quantum physics problem sheet 1€¦ · friday 18th march 2011 (answers available from 5pm on...
TRANSCRIPT
Monday 7th March 2011(Answers available from 5pm on Friday 18th March)
Quantum Physics Problem Sheet 1
Note that these questions are ordered by topic, not by difficulty. Do not get disheartened if you findsome of the questions near the beginning of the sheet hard.
Units and Magnitudes
1. Metallic aluminium has atomic weight 27 and density 2700 kg m−3. Estimate the radiusof an aluminium atom in Ångstroms.
2. Electrons (mass m = 0.511 MeV/c2) in a television tube are accelerated through a poten-tial difference of 25 kV. Using relativistic formulae and units (MeV/c2 for energies, MeV/cfor momenta, and so on), calculate the final speed of the electrons as a fraction of the speedof light. Are relativistic effects important?
3. The ionisation energy of a hydrogen atom is 13.6 eV. By equating the thermal energyscale, kBT , to the ionisation energy, estimate the temperature of the universe when thefirst neutral H atoms formed.
Travelling Waves
4. Without referring to your lecture notes, show that the wave
ψ(x, t) = a cos(−kx− ωt+ φ)
has wavelength λ = 2π/k and frequency ν = ω/2π. Show that the wave crests travel withspeed ω/k in the −x direction.
5. The dispersion relation of large deep ocean waves is ω =√gk. What is the phase velocity
(the velocity of the wave crests) when the wavelength is 10 m?
Complex Representation of Waves, Interference and Diffraction
6. Use complex numbers to show that cos θ + sin θ =√
2 cos(θ − π/4). [Hint: sin θ =Re(−ieiθ).]
1
7. Consider the function
ψ(x, t) = a cos(kx− ωt) + a cos(−kx− ωt)
obtained by superposing right- and left-going travelling waves of equal amplitude. Using(i) real numbers only, and (ii) complex numbers, show that ψ(x, t) is a standing wave andcalculate its intensity I(x). Show that the position average of I(x) is 2a2. Why is thisresult expected?
8. This question takes you through the theory of diffraction — the way that waves spread outas they emerge from a narrow opening. Although quite difficult, it provides good practicewith the complex representation of waves and will prove important when we discuss theuncertainty principle. It is worth the effort.
θ
y=0
y=−d/2
y=d/2
y
ζ direction
ζ=0
Parallel monochromatic plane waves of wavelength λ = 2π/k and frequency ν = ω/2πpass through a narrow slit stretching from y = −d/2 to y = d/2. To work out the waveemerging at angle θ, it is helpful to imagine dividing the slit into huge numbers of tinysegments, each of height ∆y. The amplitude of the wave emerging from each segment isproportional to the height of that segment, and so the wave emerging from the segment aty = 0 may be written as
Aei(kζ−ωt)∆y ,
where ζ measures the distance from the centre of the slit in the θ direction and the complexconstant A encodes the overall amplitude and phase of the wave.
(i) Write down the wave emerging from the segment ∆y at height y.
(ii) Show that the total wave emerging in the ζ direction is
ψ(ζ, t) = Aei(kζ−ωt)∫ d/2
−d/2eiky sin θdy .
2
(iii) Evaluate the integral and hence show that the intensity emerging at angle θ is
I(θ) =|A|2d2 sin2
(kd sin θ
2
)(kd sin θ
2
)2 .
(iv) Sketch I as a function of 12kd sin θ. What is the path-length difference between the
top and the bottom of the slit when θ corresponds to the first zero of the diffractionpattern?
Physical Constants
me ≈ 9.11× 10−31 kg ≈ 511 keV/c2
atomic mass unit ≈ 1.66× 10−27 kgh ≈ 6.63× 10−34 Jsh̄ ≈ 1.05× 10−34 Jsc ≈ 3.00× 108 ms−1
e ≈ 1.60× 10−19 Cg ≈ 9.8 ms−1
NA ≈ 6.02× 1023
R ≈ 8.314 JK−1
kB ≈ 1.38× 10−23 JK−1
Numerical Answers
1. Any radius between 1.25 Å and 1.6 Å acceptable.
2. v/c ≈ 0.30.
3. 160, 000 K.
