation, we substitute Eq. (11) into Eq. (94). The result is

    x = k2(x vt) + kvt, (14)

    from which we find that

    t = kt+1 k2kv

    x. (15)

    To find k, we use the second postulate. At the instant t = t = 0, the origins of the two

    frames of reference S and S are in the same place. Suppose that a flare is set off at the

    common origin at t = t = 0, and the observers in each system proceed to measure the speed

    with which the light spreads out. Both observers must find the same speed c, which means

    that the light propagation in the frames S and S is governed by the equations

    x = ct (16)

    and

    x = ct, (17)

    7

  • respectively. Substituting Eqs. (11) and (15) into Eq. (17) yields

    k(x vt) = ckt+ 1 k2

    kvcx. (18)

    Solving for x, we find

    x =ckt+ vkt

    k 1k2kv

    c= ct

    1 + v/c

    1 (1/k2 1)(c/v) . (19)

    The use of Eq. (17) gives1 + v/c

    1 (1/k2 1)(c/v) = 1. (20)

    Hence, we obtain

    k =1

    1 v2/c2 . (21)

    Thus, the transformation equations are

    x =x vt1 v2/c2

    y = y

    z = z

    t =t (vx/c2)1 v2/c2 .

    (22)

    The above transformation is called the Lorentz transformation.

    The inverse Lorentz transformation is

    x =x + vt1 v2/c2

    y = y

    z = z

    t =t + (vx/c2)1 v2/c2 .

    (23)

    The Lorentz equation reduce to the ordinary Galilean equation when the relative velocity

    v of S and S is small compared to the velocity of light c. Therefore, the relativistic effects

    to be explored in the remainder of this section are usually small except for the case where

    enormous velocities are encountered.

    According to the Lorentz transformation, measurements of time and position depend on

    the frame of reference of the observer, so that two events occuring simultaneously in one

    frame at different places need not be simultaneous in another.

    8

  • Simultaneity

    The relative character of time as well as space has many implications. Notable, events

    that seem to take place simultaneously to one observer may not be simultaneous to another

    observer in relative motion, and vice versa.

    Consider two eventsthe setting off of a pair of flaresthat occur at the same time t0 to

    somebody at two different locations x1 and x2. What does the pilot of a spacecraft in flight

    see? To him, the flare at x1 and t0 appears at the time

    t1 =t0 vx1/c21 v2/c2 , (24)

    while the flare at x2 and t0 appears at the time

    t2 =t0 vx2/c21 v2/c2 . (25)

    Since x1 6= x2, we have t1 6= t2. Hence two events that occur simultaneously to one observerare separated by a time interval

    t2 t1 =v(x1 x2)/c2

    1 v2/c2 (26)

    to another observer moving at the speed v relative to the first observer. Thus, simultaneity

    is a relative concept.

    E. Length contraction

    A rod is lying at rest along the x axis of a frame of reference S . The coordinates of its

    ends are x1 and x2. The length L0 of the rod is

    L0 = x2 x1. (27)

    Suppose that we measure the length of the rod from a frame of reference S, parallel to which

    the rod is moving with the velocity v. Will the length L measured in S be the same as the

    length L0 measured in S?

    The length L of the rod in the frame S is determined as

    L = x2 x1, (28)

    9

  • where x1 and x2 are the coordinates of the rod ends measured at the same time t. According

    to the inverse Lorentz transformation, we have

    x1 =x1 vt1 v2/c2

    x2 =x2 vt1 v2/c2 .

    (29)

    Hence, we obtain

    x2 x1 =x2 x11 v2/c2 . (30)

    The use of the definitions (27) and (28) yields

    L0 =L

    1 v2/c2 (31)

    or, equivalently,

    L = L01 v2/c2. (32)

    According to the above equation, the length of an object in motion with respect to an

    observer appears to be shorter than when it is at rest with respect to him. This phenomenon

    is called the Lorentz-FitzGerald contraction or the length contraction.

    The length of an object is a maximum when measured in a reference frame in which the

    object is at rest.

    The relativistic length contraction is negligible for ordinary speeds, but it is an important

    effect at speeds close to the speed of light. A speed of 3000 km/s seems enormous to us, but

    it results in a shortening in the direction of motion by a factor of only

    L

    L0=1 v2/c2 =

    1

    (3000

    3 105)2

    = 0.99995 = 99.995%. (33)

    On the other hand, a body traveling at 0.8 the speed of light is shortened by a factor of

    L

    L0=1 v2/c2 =

    1

    (0.8c

    c

    )2= 0.6 = 60%. (34)

    The Lorentz-FitzGerald contraction occurs only in the direction of the relative motion:

    if v is parallel to x, the y and z dimensions of a moving object are the same in both S and

    S .

    The Lorentz-FitzGerald contraction is a real physical phenomenon and is different from

    the visual effects.

    10

  • F. Time dilation

    Time intervals, too, are affected by relative motion. Clocks moving with respect to an

    observer appear to tick less rapidly than they do when at rest with respect to him. This

    effect is called time dilation.

    Lets consider an event happening at a point x in the frame S . An observer in S

    measures the time. He finds that the event happens from the time t1 to the time t2. The

    duration of the event is

    T0 = t2 t1. (35)

    Meanwhile, an observer in S also measures the time. He finds that the above event happens

    from t1 to t2, where

    t1 =t1 + (v/c

    2)x1 v2/c2 (36)

    and

    t2 =t2 + (v/c

    2)x1 v2/c2 . (37)

    To the observer in S, the duration of the event is

    T = t2 t1 = t2 t11 v2/c2 . (38)

    Hence, we obtain

    T =T0

    1 v2/c2 . (39)

    Clearly, T > T0. Thus, a clock moving with respect to an observer appears to tick less

    rapidly than it does when at rest with respect to him. In other words, a moving clock runs

    more slowly than a stationary clock.

    Example 1: mesons

    We show an interesting manifestation of both the time dilation and the length contraction

    in the decay of unstable particles called mesons. A meson decays into an electron an

    average of T0 = 2 106 s after it comes into being. mesons are created high in theatmosphere by fast cosmic-ray particles arriving at the earth from space. mesons reach

    sea level in profusion. The typical speed of mesons is v = 2.994 108 m/s, which is 0.998of the velocity of light c. In the mean lifetime T0, mesons can travel a distance of only

    L = vT0 = (2.994 108 m/s) (2 106 s) = 600 m. (40)

    11

  • However, mesons are actually created at attitudes more than 10 times greater than the

    distance L. How to explain this paradox?

    a) Lets examine the problem from the frame of reference of an observer on the ground.

    The lifetime of the meson in our reference frame has been extended, due to the relative

    motion, to the value

    T =T0

    1 v2/c2 =2 1061 0.9982 s =

    2 1060.063

    m = 32 106 s. (41)

    This value is almost 16 times greater than when it is at rest with respect to us. In 32106s, a meson can travel a distance

    L0 = vT = (2.994 108 m/s) (32 106 s) = 9600 m. (42)

    This distance is larger than the attitude at which the meson is created.

    b) Lets examine the problem from the frame of reference of the meson. In this frame,

    the meson is at rest, its lifetime is T0 = 2 106 s, the earth ground is moving toward themeson. Compared to the distance L0 = 9600 m in the frame of the ground, the distance L

    appears to be shortened by the factor1 v2/c2 = 0.063, that is,

    L

    L0=1 v2/c2. (43)

    Hence, we have

    L = L01 v2/c2 = 9600

    1 0.9982 m = 9600 0.063 m = 600 m. (44)

    The travel time in the frame of reference of the meson is L/v = (600 m)/(3 108 m/s)= 2 106 s. This time is the same as the lifetime of the meson. Thus, the two points ofview give identical results.

    Example 2: Twin paradox

    Consider the famous relativistic effect known as the twin paradox. This paradox involves

    two identical clocks, one remains on earth and the other one is taken on a trip into space at

    the speed v and eventually is brought back. It is customary to replace the clocks with the

    pair of twins Dick and Jane. Dick is 20 years old when he takes off a space trip at a speed

    of 0.8c to a star 20 light-years away. His trip takes 50 years. To Jane, who stays behind,

    the pace of Dicks life is slower than her pace by a factor of1 v2/c2 =

    1 0.82 = 0.6 = 60%. (45)

    12

  • To Jane, Dicks heart beats only 3 times for every 5 beats of her heart; Dick thinks only

    3 thoughts for every 5 thoughts of hers. Finally, Dick returns after 50 years according to

    Janes calendar, but to Dick the trip has taken only 30 years. Dick is therefore 50 years old

    whereas Jane is 70 years old.

    Where is the paradox? If we consider the situation from the point of view of Dick in the

    spacecraft, Jane on the earth is in motion relative to him at a speed of 0.8c. Should not

    Jane then be 50 years old when the spacecraft returns, while Dick is then be 70the precise

    opposite of what was conducted above?

    But the two situations are not equivalent. Dick changed from one inertial frame to a

    different one when he started out, when he reversed direction to head home, and when he

    landed on the earth. Jane, however, remained in the same inertial frame during Dicks trip.