5. 3.96 ms−1.
3
Friday 18th March 2011(Answers available from 5pm on Friday 25th March)
Quantum Physics Problem Sheet 2
Note that these questions are ordered by topic, not by difficulty. Do not get disheartened if you findsome of the questions near the beginning of the sheet hard.
Photons
1. The energy flux of starlight reaching us from a sixth-magnitude star (approximately thefaintest that can be seen by the naked eye) is 1.4×10−10 Wm−2. If you are looking at sucha star, how many photons enter your eye per second? On average, how many photons areinside your eye at any one time? Do these estimates provide any evidence that the humaneye can detect single photons? (Assume that the diameter of your dark-adapted pupil is0.7 cm, the length of your eye is 4 cm, and the wavelength of the light is 500 nm.)
2. Electrons in an X-ray tube are accelerated by a potential difference of 30 kV. What is theminimum wavelength of the resulting X-rays?
3. In a photoelectric effect experiment, a current was observed when the metallic cathodewas illuminated with light of wavelength 310 nm, but none was observed with longerwavelength light.
(i) Estimate the work function of the cathode in eV.
(ii) What is the energy (in eV) of a photon of wavelength 200 nm?
(iii) Find the stopping potential (in Volts) at 200 nm. Hence find the maximum kineticenergy (in eV) of the electrons emitted at this wavelength.
4. In a photoelectric effect experiment, the cathode was illuminated with EM radiation ofthree different frequencies and the stopping potential V0 was measured for each.
Frequency ν (1015 Hz) Stopping Potential V0 (V)0.75 1.01.00 2.01.25 3.0
(i) Make a sketch of stopping potential against frequency and hence estimate the workfunction of the metal in eV.
(ii) From the slope of your line, estimate the value of Planck’s constant h. How close isthis to the accepted value?
1
5. X-rays of energy 20 keV are Compton scattered by a thin metal foil through an angle of60o. Find (i) the wavelength of the scattered photons, and (ii) the energy (in eV) lost byeach photon. Would the kinetic energy of the recoiling electrons be sufficient to allowsome of them to escape from the metal?What value of the scattering angle θ would give the largest change in wavelength? Whatis the maximum possible wavelength of the scattered photon?
6. (Q40-63 from Young and Freedman) Nuclear fusion reactions at the centre of the sun pro-duce gamma-ray photons with energies of order 1 MeV. By contrast, what we see emanat-ing from the sun’s surface are visible photons with wavelengths of order 500 nm. Modelsof the solar interior explain this wavelength difference by suggesting that every photon isCompton scattered about 1026 times during its journey from the centre of the sun to thesurface.
(i) Estimate the average increase in wavelength of a solar photon per Compton-scatteringevent.
(ii) Find the angle in degrees through which the photon is scattered in the scattering eventdescribed in part (i). (Hint: a useful approximation is cosφ ≈ 1 − φ2/2, which isvalid when φ is � 1 radian.)
(iii) It is estimated that a photon takes about 106 years to travel from the core to thesurface of the sun. Find the average distance that light can travel within the interiorof the sun without being scattered.
As you can see, the sun is very opaque, and the radiation has plenty of opportunity to reach equi-librium with the matter before emerging. This explains why the sun emits black-body radiation.
Physical Constantsh ≈ 6.63 × 10−34 Jsh̄ ≈ 1.05 × 10−34 Jsc ≈ 3.00 × 108 ms−1
e ≈ 1.60 × 10−19 Cme ≈ 9.11 × 10−31 kg
Numerical Answers
1. Number of photons entering eye per second ≈ 13, 500; expected number inside eye at any onetime ≈ 1.8 × 10−6.
2. 4.14 × 10−11 m.
3. (i) 4.00 eV; (ii) 6.22 eV; (iii) Stopping potential 2.22 V; maximum KE 2.22 eV.
4. (i) 2.0 eV.
5. (i) 6.34 × 10−11 m; (ii) 378 eV (with large and difficult to estimate numerical uncertainty).Angle for largest change in wavelength is θ = 180o; maximum possible scattered wavelength is6.71 × 10−11 m.
6. (i) 5 × 10−33 m; (ii) 3.68 × 10−9 degrees; (iii) 9.46 × 10−5 m.