    Therefore, the time dilation formula applies to Janes observations of Dick, but not to Dicks

    observations of Jane.

    To look at Dicks trip from his perspective, we must take into account that the distance

    L he covers is shortened to

    L = L01 v2/c2 = 20 light-years 0.6 = 12 light-years. (46)

    Hence, to Dick, his trip took

    2L/v = 2 12 light-years/0.8c = 30 light-years. (47)

    Thus, the aging of the twins is nonsymmetric. This effect has been verified experimentally.

    G. Doppler effect

    1. Doppler effect in sound

    We are familiar with the increase in frequency of a sound when its source approaches us

    (or we approach the source) and the decrease in frequency when the source recedes from us

    (or we recede from the source). These changes in frequency constitute the Doppler effect.

    The origin of this effect is straightforward. Indeed, successive waves emitted by a source

    moving toward an observer are closer together than normal because of the advance of the

    source. Consequently, the separation between the waves, i.e. the wavelength, is shorter, and

    hence the corresponding frequency is higher.

    13

  • The relation between the source frequency 0 and the observed frequency is

    = 01 + v/c

    1 V/c, (48)

    where c is the speed of sound, v is the speed of the observer, and V is the speed of the source.

    The signs of v and V are plus for approaching and minus receding. Transverse motion does

    not cause a Doppler shift in sound.

    2. Doppler effect in light

    We consider a light source as a clock that ticks 0 times per second and emits a wave of

    light with each tick.

    a) Transverse Doppler effect in light

    Assume that the observer is moving perpendicular to a line between him and the light

    source. In the reference frame of the source, the proper time between two adjacent ticks is

    T0 = 1/0. In the reference frame of the observer, the time between two adjacent ticks is

    T = T0/1 v2/c2. The frequence measured in the reference frame of the observer is

    =1

    T=

    1

    T0

    1 v2/c2. (49)

    Hence, we have

    = 01 v2/c2. (50)

    The observed frequency is always lower than the source frequency 0. The same formula

    is true when the source is moving perpendicular to the line between the source and the

    observer. Thus, transverse motion does cause a Doppler shift in light, unlike in the case of

    sound.

    b) Longitudinal Doppler effect in light

    Assume that the observer is receding from the source. We call S the reference frame

    of the source, and call S the reference frame of the observer. Assume that light source is

    positioned at x = 0, the first tick occurs at t=0, and the second ticks occurs at t = T0 = 1/0.

    Assume that the observer is positioned at x = 0. In the frame S , the first tick is received

    by the observer at t = 0. The second tick occurs at

    x =x vt1 v2/c2 =

    vT01 v2/c2

    t =t (vx/c2)1 v2/c2 =

    T01 v2/c2 .

    (51)

    14

  • Since x < 0, the second light wave takes a time |x|/c to reach the observer. Therefore, theobserver receives the second wave at the time t + |x|/c, that is, at the time

    T =T0

    1 v2/c2 +(v/c)T01 v2/c2 = T0

    1 + v/c1 v2/c2 = T0

    1 + v/c

    1 v/c. (52)

    Hence, the observed frequency = 1/T is

    = 0

    1 v/c1 + v/c

    . (53)

    The observed frequency is lower than the source frequency. The same formula is true for

    the motion of the source away from the observer.

    In the case where the observer is approaching the source, we have

    = 0

    1 + v/c

    1 v/c. (54)

    In this case, the observed frequency is higher than the source frequency. The same formula

    is true for the motion of the source toward the observer.

    The expanding universe

    The Doppler effect in light is an important tool in astronomy. Stars emit light of certain

    characteristic frequencies called spectral lines. Motion of a star toward or away from the

    earth results in a Doppler shift in these frequencies. The spectral lines of distant galaxies

    of stars are all shifted toward the lower-frequency end and hence are called red shifts. Such

    shifts indicate that galaxies are receding from us and from each other, that is, the universe

    is expanding.

    H. The relativity of mass

    We have seen that fundamental physical quantities such as length and time have meaning

    only when the reference frame in which they are measured is specified. We now show the

    mass of a body also depends on the reference frame.

    For this purpose, we consider an elastic collision between two particles A and B, see Fig.

    5. In this collision, kinetic energy is conserved. The properties of A and B are identical

    when determined in the reference frames in which they are at rest. We observe the collision

    in two different reference frames S and S which are in uniform relative motion. The frame

    S is moving in the +x direction with respect to S at the velocity v.

    15

  • =2L

    FIG. 5: An elastic collision as observed in two different frames of reference. The balls are initially

    2L apart, which is the same distance in both frames since S moves only in the x direction.

    Before the process, particle A had been at rest in frame S, at the point (x = 0, y = L)and particle B in frame S , at the point (x = 0, y = L). Then, at the same instant

    tA = tB = L/V , A was thrown in the y direction at the speed VA while B is thrown in the

    16

  • y direction at the speed V B, where

    VA = VB = V. (55)

    Hence the behavior of A as seen from S is exactly the same as the behavior of B as seen

    from S . At the time t = tA+L/V = 0 and t = tB +L/V = 0, the particles A and B reach

    the origins O and O, respectively. Since O = O at the time t = t = 0, the two particles

    collide with each other at this time. When tho two particles collide, A rebounds in the ydirection at the speed VA as seen from S, while B rebounds in the +y direction at the speed

    V B as seen from S. The round-trip time T0 for A as measured in the frame S and for B as

    measured in S is therefore

    T0 =2L

    VA=

    2L

    V B. (56)

    Due to the time dilation effect, the round-trip time T for B as measured in S is longer than

    T0 by the factor 1/1 v2/c2, that is,

    T =T0

    1 v2/c2 . (57)

    Hence, the speed VB of B as measured in S is

    VB =2L

    T=

    2L1 v2/c2T0

    = VA1 v2/c2. (58)

    Since the collision is elastic, the total momentum in conserved, that is,

    mAVA mBVB = mAVA +mBVB, (59)

    where mA and mB are the masses of A and B as measured in S. This leads to

    mAVA = mBVB. (60)

    Inserting Eq. (58) into Eq. (60), we find

    mA = mB1 v2/c2. (61)

    In the above example, both A and B are moving in S. In order to obtain a formula for the

    mass m of a moving body in terms of its mass m0 when measured at rest, we consider the

    limit where VA and VB are very small. Then, we have mA = m0 and mB = m and so

    m =m0

    1 v2/c2 . (62)

    17

  • We call m0 the rest mass or the proper mass. Thus, the mass m of a moving body is larger

    than its rest mass m0 by the factor 1/1 v2/c2. The increase of the mass due to motion

    is a relativistic effect. We call m the relativistic mass of the body.

    The relativistic momentum is defined as

    p = mv =m0v

    1 v2/c2 . (63)

    In the theory of special relativity, the Newtons second law of motion is expressed by the

    formula

    F =dp

    dt=

    d

    dt(mv) = m0

    d

    dt

    [v

    1 v2/c2

    ]. (64)

    According to the above formula, we have

    F =d

    dt(mv) = m

    d

    dtv + v

    d

    dtm = ma+ v

    d

    dtm. (65)

    Here a = dv/dt is the acceleration. If v varies with time, m also varies with time. In this

    case, we have F 6= ma.

    I. Mass and energy

    The most famous formula Einstein obtained from the postulates of special relativity is

    the relationship between mass and energy.

    We recall from elementary physics that the work done on an object by a constant force F

    that acts through a distance s is Fs. Here we have assumed that F is in the same direction

    as s. If no other forces act on the object and the object starts from rest, all work done on it

    becomes kinetic energy K, so K = Fs. In the general case where F need not be constant,

    the formula for kinetic energy is the integral

    K =

    s0

    F ds. (66)

    Using the relativistic form of the second law of motion

    F =d(mv)

    dt, (67)

    18

  • Eq. (66) becomes

    K =

    s0

    d(mv)

    dtds =

    v0

    v d(mv) =

    v0

    v d

    (m0v

    1 v2/c2

    )

    =m0v

    21 v2/c2 m0

    v0

    v1 v2/c2 dv

    =m0v

    21 v2/c2 +m0c

    21 v2/c2

    v0

    =m0c

    21 v2/c2 m0c

    2.

    (68)

    The above equation may be rewritten as

    mc2 = K +m0c2. (69)

    We interpret mc2 as the total energy of the body. It follows from Eq. (69) that, when the

    body is at rest, that is, when K = 0, the body nevertheless possesses the energy m0c2. The

    energy

    E0 = m0c2 (70)

    is called the rest energy. The total energy can be written as

    E = mc2 =m0c

    21 v2/c2 . (71)

    In terms of E and E0, Eq. (69) becomes

    E = E0 +K. (72)

    Equation (71) says that mass and energy are not independent. Mass can be created or

    destroyed, but when this happens, an equivalent amount of energy simultaneously vanishes

    or comes into being, and vice versa. Mass and energy are different aspects of the same thing.