2
Friday 25th March 2011(Answers available from 5pm on Monday 16th May)
Quantum Physics Problem Sheet 3
Note that these questions are ordered by topic, not by difficulty. Do not get disheartened if you findsome of the questions near the beginning of the sheet hard.
De Broglie Waves
1. Neutron diffraction is often used to study the atomic positions and atomic-scale magneticfields in solids. Suggest a reasonable value for the de Broglie wavelength of the neutronsused in such experiments. Find the kinetic energy (in eV) of neutrons of this wavelength.Use the equation E = 3kBT/2 to translate the kinetic energy into an equivalent tempera-ture.
2. Electrons of energy 100 eV pass through a narrow slit of width 1µm. What is the distancebetween the zeros of intensity on either side of the central peak of the electron diffractionpattern 1 m away from the slit? (Hint: you worked out the form of the diffraction patternin Q8 of Problem Sheet 1.)
3. An electron and a positron (which has the same mass as an electron but the oppositecharge) annihilate at rest to produce two photons.
(i) The two photons have the same wavelength and travel in opposite directions. Why?
(ii) Calculate the photon momentum and wavelength.
4. Show that the de Broglie wavelength of a particle of mass m and kinetic energy 3kBT/2is
λ =h√
3mkBT.
The molar volume of liquid 4He is 27.6 cm3. Assuming that each atom occupies a cubeof side d, calculate the temperature at which λ = d. Discuss the temperature range overwhich the wave-like properties of the atoms in liquid 4He are likely to be important.(Incidentally, liquid 4He becomes a superfluid, able to flow without friction, below 2.17 K.The above calculation suggests that superfluidity is almost certainly a QM phenomenon.)
5. When liquid 4He freezes, every atom is confined to a “box” (its lattice site in the crystal).Since liquids and solids normally have similar densities, the box size is similar to the value
1
of d calculated in Q4. Assuming that the de Broglie wave of wavelength λ associated withthe 4He atom has to equal zero at the box walls, show (perhaps by drawing a diagram) thatd = nλ/2, where n is any integer > 0. Hence calculate the smallest possible momentumand kinetic energy of the confined atom. At what temperature T would the thermal kineticenergy 3kBT/2 equal the quantum mechanical KE of confinement?
Note: The idea behind this question is important and very general. Since the de Brogliewavelength of a confined particle has to “fit in” to the confining box, the particle musthave a non-zero momentum and KE. The particle must therefore be moving — rattlingbackwards and forwards inside the box — even at zero temperature. This confinement-induced motion is called zero-point motion, and the corresponding kinetic energy iscalled zero-point energy. The orbits of electrons in atoms can be viewed as a type ofzero-point motion.
In most solids, the atoms are heavy enough and the chemical bonds strong enough thatthe zero-point motion of the atoms (as opposed to the electrons) is unimportant. In4He , however, where the atoms are comparatively light and the bonding is very weak,the zero-point motion alone is sufficient to melt the solid — there is no need to heat itup! This is why liquid 4He remains liquid right down to T = 0 K. To solidify 4He , itis necessary to apply pressure.
The Bohr Atom
6. Light of frequency 7.316×1014 Hz is emitted in a downward transition to the n = 2 energylevel of a hydrogen atom. What was the initial energy level?
7. Generalise the derivation of the Bohr model given in lectures to obtain a formula for theenergy levels of ions such as He+ or Al12+, which have atomic number Z > 1 (Z = 2 forHe; Z = 13 for Al) but only one orbiting electron.
8. (i) What is the shortest wavelength photon that could be emitted by a hydrogen atomthat did not end up in the ground state n = 1 level.
(ii) A spectral line at 121.5 nm is observed in both the H atom emission spectrum and,at exactly the same wavelength, in the spectrum from the singly ionized He+ ion.Estimate the initial and final (ni and nf) levels of this transition in the H and He+
spectra.
9. An electron and a positron (same mass, opposite charge) can form a short-lived boundstate called a positronium atom, in which the two particles orbit their centre of mass. Usethe Bohr model to estimate the binding energy and the distance between the electron andthe positron in the positronium ground state. (Hint: consider a single particle of massmreduced = (1/m+ 1/m)−1 orbiting a fixed origin.)