    The conversion factor between the unit of mass (kg) and the unit of energy (J) is c2, so

    1 kg of matter has an energy of mc2 = 1 kg (3 108 m/s)2 = 9 1016 J. This energy isenough to send a payload of a million tons to the moon.

    Kinetic energy at low speeds

    The relativistic formula for the kinetic energy is

    K = mc2 m0c2 = m0c2

    1 v2/c2 m0c2. (73)

    19

  • Consider the case where v c. We use the approximation (1 + x)n = 1+ nx for small x toexpand the first term in the above formula. Then we obtain the expression

    K = mc2 m0c2 = 12m0v

    2, (74)

    in agreement with classical mechanics.

    Energy and momentum

    Total energy and momentum are conserved in an isolated system, and the rest energy

    of a particle is invariant. Hence these quantities are in some sense more fundamental than

    kinetic energy and velocity. Lets look into how the total energy, rest energy, and momentum

    of a particle are related.

    Total energy is

    E =m0c

    21 v2/c2 . (75)

    Square of E is

    E2 =m20c

    4

    1 v2/c2 . (76)

    Momentum is

    p =m0v

    1 v2/c2 . (77)

    We have

    p2c2 =m20v

    2c2

    1 v2/c2 . (78)

    When we subtract p2c2 from E2, we obtain

    E2 p2c2 = m20c

    4 m20v2c21 v2/c2 = m

    20c

    4. (79)

    Hence, we have the relation

    E2 = m20c4 + p2c2. (80)

    Massless particles

    In classical mechanics, a particle must have rest mass in order to have energy and mo-

    mentum. However, this requirement does not hold true in relativistic mechanics.

    When m0 = 0 and v < c, we find from Eqs. (75) and (77) that E = p = 0. Such a

    particle is meaningless. However, when m0 = 0 and v = c, we have E = 0/0 and p = 0/0,

    indicating that E and p can have any values. Thus, massless particles may exist if they

    20

  • travel with the speed of light. The relation between the total energy E and the momentum

    p of a massless particle is

    E = pc. (81)

    An example of massless particles is the photon.

    J. Velocity addition

    Special relativity postulates that the speed of light c in free space has the same value for

    all observers, regardless of their relative motion. Common sense tells us that if we throw a

    ball forward at 10 m/s from a car moving at 30 m/s, the balls speed relative to the road

    will be 40 m/s. What if we switch on the cars headlights when its speed is v? The same

    reasoning suggests that their light ought to have a speed of c + v relative to the road. But

    this violates the above postulate. Common sense is not reliable in dealing with light or with

    a body moving with a speed comparable to the speed of light. To get the correct results for

    velocity addition, we must use the Lorentz transformation.

    Consider something moving relative to both S and S . An observer in S measures its

    three velocity components to be

    Vx =dx

    dt, Vy =

    dy

    dt, Vz =

    dz

    dt. (82)

    Meanwhile, to an observer in S they are

    V x =dx

    dt, V y =

    dy

    dt, V z =

    dz

    dt. (83)

    By differentiating the Lorentz transformation equations for x, y, z, and t, we obtain

    dx =dx vdt1 v2/c2

    dy = dy

    dz = dz

    dt =dt (v/c2)dx

    1 v2/c2 .

    (84)

    21

  • So we have

    V x =dx

    dt=

    dx vdtdt (v/c2)dx =

    dxdt v

    1 (v/c2)dxdt

    =Vx v

    1 (vVx/c2) ,

    V y =dy

    dt=dy1 v2/c2

    dt (v/c2)dx =dydt

    1 v2/c2

    1 (v/c2)dxdt

    =Vy1 v2/c2

    1 (vVx/c2) ,

    V z =dz

    dt=dz1 v2/c2

    dt (v/c2)dx =dzdt

    1 v2/c2

    1 (v/c2)dxdt

    =Vz1 v2/c2

    1 (vVx/c2) .

    (85)

    Thus, the formulae for velocity addition are

    V x =Vx v

    1 (vVx/c2) ,

    V y =Vy1 v2/c2

    1 (vVx/c2) ,

    V z =Vz1 v2/c2

    1 (vVx/c2) .

    (86)

    The inverse formulae are

    Vx =V x + v

    1 + (vV x/c2),

    Vy =V y1 v2/c2

    1 + (vV x/c2),

    Vz =V z1 v2/c2

    1 + (vV x/c2).

    (87)

    Consider a ray of light emitted in the moving frame S in its direction of motion relative

    to S. In this case, we have V x = c. An observer in S will measure the speed

    Vx =V x + v

    1 + (vV x/c2)=

    c+ v

    1 + (vc/c2)= c. (88)

    Thus both observers in S and S find the same value for the speed of light.

    Example

    Spacecraft A is moving at a speed of 0.9c with respect to the earth. If spacecraft B is

    to pass A at a relative speed of 0.5c in the same direction, what speed must B have with

    respect to the earth.

    Solution

    Conventional mechanics says that the speed of B ought to be 1.4c, that is, larger than

    the speed of light. However, according to the special relativity, the necessary speed of B is

    only

    Vx =V x + v

    1 + (vV x/c2)=

    0.5c+ 0.9c

    1 + (0.5c)(0.9c)/c2)= 0.97c. (89)

    This speed is less than c.

    22

  • II. WAVE PROPERTIES OF PARTICLES

    A. De Broglie waves

    A photon of light of frequency has the energy E = h. According to the special relativity

    theory, the energy E is related to the momentum p as E = pc. Hence, the momentum of a

    photon is

    p =h

    c. (90)

    Here, h = 6.626 1034 J s is the Plancks constant and c = 2.998 108 m/s is the speedof light in free space.

    On the other hand, the wavelength of a photon is = c/. When we use Eq. (90), we

    find the following relation between the wavelength and momentum of a photon:

    =h

    p. (91)

    The momentum p describes the particle property of the photon. The wavelength de-

    scribes the wave property of the photon. Each photon has both particle and wave properties.

    De Broglie suggested that Eq. (91) is completely general: It applies not only to photons

    but also to material particles. A material particle, i.e., a moving body, can behave as it

    has a wave nature. The matter waves are called the de Broglie waves. The wavelength of

    a particle is given by Eq. (91) and is called the de Broglie wavelength. The greater the

    particles momentum, the shorter its de Broglie wavelength.

    The momentum of a particle of mass m and velocity v is p = mv. The de Broglie

    wavelength is therefore given by

    =h

    mv. (92)

    In the above equation, m is the relativistic mass, which is related to the rest mass m0 as

    m =m0

    1 v2/c2 . (93)

    The wave and particle properties of moving bodies can never be observed at the same time.

    In some situations a moving body resembles a wave and in others it resembles a particle.

    Which set of properties is most conspicuous depends on how its de Broglie wavelength

    compares with its dimensions and the dimensions of whatever it interacts with.

    To illustrate this statement, we show two examples.

    23

  • Example 1

    Find the de Broglie wavelength of a dust particle with a mass of 1015 kg and a velocity

    of 1 mm/s.

    Solution

    Since v c, we can let m = m0. Hence

    =h

    mv=

    6.6 1034 J s(1015 kg) (103 m/s) = 6.6 10

    16 m. (94)

    Assume that the diameter of the dust particle is 1 m. Then, the de Broglie wavelength

    of the dust particle is very small compared with its dimensions. Therefore, we do not expect

    to find any wave aspects in its behavior.

    Example 2

    Find the de Broglie wavelength of an electron with a velocity of 107 m/s. The rest mass

    of an electron is m0 = 9.1 1031 kg.Solution

    Since v c, we can let m = m0. Hence

    =h

    mv=

    6.6 1034 J s(9.1 1031 kg) (107 m/s) = 7.3 10

    11 m. (95)

    The radius of the hydrogen atom is 5.3 1011 m. The de Broglie wavelength of theelectron (with a velocity of 107 m/s) is comparable with the dimensions of atoms. Therefore,

    the wave character of moving electrons is the key to understanding atomic structure and

    behavior.

    B. Wave function: de Broglie waves are waves of probability amplitude

    In water waves, the quantity that varies periodically is the height of the water surface. In

    sound waves, it is pressure. In light waves, electric and magnetic fields vary. What quantity

    varies in the case of de Broglie matter waves?

    The quantity whose variations make up matter waves is called the wave function and is

    denoted by the symbol . This quantity is a function of space and time, i.e., = (r, t).

    The value of the wave function at the particular point r = (x, y, z) in space at the time

    t is related to the likelihood of finding the body there at the time. More precisely, |(r, t)|2,

    24

  • the square of the absolute value of the wave function, is the probability of finding the body at

    the point r at the time t. The value of the wave function (r, t) is the probability amplitude.

    It can be negative, and is not an observable quantity.

    C. Describing a wave

    Consider a string stretched along the x direction. We shake the string at x = 0 up and

    down along the y direction. Assume that the vibrations are harmonic in character. The

    displacement of the string at x = 0 can be written as

    y = A cos 2pit. (96)

    Here is the frequency of the vibrations and A is their amplitude.