2
10. A particle of massm is bound in a spherical potential of the form V (r) = Cr. By adaptingthe derivation of the Bohr model to this case, show that:
(i) the quantised orbital radii are
rn =
(h̄2n2
mC
)1/3
(n any integer ≥ 1) ;
(ii) the quantised energy levels are
En =3
2
(C2h̄2n2
m
)1/3
(n any integer ≥ 1) .
Physical Constants
me ≈ 9.11× 10−31 kg ≈ 511 keV/c2
mn ≈ 1.67× 10−27 kgatomic mass unit ≈ 1.66× 10−27 kg
h ≈ 6.63× 10−34 Jsh̄ ≈ 1.05× 10−34 Jsc ≈ 3.00× 108 ms−1
e ≈ 1.60× 10−19 CkB ≈ 1.38× 10−23 JK−1
NA ≈ 6.02× 1023
RH ≈ 1.097× 107 m−1
ε0 ≈ 8.85× 10−12 C2J−1m−1
Numerical Answers
1. Any wavelength within an order of magnitude of the interatomic spacing OK. Choosing λ =10−10 m gives KE of 0.083 eV and temperature of 640 K.
2. 2.46× 10−4 m.
3. (ii) p = 2.73× 10−22 kg m s−1; λ = 2.43× 10−12 m.
4. T = 12.5 K.
5. pmin = 9.26× 10−25 kg m s−1; KEmin = 6.46× 10−23 J; T = 3.12 K.
6. Initial energy level ni = 6.
8. (i) 3.65× 10−7 m; (ii) For H, ni = 2 and nf = 1; for He, ni = 4 and nf = 2.
9. Ebinding ≈ 6.8 eV; r ≈ 1.06 Å.
3
Monday 16th May 2011(Answers available from 5pm on Friday 27th May)
Quantum Physics Problem Sheet 4
Note that these questions are ordered by topic, not by difficulty. Do not get disheartened if you findsome of the questions near the beginning of the sheet hard.
Working with Wavefunctions
1. A particle of mass m is confined within a potential well of the form:
V (x) =
{0 |x| < d/2 ,∞ otherwise .
The (unnormalised) ground-state wavefunction is
ψ(x, t) =
{cos (πx/d) e−i(h̄π
2/2md2)t |x| < d/2 ,0 otherwise .
(i) A measurement is made of the particle position. Show that the probability that themeasured value lies between x and x + dx is independent of time. Where is thepartice most likely to be found?
(ii) Normalise ψ(x, t).
(iii) Find 〈x〉 and 〈x2〉. Hence obtain the rms width ∆x of the probability density func-tion.
[The following result may be used without proof:
∫ π/2
−π/2θ2 cos2 θ dθ =
π
4
(π2
6− 1
)]
2. The normalised wavefunction ψ(r, t) of a particle moving in three dimensions has thefollowing probability interpretation:
|ψ(r, t)|2 dV ={probability that a measurement of the position of the particle
yields a value in the volume element dV at r.
The (unnormalised) ground-state wavefunction of the electron in a hydrogen atom ise−r/a0e−iE0t/h̄, where a0 ≈ 0.53 Å is the Bohr radius and E0 ≈ −13.6 eV is the ground-state energy.
1
(i) Show that the electron probability density is independent of time. Where is the prob-ability density largest?
(ii) Write down the normalisation condition that must be satisfied by any wavefunctionsatisfying the probabiliy interpretation described above. Show that the normalisedground-state wavefunction is
ψ(r, t) =1√πa3
0
e−r/a0e−iE0t/h̄ .
(Hint: you will have to evaluate the integral∫∞
0 e−2r/a04πr2dr. You may assume that∫∞0 ξne−ξdξ = n!)
(iii) A measurement is made of the electron position r. Write down an expression forthe probability that the electron is found in the spherical shell with inner and outerradii r and r + dr. Hence show that the most probable distance of the electron fromthe nucleus is a0. Is this result consistent with the position of maximum probabilitydensity found in part (i)?
(iv) Show that the mean distance of the electron from the nucleus is 3a0/2 and that therms distance is
√3 a0.