    A wave of vibrations propagates along the x direction with a wave velocity vp. Wave

    formula:

    y = A cos 2pi

    (t x

    vp

    ). (97)

    The wave velocity vp is called the phase velocity. Since the wave velocity vp is given by

    vp = , we have

    y = A cos 2pi(t x

    ). (98)

    The angular frequency is defined by the formula

    = 2pi. (99)

    The wave number k is defined by the formula

    k =2pi

    =

    vp. (100)

    In terms of and k, the wave formula can be written as

    y = A cos(t kx). (101)

    In the three-dimensional space, k becomes a vector k normal to the wave fronts and x is

    replaced by the radius vector r. Then, kx is replaced by the scalar product k r = kr.In the case of de Broglie waves, the momentum of the particle is

    p = h/ = hk/2pi = hk. (102)

    25

  • Here, h = h/2pi = 1.054 1034 J s is the reduced Plancks constant.Exercise 1:

    A photon and a particle have the same wavelength. (a) Compare their linear momenta.

    (b) Compare the photons energy and the particles total energy. (c) Compare the photons

    energy and the particles kinetic energy.

    Answer:

    (a) They have the same linear momenta: p = h/.

    (b) The photons energy is Eph = h = hc/ = pc. The particles total energy is

    Ep =p2c2 + E20 . Thus the photons energy is smaller than the particles total energy.

    (c) The particles kinetic energy is K = Ep E0 =p2c2 + E20 E0 < pc. Thus the

    particles kinetic energy is smaller than the photons energy.

    Exercise 2:

    Show that the de Broglie wavelength of a particle of rest mass m0 and kinetic energy K

    is = hc/K(K + 2m0c2).

    Answer: Be definition, we have = h/p = hc/pc. To calculate pc, we use the kinetic

    energy K = E E0 =p2c2 + E20 E0 =

    p2c2 +m20c

    4 m0c2. The latter yields pc =K(K + 2m0c2). Hence, we obtain

    = hc/K(K + 2m0c2). (103)

    Note that, when v c, we have K = m0c2(1/1 v2/c2 1) m0c2. Hence, we obtain

    = h/2m0K. This formula can be derived by another way using the nonrelativistic

    formulae = h/p and K = p2/2m0.

    Exercise 3:

    Show that if the total energy of a moving particle greatly exceeds its rest energy, its de

    Broglie wavelength is nearly the same as the wavelength of a photon with the same total

    energy.

    Answer: For a particle, we have E =p2c2 + E20 . Hence, pc =

    E2 E20 . When

    E E0, we obtain pc = E. In this case, the de Broglie wavelength of the particle isp = h/p = hc/E. Meanwhile, for a photon, we always have E = pc and hence, ph =h/p = hc/E. Thus p = ph.

    26

  • D. Phase and group velocities of de Broglie waves

    How fast do de Broglie waves travel? Since a de Broglie wave is associated with a moving

    body, one may expect that this wave has the same velocity as that of the body. Let us see

    if this is true.

    We call the de Broglie wave velocity vp. To find vp, we can apply the usual formula

    vp = . (104)

    The wavelength is simply the de Broglie wavelength, i.e.,

    =h

    mv. (105)

    To find the frequency , we equate the quantum expression E = h with the relativistic

    expression E = mc2. Then we obtain

    =mc2

    h. (106)

    The de Broglie wave velocity is therefore

    vp = =

    (mc2

    h

    )(h

    mv

    )=c2

    v. (107)

    The wave velocity vp is the phase velocity. The particle velocity v is the group velocity.

    Because the particle velocity v must be less than the velocity of light c, the de Broglie wave

    velocity is always larger than c. To understand this result, we must look into the distinction

    between phase velocity and group velocity.

    We consider a harmonic wave

    y = A cos(t kx). (108)

    The de Broglie waves associated with a moving body cannot be represented simply by a for-

    mula resembling Eq. (108). Instead, the wave representation of a moving body corresponds

    to a wave packet, or wave group. A wave group is a superposition of individual waves of

    different wavelengths. The interference of the individual waves with one another results in

    the variation in amplitude that defines the group shape. If the velocities of the individual

    waves are the same, the velocity of the wave group is the common phase velocity. However,

    27

  • FIG. 6: A wave group.

    if the phase velocity varies with wavelength, an effect called dispersion, the different indi-

    vidual waves do not proceed together. As a result, the wave group has a velocity different

    from the phase velocities of the individual waves. This is the case with de Broglie waves.

    As an example, we consider the case where the wave group consists of two waves that

    have the same amplitude A but differ by a small amount in angular frequency and a

    small amount k in wave number:

    y1 = A cos(t kx),y2 = A cos[( +)t (k +k)x]. (109)

    The wave group is then given by

    y = y1 + y2

    = 2A cos1

    2( tk x) cos 1

    2[(2 +)t (2k +k)x]. (110)

    Since and k k, we find

    y = 2A cos1

    2( tk x) cos(t kx). (111)

    The above equation represents a wave of angular frequency and wave number k whose

    amplitude is modulated by an angular frequency /2 and a wave number k/2.

    The effect of the modulation is to produce successive wave groups. The phase velocity vp

    is

    vp =

    k, (112)

    and the velocity of the successive wave groups is

    vg =

    k. (113)

    28

  • When and k have continuous spreads instead of the two values in the above discussion,

    the group velocity is

    vg =d

    dk. (114)

    We now use Eqs. (112) and (114) to calculate the phase and group velocities of de Broglie

    waves. The angular frequency of the de Broglie waves associated with a body of rest mass

    m0 moving with the velocity v is

    = 2pi =2pimc2

    h=

    2pim0c2

    h1 v2/c2 . (115)

    The wave number of the de Broglie waves is

    k =2pi

    =

    2pimv

    h=

    2pim0v

    h1 v2/c2 . (116)

    The group velocity of the de Broglie waves is

    vg =d

    dk=d/dv

    dk/dv. (117)

    It follows from Eqs. (115) and (116) that

    d

    dv=

    2pim0v

    h(1 v2/c2)3/2 ,dk

    dv=

    2pim0h(1 v2/c2)3/2 . (118)

    Hence, we find

    vg = v. (119)

    Thus, the group velocity of the de Broglie waves is the velocity of the moving body.

    The phase velocity of the de Broglie waves is, as found earlier,

    vp =

    k=c2

    v. (120)

    The fact that vp > c does not violate the special relativity theory because vp is the motion

    of the phase of the wave group, not the motion of the individual waves that make up the

    group, and consequently, not the motion of the body.

    Exercise 1:

    An electron has a de Broglie wavelength of 2 1012 m. Find its kinetic energy and thephase and group velocities of its de Broglie waves.

    29

  • Solution: First, we calculate pc:

    pc =hc

    =

    (6.6 1034)(3 108)2 1012 = 1 10

    13 J. (121)

    The rest energy of the electron is E0 = m0c2 = (9.1 1031)(3 108)2 = 0.8 1013 J. The

    kinetic energy of the electron is

    K = E E0 =p2c2 + E20 E0 =

    (1 1013)2 + (0.8 1013)2 0.8 1013

    = 0.48 1013 J. (122)

    In the units of eV (1 eV= 1.6 1019 J), we have K = 3 105 eV = 300 keV.To find the electron velocity v, we use the formula

    v

    c=mcv

    mc2=pc

    E=

    pcp2c2 + E20

    =1 1013

    (1 1013)2 + (0.8 1013)2 = 0.78. (123)

    Hence, the group velocity is vg = v = 0.78 c and the phase velocity is vp = c2/v = 1.28 c.

    Exercise 2:

    A proton and an electron have the same velocity. Compare the wavelengths and phase

    and group velocities of their de Broglie waves.

    Answer: (a) = h/p = h/mv. The electron has a smaller mass and consequently a longer

    wavelength compared to those of the proton.

    (b) Since vg = v, the de Broglie waves of the proton and electron have the same group

    velocity.

    (c) Since vp = c2/v, the de Broglie waves of the proton and electron have the same phase

    velocity.

    Exercise 3:

    (a) A proton and an electron have the same kinetic energy. (b) Compare the wavelengths

    and phase and group velocities of their de Broglie waves.

    Answer: (a) = h/p and pc =K(K + 2m0c2) (or K = p2/2m0 in nonrelativistic

    considerations). The electron has a smaller mass than the proton. Therefore, we have

    elec > proton.

    (b) From K = E E0 = m0c2(1/1 v2/c2 1), we find

    v2

    c2= 1

    (1 K

    K +m0c2

    )2.

    30

  • Since melec < mproton, we have velec > vproton. From the nonrelativistic formula K = m0v2/2,we get the same conclusion. Since vg = v and vp = c

    2/v, the electron has a larger group

    velocity and a smaller phase velocity than the proton does.

    Exercise 4:

    Verify the statement that, if the phase velocity is the same for all wavelengths of a certain

    wave phenomenon, the group and phase velocities are the same.

    Answer: vp = /k and vg = d/dk. If vp is a constant then vg = vp.

    Exercise 5:

    (a) Show that the phase group velocity of a particle of rest mass m0 and de Broglie

    wavelength is vp = c1 + (m0c/h)2. (b) Compare the group and phase velocities of an

    electron with the de Broglie wavelength of 1 1013 m.Answer: (a) We have = h/p and p = m0v/

    1 v2/c2. Hence, we find

    v = c/1 + (m0c/h)2. (124)

    Since vg = v and vp = c2/v, we obtain

    vp = c1 + (m0c/h)2. (125)

    (b) For m0 = 9.11031 kg, h = 6.61034 Js, c = 3108 m/s, = 11013 m, we find1 + (m0c/h)

    2 = 1.0016. Consequently, vp/vg = 1.0016, vp = 1.0008 c and vg = 0.9992 c.

    31

  • FIG. 7: Scheme of the Davission-Germer experiment.

    III. PARTICLE DIFFRACTION

    A wave effect with no analog in the behavior of Newtonian particles is diffraction. In

    1927, Davisson and Germer demonstrated that electron beams are diffracted when they are

    scattered by the regular atomic arrays of crystal.

    Davisson and Germer studied the scattering of electrons from a solid using an apparatus

    sketched in Fig. 7. The energy of the electrons in the primary beam, the angle at which

    they reach the target, and the position of the detector could all be varied.

    Classical physics predicts that the scattered electrons will emerge in all directions with

    only a moderate dependence of their intensity on scattering angle and even less on the energy

    of the primary electrons. Using a block of nickel as the target, Davisson and Germer verified

    these predictions. To prevent the crystal from being oxidized, the apparatus was kept in the

    vacuum.

    In the midst of their work an accident occurred that allowed air to enter their apparatus

    and oxidize the metal surface. To reduce the oxide, the target was baked in a hot oven. After

    this treatment, the target was returned to the apparatus and the measurements resumed.

    Then the results were very different. Instead of a continuous variation of scattered electron

    intensity with angle, distinct maxima and minima were observed whose positions depend on

    the electron energy, see Fig. 8.

    Two questions arise: What is the reason for this new effect? Why did it not appear until

    the nickel target was baked?

    De Broglies hypothesis suggested that electron waves were being diffracted by the target.

    32

  • FIG. 8: Results of the Davission-Germer experiment.

    This was realized when heating a block of nickel at high temperature causes the many

    individual crystals in the target to form a single large crystal. This crystal acts like a set

    of many planes of atoms, called Bragg planes. The spacing of the planes in this crystal is

    denoted by d = 0.091 nm. The angle of incidence and scattering relative to the Bragg planes

    is denoted by . The Bragg equation for maxima in the diffraction pattern is

    n = 2d sin . (126)

    When the kinetic energy of electrons is 54 eV, the angle is = 65. For n = 1 and with

    sin 65 = 0.906, the de Broglie wavelength is estimated to be

    = 2d sin = 2 0.091 nm sin 65 = 0.165 nm. (127)

    Now we use de Broglies formula = h/mv to find the expected wavelength of the

    electrons. The electron kinetic energy of 54 eV is small compared with its rest energy m0c2

    of 0.51 MeV, so we can ignore relativistic considerations. Therefore, the electron kinetic

    energy is given by

    K =mv2

    2. (128)

    Hence the electron momentum is

    mv =2mK =

    (2)(9.1 1031 kg)(54 eV)(1.6 1019J/eV) = 41024 kg m/s. (129)

    The electron wavelength is therefore

    =h

    mv=

    6.6 1034 Js4 1024 kg m/s = 0.165 nm. (130)

    33

  • FIG. 9: The diffraction of the de Broglie waves by the target is responsible for the results of the

    experiment.

    Note that the realistic situation is more complicated than the above analysis. For ex-

    ample, a complication arises from the fact that the energy of an electron increases when it

    enters a crystal. Another complication is that there are several families of Bragg planes. The

    interference between the waves diffracted from different families may prevent the observation

    of maxima even when the Bragg condition is satisfied.

    Exercise:

    What effect on the scattering angle in the Davisson-Germer experiment does increasing

    the electron energy have?

    Answer: leads to a decrease of and an increase of the scattering angle .

    Exercise:

    A beam of 50-keV electrons is directed at a crystal and diffracted electrons are found at

    an angle of 50 relative to the original beam. What is the spacing of the atomic planes of

    the crystal? A relativistic calculation is needed for . Here m0c2 = 0.5 MeV.

    Answer: We have

    =hc

    K(K + 2m0c2)

    =(6.6 1034 Js)(3 108 m/s)

    1.6 1019 J/eV(50 103 eV)(50 103 eV + 2(0.5 106 eV))=

    (6.6 1034)(3 108)1.6 1019 [104 5 (5 + 100)] m = 19.81.6 23 1011 m

    = 0.5 1011 m = 0.005 nm. (131)

    34

  • The spacing between the atomic planes is

    d =

    2 sin =

    0.005 nm

    2 sin 65=

    0.005 nm

    2 0.906 = 2.76 103 nm. (132)

    Exercise:

    A beam of 5.4-keV electrons is directed at a crystal and diffracted electrons are found at

    an angle of 50 relative to the original beam. What is the spacing of the atomic planes of

    the crystal? We can ignore relativistic considerations.

    Answer: We have

    =h

    2m0K

    =6.6 1034 Js

    (2)(9.1 1031 kg)(5.4 103 eV)(1.6 1019 J/eV)

    =6.6 1034 m15.7 1046 =

    6.6 1034 m4 1023 = 1.65 10

    11 m = 0.0165 nm. (133)

    The spacing between the atomic planes is

    d =

    2 sin =

    0.0165 nm

    2 sin 65=

    0.0165 nm

    2 0.906 = 9.1 103 nm. (134)

    Particle in a box

    The wave nature of a moving particle leads to some remarkable consequences when the

    particle is restricted to a certain region of space instead of being able to move freely.

    The simplest case is that of a particle in a box. We assume that the particle can move

    only along one direction of the box, bouncing back and forth between the walls. We assume

    that the walls are infinitely hard, so the particle does not lose energy each time it strikes a

    wall. We also assume that the velocity of the particle is sufficiently small that we can ignore

    relativistic considerations.

    From a wave point of view, a particle trapped in a box is like a standing wave. The

    possible de Broglie wavelengths of the particle are therefore determined by the width L of

    the box. The longest wavelength is = 2L, the next is = L, then = 2L/3, and so forth.

    The general formula is

    =2L

    n(n = 1, 2, 3, . . . ). (135)

    Because mv = h/, the restrictions on imposed by the box width L are equivalent to

    limits on the momentum of the particle and, in turn, to limits on its kinetic energy. The

    35

  • kinetic energy of the particle is

    K =mv2

    2=

    h2

    2m2. (136)

    Since the permitted wavelengths are n = 2L/n and the particle has no potential energy in

    this model, the permitted energies of the particle are

    En =n2h2

    8mL2(n = 1, 2, 3, . . . ). (137)

    Each permitted energy is called an energy level. The integer number n that specified an

    energy level En is called its quantum number.

    We can draw three general conclusions:

    1. A trapped particle cannot have an arbitrary energy, as a free particle can. The energies

    are quantized (discrete) and can be characterized by a quantum number.

    2. A trapped particle cannot have zero energy. The zero energy means v = 0 and

    therefore =. There is no way to trap a wave with an infinite wavelength in a box.3. Because the Planck constant h is very smallonly 6.63 1034 J s quantization of

    energy is conspicuous only when m and L are also small. This is why we are not aware

    of energy quantization in our own experience. The smaller the confinement, the larger the

    energy required for confinement.

    If a particle is confined into a rectangular volume (a three-dimentional box), the permitted

    energies are

    En1n2n3 =(n21 + n

    22 + n

    23)h

    2

    8mL2(n = 1, 2, 3, . . . ). (138)

    Exercise: The lowest possible energy of a particle in a box is 1 eV. What are the next

    two higher energies the particle can have?

    Answer: 4 eV and 9 eV.

    36

  • FIG. 10: (a) A narrow de Broglie wave group. (b) A wide wave group.

    IV. UNCERTAINTY PRINCIPLE

    Look at the wave group of Fig. 6. The particle that corresponds to this wave group can

    be found anywhere within the group at a given time. The probability of finding the particle

    is given by ||2.When the wave group is narrower, the particles position can be specified more precisely,

    see Fig. 10(a). However, the wavelength of the waves in a narrow packet is not well defined.

    The reason is that the range of wavelengths of individual waves is large. This means that,

    since = h/mv, the particles momentum is not a precise quantity. If we make a series of

    momentum measurements, we will find a broad range of values.

    When the wave group is wider, the particles wavelength can be specified more precisely,

    see Fig. 10(b). Therefore, the momentum can be measured more precisely. However, since

    the wave group is wide, the position of the particle is not well defined. If we make a series

    of position measurements, we will find a broad range of values.

    Thus we have the uncertainty principle (discovered by Heisenberg in 1927):

    It is impossible to know both the exact position and exact momentum of an object at the

    same time.

    We present a mathematical expression for this principle. A moving body corresponds to

    a single wave group, not a series of them. An isolated wave group is a superposition of an

    infinite number of wave trains with different frequencies, wave numbers, and amplitudes, see

    37

  • FIG. 11: An isolated wave group is the result of superposing an infinite number of waves with

    different wave lengths.

    Fig. 11.

    At a certain time t, when (x) is a real function, the wave group (x) can be represented

    by the Fourier integral

    (x) =

    0

    g(k) cos kx dk. (139)

    More precisely and more generally, we have the formula

    (x) =

    g(k)eikx dk. (140)

    The function g(k) describes the amplitudes and wavelengths of the harmonic waves that

    contribute to the wave group. The narrower the wave group, the broader the range of

    wavelengths involved, see Fig. 12.

    We assume that the wave function (x) is spread in the interval x, and that the

    Fourier transform g(k) is spread in the interval k. The spreads x and k are defined as

    the standard deviations of x and k, respectively. These spreads are related to each other.

    Due to the properties the Fourier transformation, we always have

    xk 12. (141)

    To understand Eq. (141) qualitatively, we consider a simple example. Assume that

    g(k) consists of only three individual components. The wave vectors of these individual

    components are k0, k0k/2, and k0+k/2. Their amplitudes are proportional to 1, 1/2,and 1/2, respectively. We then have

    (x) = g0

    {cos(k0x) +

    1

    2cos

    [(k0 k

    2

    )x

    ]+1

    2cos

    [(k0 +

    k

    2

    )x

    ]}= g0 cos(k0x)

    [1 + cos

    (k

    2x

    )]. (142)

    38

  • FIG. 12: The wave functions and Fourier transforms for (a) a pulse, (b) a wave group, (c) a wave

    train, and (d) a Gaussian distribution.

    The above function is maximum at x = 0 and goes to zero at x = 2pi/k. The width ofthis function is x = 4pi/k. Thus we have

    xk = 4pi 12. (143)

    If the wave function (x) is a Gaussian function, then the Fourier transform g(k) is also

    a Gaussian function, and we have xk = 1/2. Indeed, we take

    (x) = N exp

    ( x

    2

    4a2

    ). (144)

    The width of (x) is x = a. We find

    g(k) = N exp(a2k2) , (145)

    which is a Gaussian function. We can prove this with the help of the formula

    ea2k2 cos kx dk =

    pi

    ae

    x2

    4a2 . (146)

    The width of g(k) is k = 1/(2a). Thus we have xk = 1/2.

    The Fourier transform g(k) is related to the probability of finding the particle with the

    momentum p = hk. More precisely, |g(k)|2 is proportional to the probability for the particleto have the momentum p = hk. When we use the relation p = hk and Eq. (141), we

    find

    xp h2. (147)

    The above equation is a mathematical expression of the uncertainty principle. The spreads

    x and k are the measures of the uncertainties in the position and momentum, respectively,

    39

  • of the particle. Equation (147) says that the product of the uncertainty x in the position

    of an object at some instant and the uncertainty p in its momentum component in the x

    direction at the same time is equal or greater than h/2.

    If we arrange the particle in such a way that x is small, then p will be large. If we

    reduce p in some way, then x will be large.

    Average value and standard deviation

    Consider an observable A. Assume that the probability for A to take the value is

    described by the probability density (). The mean value of A is

    A =

    () d. (148)

    We introduce the notation

    F (A) =

    F ()() d. (149)

    In particular, we have

    An =

    n() d. (150)

    The standard deviation A of A is defined by the formula

    (A)2 = (A A)2 = A2 A2. (151)

    Compatible observables

    If two observables A and B are compatible, they can be described by a joint probability

    density (, ). We can define

    F1(A)F2(B) =

    F1()F2()(, ) dd. (152)

    Proof of the uncertainty principle

    Consider the position x and the momentum p of a particle. In quantum mechanics, x

    and p are not compatible, that is, the averages xp and px cannot be defined in theconventional way. Moreover, we have

    xp 6= px. (153)

    40

  • More precisely, we have

    xp px = ih. (154)

    Define x = x x and p = p p. It follows from the above equation that

    xp px = ih. (155)

    As known, CC 0. For any real variable , we always have

    (x+ ip)(x ip) 0. (156)

    We calculate the left-hand side and find

    (x+ ip)(x ip) = a2 + b + c, (157)

    where

    a = p2b = i(xp px)c = x2. (158)

    Since a2 + b + c 0 for any real , the following condition should be satisfied:

    b2 4ac. (159)

    On the other hand, we have a = (p)2, b = h, and c = (x)2. Hence, we find

    h2 4(p)2(x)2, (160)

    that is,

    xp h2. (161)

    Uncertainty principle from the particle approach

    Suppose we look at an electron using light of wavelength . Each photon of this light

    has the momentum h/. We can see the electron only if one of these photons bounces off

    the electron. The electrons original momentum will be changed. The exact amount of the

    41

  • change p cannot be predicted but will be of the same order of magnitude as the photon

    momentum h/. Consequently, the uncertainty in the electrons momentum is

    p h. (162)

    On the other hand, light is a wave phenomenon as well as a particle phenomenon. We

    cannot determine the position of the electron with an accuracy better than the wavelength.

    Consequently, we have

    x . (163)

    Combining Eqs. (162) and (163) gives

    xp h. (164)

    This result is consistent with the formula xp h/2.

    Uncertainty principle for energy and time

    Another form of the uncertainty principle concerns energy and time.

    Consider the measurement of the energy E emitted during the time interval t in an

    atomic process. Assume that the energy is in the form of electro-magnetic waves. The

    energy is E = h. Therefore, the uncertainty in energy is

    E = h. (165)

    To measure the frequency , we account the number of waves N for the interval t and

    divide this number by the time interval, that is, = N/t. Assume that the uncertainty in

    number of waves in the wave group is one. Then, the uncertainty in frequency is

    1t

    . (166)

    It follows from Eqs. (165) and (166) that

    Et h. (167)

    A more rigorous treatment gives

    Et h2. (168)

    42

  • Thus the product of the uncertainty in an energy measurement and the uncertainty in the

    time at which the measurement is made is greater than or equal to h/2.

    Consider a conservative system. For this system, the greater the energy uncertainty, the

    more rapid the time evolution. More precisely, if t is a time interval at the end of which

    the system has evolved to an appreciable extent and if E denotes the energy uncertainty,

    t and E satisfy the relation

    Et h2. (169)

    The above equation is a mathematical expression of the uncertainty principle for energy and

    time.

    The proof is given below. Consider a wave packet. The energy uncertainty E is asso-

    ciated with the momentum uncertainty p via the formula

    E =dE

    dpp. (170)

    Since E = h and p = hk, we have

    dE

    dp=d

    dk= vg. (171)

    Hence

    E = vgp. (172)

    Now the characteristic evolution time t is the time taken by this wave packet to pass a

    point in space. If x is the spatial extension of the wave packet, we have

    t =x

    vg. (173)

    From this we obtain

    Et = px h2. (174)

    Example 1

    A measurement establishes the position of a proton with an accuracy of 1 1011 m.Find the uncertainty in the position of the proton 1 second later. The rest mass of a proton

    is m0 = 1.672 1027 kg. Assume v c.Solution

    The uncertainty in the protons position at t = 0 is x0 = 1 1011 m. According toEq. (147), the uncertainty in its momentum at this time is

    p h2x0

    . (175)

    43

  • Since v c, the momentum is p = mv = m0v. Therefore, we have p = m0v. Hence, theuncertainty in the protons velocity is

    v h2m0x0

    . (176)

    After the time t, the position of the proton cannot be known more accurately than

    x = tv ht2m0x0

    . (177)

    Hence x is inversely proportional to x0. This means that the more we know about the

    protons position at a given time, the less we know about its later position.

    The value of x at t = 1 s is

    x (1.054 1034 J s) (1 s)

    2 (1.672 1027 kg) (1 1011 m) = 3.15 103 m. (178)

    Exercise: (a) Discuss the prohibition of E = 0 for a trapped particle in a box in terms of

    the uncertainty principle. (b) How does the minimum momentum of such a particle compare

    with the momentum uncertainty required by the uncertainty principle if we take x = L?

    Answer: (a) Since the particle is trapped in the box, x is not infinite. Therfore, p

    cannot be zero and consequently p cannot be zero. This is why the particle cannot have

    E = 0. (b) If we take x = L then p h/2x = h/2L. On the other hand, the firstpermitted value of is 2L. Therefore, the minimum momentum is pmin = h/ = h/2L =

    pih/L > h/2L. Thus the minimum momentum of the trapped particle is larger than the

    momentum uncertainty required by the uncertainty principle.

    Exercise: Compare the uncertainties in the velocities of an electron and a proton in a

    small box.

    Answer: Take x = L. Then (p)min = h/2x = h/2L. Since p = mv, we have

    (v)min = h/2mL. Hence, the uncertainty in the velocity of the electron is larger than that

    of the proton.

    Exercise: Verify that the uncertainty principle can be written as L h/2, where Lis the angular momentum and is the angular position.

    Answer: Consider the rotational motion of a particle along a circle of radius a. We

    have L = mvr and = x/r. Hence L = mrv = rp and = x/r. Therefore,

    L = px h/2.Exercise: A hydrogen atom is 5.3 1011 m in radius. Use the uncertainty principle to

    estimate the minimum kinetic energy an electron can have in this atom.

    44

  • Answer: Here we have x = 5.3 1011 m. The uncertainty in momentum is

    p h2x

    = 1 1024kgm/s. (179)

    The momentum of the electron must be at least comparable to its uncertainty. Consequently,

    the kinetic energy of the electron is

    K =p2

    2m (1 10

    24)2

    (2)(9.1 1031) J 5 1019 J = 3 eV. (180)

    45

  • V. ATOMIC SPECTRA

    When an atomic gas is excited by passing an electric current through it, the emitted

    radiation has a spectrum which contains specific wavelengths only.

    Each element has a characteristic line spectrum.

    The number, strength, and exact wavelengths of the lines in the spectrum of an element

    depend on temperature, pressure, the presence of electric and magnetic fields, and the motion

    of the source.

    Spectroscopy is therefore a useful tool for analyzing the composition and the state of a

    source.

    A. Spectral series

    It has been experimentally found that the spectral lines of an element fall into sets called

    spectral series.

    For hydrogen, the Lyman series contains the wavelengths given by the formula

    1

    = R

    (1

    12 1n2

    )with n = 2, 3, 4, . . . , (181)

    and is in the ultraviolet region (40010 nm). Here, R = 1.097 107 m1 is the Rydbergconstant.

    The Balmer series contains the lines

    1

    = R

    (1

    22 1n2

    )with n = 3, 4, 5, . . . , (182)

    and is in the visible region (800400 nm).

    In the infrared region (from 800 nm to 1 mm), three series have been found. They are

    Paschen:1

    = R

    (1

    32 1n2

    )with n = 4, 5, 6, . . . , (183)

    Brackett:1

    = R

    (1

    42 1n2

    )with n = 5, 6, 7, . . . , (184)

    and

    Pfund:1

    = R

    (1

    52 1n2

    )with n = 6, 7, 8, . . . (185)

    The existence of spectral lines and series poses a test for any theory of atomic structure.

    46

  • FIG. 13: Spectral series of hydrogen.

    Exercise: What is the shortest wavelength present in the Brackett series of spectral lines?

    Answer: The Brackett series is 1/ = R(1/42 1/n2) with n = 5, 6, 7, . . . . The shortestwavelength in this series is = 16/R.

    Exercise: What is the shortest wavelength present in the Paschen series of spectral lines?

    Answer: The Paschen series is 1/ = R(1/32 1/n2) with n = 4, 5, 6, . . . . The shortestwavelength in this series is = 9/R.

    B. The Bohr atom

    The first theory of the atom to meet with any success was put forward in 1913 by Niels

    Bohr. This theory can be formulated in terms of the de Broglie waves as shown below.

    Consider an electron in orbit around a hydrogen nucleus. The de Broglie wavelength of

    this electron is

    =h

    mv. (186)

    47

  • To determine v, we recall that the centripetal force is

    Fc =mv2

    r. (187)

    This force is provided by the electric force

    Fe =1

    4pi0

    e2

    r2. (188)

    The condition for a stable orbit is Fc = Fe, i.e.,

    mv2

    r=

    1

    4pi0

    e2

    r2. (189)

    The electron velocity is therefore found to be

    v =e

    4pi0mr. (190)

    Hence, the orbital electron wavelength is

    =h

    e

    4pi0r

    m. (191)

    We assume that the motion of the electron in the hydrogen atom is analogous to the

    vibrations of a wire loop. We know that, in a wire loop, the loops circumference is an

    integer number of the wavelength of the resonant mode. Therefore, we assume that an

    electron can circle a nucleus only if its orbit contains an integer number of the de Broglie

    wavelength. Thus, the condition for orbital stability is

    n = 2pir, with n = 1, 2, 3, . . . (192)

    The integer n is called the quantum number of the orbit. When we substitute Eq. (192)

    into Eq. (191), we find that the radii of the orbits are

    r = rn = n2 h

    20pime2

    , with n = 1, 2, 3, . . . (193)

    The radius of the innermost orbit is called the Bohr radius of the hydrogen atom and is

    denoted by a0:

    a0 = r1 =h20pime2

    = 5.292 1011 m. (194)

    Here we have used the parameters e = 1.61019 C, 0 = 8.851012 F/m, m0 = 9.11031kg, and h = 6.6 1034 Js.

    48

  • FIG. 14: Vibrations of a wire loop.

    The other radii are given in terms of a0 by the formula

    rn = n2a0, with n = 1, 2, 3, . . . (195)

    Exercise 1: In the Bohr model, the electron is in constant motion. How can such an

    electron have a negative energy?

    Answer: The potential energy of the electron interacting with the proton is U =

    e2/4pi0r, a negative value. The kinetic energy of the electron is K = mv2/2 = e2/8pi0r.Since the absolute of the potential energy is larger than the kinetic energy, the total energy

    E = K + U of the electron is negative.

    Exercise 2: Derive a formula for the speed of an electron in the nth orbit of a hydrogen

    atom using the Bohr model.

    Answer: In the Bohr model, rn = n2a0, where a0 = h

    20/pime2 is the Bohr radius. From

    the condition for orbital stability n = 2pir, we have = 2pina0. From the de Broglie

    relation = h/mv, we obtain v = h/nma0.

    Exercise 3: An electron at rest is released far away from a proton and moves toward the

    proton. (a) Show that the de Broglie wavelength is proportional tor, where r is the

    distance of the electron from the proton. (b) Find when r = a0. How does this compare

    with the wavelength of an electron in the ground-state Bohr orbit? (c) In order for the

    electron to be captured by the proton to form a ground-state hydrogen atom, energy must

    be lost by the system. How much energy?

    Answer: (a) It follows from the energy conservation law E = K + U = 0 that K =

    49

  • p2/2m = U = e2/4pi0r. Hence p = em/2pi0r. Therefore, = h/p =

    2pi0h2r/me2 =

    pi2a0r.

    (b) When r = a0, we have = pi2 a0. The wavelength of an electron in the ground-state

    Bohr orbit is 0 = 2pia0. As we can see, < 0.

    (c) The energy to be released is E1 = U/2 = (1/8pi0)(e2/r1) = e2/8pi0a0 =me4/820h

    2 = 2.18 1018 J = 13.6 eV.Exercise 4: Compare the uncertainty in the momentum of an electron confined to a region

    of linear dimension a0 with the momentum of an electron in the ground-state Bohr orbit.

    Answer: p h/2a0. According to exercise 2, the momentum of an electron in theground-state Bohr orbit is p1 = mv1 = h/a0. Thus the uncertainty in the momentum of the

    electron is less than the momentum of the electron in the ground-state Bohr orbit.

    C. Energy levels and spectra

    The total electron energy E is the sum of its kinetic energy

    mv2

    2(196)

    and potential energy

    e2

    4pi0r. (197)

    This means that

    E =mv2

    2 e

    2

    4pi0r. (198)

    Due to Eq. (189), we find

    E = e2

    8pi0r. (199)

    Different permitted orbits have different energies: Due to Eq. (189), we find

    En = e2

    8pi0rn. (200)

    When we use Eq. (193) for rn, we see that

    En = me4

    820h2

    1

    n2=E1n2

    , with n = 1, 2, 3, . . . (201)

    Here, E1 = 2.18 1018 J = 13.6 eV.The energies specified by Eq. (201) are called the energy levels of the hydrogen atom.

    These levels are all negative. This means that the electron does not have enough energy to

    50

  • escape from the nucleus. An electron in the hydrogen atom can have only these energies

    and no others.

    The lowest level E1 is called the ground state, and the higher levels E2, E3, E4, . . . are

    called excited states.

    As the quantum number n increases, the level energy En tends to 0. In the limit of

    n =, we have E = 0, that is, the electron is no longer bound to the nucleus to form anatom. The work needed to remove an electron from its ground state is called its ionization

    energy. It is equal to E1. In the case of hydrogen, the ionization energy is 13.6 eV.

    D. Origin of line spectra

    According to the Bohr model, electrons cannot exist in an atom except in certain specific

    energy levels. When an electron in an excited state drops to a lower state, the lost energy is

    emitted as a single photon of light. If the energy of the initial state of the electron is Ei and

    the energy of the final state of the electron is Ef , then the energy of the emitted photon is

    h = Ei Ef . (202)

    Here, is the frequency of the photon. According to Eq. (201), we have

    Ei Ef = E1(

    1

    n2i 1n2f

    ), (203)

    where ni is the quantum number of the initial state and nf is the quantum number of the

    final state. Therefore, we have

    = E1h

    (1

    n2f 1n2i

    ). (204)

    Since = c/, we find the formula for the spectrum:

    1

    = E1

    hc

    (1

    n2f 1n2i

    ). (205)

    The Lyman series corresponds to the case where nf = 1:

    Lyman (nf = 1):1

    = E1

    hc

    (1

    12 1n2

    ), with n = 2, 3, 4, . . . (206)

    Balmer (nf = 2):1

    = E1

    hc

    (1

    22 1n2

    ), with n = 3, 4, 5, . . . (207)

    51

  • FIG. 15: Transitions between energy levels.

    Paschen (nf = 3):1

    = E1

    hc

    (1

    32 1n2

    ), with n = 4, 5, 6, . . . (208)

    Brackett (nf = 4):1

    = E1

    hc

    (1

    42 1n2

    ), with n = 5, 6, 7, . . . (209)

    Pfund (nf = 5):1

    = E1

    hc

    (1

    52 1n2

    ), with n = 6, 7, 8, . . . (210)

    The Rydberg constant is calculated to be

    R = E1hc

    =me4

    820h3c

    = 1.097 107 m1. (211)

    Thus, the theoretical results of the Bohr model are in agreement with the experimental

    results.

    Example 1

    Find the longest wavelength present in the Balmer series of hydrogen.

    Solution

    In the Balmer series, the quantum number of the final state is nf = 2. The wavelengths

    of the lines in this series is given by the formula

    1

    = R

    (1

    22 1n2

    ), with n = 3, 4, 5, . . . (212)

    52

  • The longest wavelength corresponds to n = 3:

    1

    = R

    (1

    22 132

    )= 0.139R. (213)

    Hence, we obtain

    =1

    0.139R=

    1

    0.139 (1.097 107 m1) = 6.56 107 m = 656 nm. (214)

    Exercise 1: When radiation with a continuous spectrum is passed through a volume of

    hydrogen gas whose atoms are all in the ground state, which spectral series will be present

    in the resulting absorption spectrum?

    Answer: Lyman series: 1/ = R(1 1/n2).Exercise 2: A proton and an electron, both at rest initially, combine to form a ground-

    state hydrogen atom. A single photon is emitted in this process. What is its wavelength?

    Answer: The energy of the ground state is E1. The wavelength of the emitted photon is

    1/ = E1/ch = R = 1.097 107 m1. Hence = 0.912 107 m = 91.2 nm.Exercise 3: How many different wavelengths would appear in the emission spectrum of

    hydrogen atoms initially in the n = 5 state?

    Answer: 4 lines (n = 5 n = 1, 2, 3, 4) for the downward transitions with ni = 5.However, there are 10 lines for the downward transitions with ni = 5, 4, 3, 2.

    Exercise 4: An excited hydrogen atom emits a photon in returning to the ground state.

    Derive a formula for the quantum number ni of the initial state in terms of and R. Use

    this formula to find ni for a 102.55-nm photon.

    Answer: Lyman series: 1/ = R(1 1/n2). Hence we find n = 1/1 1/R.For a 102.55-nm photon, we find n = 1/

    1 1/[(102.55 109)(1.097 107)] ==

    1/1 1/1.125 = 1/1 0.889 = 1/0.111 = 1/0.333 = 3.

    Exercise 5: Doppler effect An excited atom of mass m and initial speed v emits a photon

    in its direction of motion. Assume that v is small compared to c, so that relativistic con-

    siderations are not required. If h/2mc2 1, use the requirement that linear momentumand energy must both be conserved to show that the frequency of the photon is higher by

    /0 = v/c than it would have been if the atom had been at rest.Answer: We have two equations

    mv = hk +mv,

    Ei +mv2

    2= Ef + h +

    mv2

    2. (215)

    53

  • Introduce the notation EiEf = h0. Elimination of v from the above two equations gives

    h0 = h

    (1 v

    c+

    h

    2mc2

    ). (216)

    Since h/2mc2 1, we neglect this term. Consequently, we obtain

    =0

    1 v/c= 0(1 + v/c). (217)

    Then, we have / = v/c.

    54

  • VI. CORRESPONDENCE PRINCIPLE

    Quantum physics is very different from classical physics in the microworld. However,

    quantum physics must give the same results as classical physics in the macroworld where

    classical physics is valid. We show that this basic requirement is true for the Bohr model of

    the hydrogen atom.

    We calculate the frequency of radiation from a hydrogen atom.

    a) Classical picture. According to the electromagnetic theory, the radiation frequency of

    a hydrogen atom is equal to the frequency of the evolution of the electron rotating around

    the nucleus, or to an integer multiple of this frequency. According to Eq. (190), the speed

    of the electron is

    v =e

    4pi0mr. (218)

    Hence, the evolution frequency of the electron is

    f =v

    2pir=

    e

    2pi4pi0mr3

    . (219)

    The radius of an stable orbit is given in terms of the quantum number n by Eq. (193) as

    r = n2h20pime2

    . (220)

    Therefore, the evolution frequency of the electron is

    f =me4

    420h3n3

    = E1h

    2

    n3. (221)

    b) Quantum picture. According to the Bohr model, when the electron drops from an

    orbit ni to an orbit nf , a photon is emitted. The frequency of the emitted photon is given

    by Eq. (204) as

    = E1h

    (1

    n2f 1n2i

    ). (222)

    Under what circumstances should the Bohr atom behave classically. Assume that the

    electron orbit is so large that we might be able to measure it directly. In this case, quantum

    effects ought to be negligible. An orbit 10 m across, for instance, meets this specification.

    The quantum number of this orbit is n = 435, a large number.

    Lets write n = ni and n p = nf . With this notation,

    = E1h

    (1

    (n p)2 1

    n2

    )= E1

    h

    2np p2n2(n p)2 . (223)

    55

  • When n is much larger than p, we can use the approximations 2npp2 2np and (np)2 n2. Hence, we find the photon frequency

    = E1h

    2p

    n3. (224)

    When p = 1, the photon frequency is exactly the same as the frequency of rotation f of

    the orbital electron given by Eq. (221). Multiples of the rotation frequency f are radiated

    when p = 2, 3, 4, . . . . Hence, both quantum and classical pictures of the hydrogen atom give

    the same results in the limit of very large quantum numbers.

    The requirement that quantum physics give the same results as classical physics in the

    limit of large quantum numbers was called by Bohr the correspondence principle.

    Since the de Broglie electron wavelength is = h/mv, we can rewrite the condition for

    orbital stability n = 2pir, see Eq. (192), in the form

    mvr =nh

    2pi. (225)

    Exercise 1: Of the following quantities, which increase and which decrease in the Bohr

    model as n increases: Frequency of evolution, electron speed, electron wavelength, angular

    momentum, potential energy, kinetic energy, total energy.

    Answer: Frequency of evolution f v/r 1/n3 decreases, electron speed v 1/r 1/n decreases, electron wavelength r/n n increases, kinetic energy K v2 1/n2decreases, potential energy U 1/r 1/n2 increases, total energy E 1/n2 increases.

    Exercise 2: Show that the frequency of the photon emitted by a hydrogen atom in going

    from the level n+1 to the level n is always intermediate between the frequencies of evolution

    of the electron in the respective orbits.

    Answer: Photon frequency is

    =E1h

    [1

    (n+ 1)2 1n2

    ]=|E1|h

    [1

    n2 1(n+ 1)2

    ]. (226)

    The frequencies of revolution is

    fn = E1h

    2

    n3=|E1|h

    2

    n3. (227)

    Since2

    (n+ 1)3 0. (298)

    In deriving the above equations we used the formula cos2 x = 1/(1 + tan2 x)

    (2) The case

    tankL

    2= k

    a(299)

    leads to sin kL2 = kk0 ,

    tankL

    2< 0. (300)

    In deriving the above equations we used the formula sin2 x = tan2 x/(1 + tan2 x)

    In Fig. 24, we illustrate the graphic solutions for the energies of the bound states of a

    particle in a square potential well. When U0 is finite, k0 is finite. In this case, the number

    of solutions is finite. The number of solutions increases with increasing potential height U0.

    Note that the wavelengths that fit into the well with a finite height are longer than those

    for the well with an infinite height. Due to this fact, the corresponding momenta are lower,

    and therefore the energy levels En are lower for each n than they are for a particle in an

    infinite well.

    78

  • XII. TUNNEL EFFECT

    In the previous section, the walls of the potential well were of finite height but they were

    assumed to be infinitely thick. As a result the particle was trapped forever though it could

    penetrate the walls.

    We now look at the situation where the potential barrier has a finite height as well as a

    finite width. What we will find is that the particle has a certain probabilitynot necessarily

    great, but not zero eitherof passing through the barrier and emerging on the other side.

    The particle lacks the energy to go over the top of