(v) Show that the probability of finding the electron on top of the nucleus, which youmay assume has radius 10−15 m, is approximately 9× 10−15. (Hint: no integration isrequired for this part.)
Momentum Measurements
3. The (unnormalised) wavefunction of a quantum mechanical particle that is being reflectedfrom an infinitely high potential barrier at x = 0 takes the form:
ψ(x, t) =
{sin(kx)e−iωt x < 0 ,0 otherwise .
where ω = h̄k2/2m. In the region x < 0, show that ψ(x, t) may be written as a sum ofright- and left-going complex travelling waves.
The momentum of the particle is measured. What values might be obtained and what aretheir relative probabilities?
The Uncertainty Principle
4. When undergoing radioactive decay, nuclei often emit electrons with energies between 1and 10 MeV. Use the position-momentum uncertainty principle to show that an electron ofenergy 1 MeV could not have been contained in the nucleus before the decay.
2
5. (i) Consider a QM simple harmonic oscillator of mass m, spring constant s and naturalfrequency ω =
√s/m. Assuming that the average displacement 〈x〉 and average
momentum 〈p〉 are both zero, show that the mean energy,
〈E〉 = 〈KE〉+ 〈PE〉 =〈p2〉2m
+1
2s〈x2〉 ,
may be expressed in the form:
〈E〉 =(∆p)2
2m+
1
2mω2(∆x)2 ,
where ∆x is the rms displacement and ∆p is the rms momentum.
(ii) Use the uncertainty principle to show that the ground-state energy of the oscillatormust be greater than or equal to 1
2h̄ω. (In fact, the ground-state energy is exactly
12h̄ω.)
6. When a short-lived excited state of an atom or molecule decays to the ground state, thewavelength of the photon emitted is uncertain. What is the minimum energy uncertainty(often known as the linewidth), measured in eV, of photons emitted from a state of lifetime2.6× 10−10 s.
7. In 1935, Yukawa proposed that the nuclear force arises through the emission of an un-known virtual particle, a pion, by one of the nucleons and its absorption by the other.
Assuming that ∆E∆t ∼ h̄, relate the lifetime of the virtual pion to its mass m. Henceshow that the distance travelled by the pion during its lifetime is unlikely to exceed h̄/mc.Given that the range of the nuclear force is approximately 1.4 × 10−15 m, estimate m inMeV/c2. (When the pion was finally discovered, its mass was found to be ∼140 MeV/c2.The high accuracy of the estimate obtained in this question is of course fortuitous.)
The Schrödinger Equation
8. A particle of mass m is confined within a potential well of the form:
V (x) =
{0 0 < x < d∞ otherwise
Write down the time-independent Schrödinger equation for this system. Verify that thewavefunctions
ψn(x) =
√
2
dsin (nπx/d) 0 < x < d
0 otherwise
where n = 1, 2, . . ., are normalised energy eigenfunctions. Find the corresponding energyeigenvalues.
3
9. Assuming that a nucleus can be modelled as an infinite potential well of width d =10−15 m, estimate the energy a nucleon (mass 1.67 × 10−27 kg) emits as it falls from then = 2 to the n = 1 level. Is this a sensible number?
10. The bond in a carbon monoxide (12C16O) molecule acts like a spring with spring con-stant s = 1857 Nm−1. Calculate the angular frequency of vibration according to classicalphysics. (Hint: use the reduced mass.) Hence work out:
(i) The wavelength of the photons emitted when a CO molecule makes a transition be-tween neighbouring vibrational states.
(ii) The vibrational zero-point energy of a CO molecule.
11. The time-independent Schrödinger equation for a simple harmonic oscillator of mass mand spring constant s is
− h̄2
2m
d2ψ(x)
dx2+
1
2sx2ψ(x) = Eψ(x) .
Verify by subsitution that the function
ψ(x) = e−αx2
satisfies this equation if α is chosen correctly. Show that the corresponding eigenvalue is12h̄ω, where ω is the classical angular frequency of the oscillator.
12. An electron of energy E encounters a potential barrier of height V > E and width a.Show that the probability that the electron tunnels across the barrier is approximatelyexp(−2γa), where γ =
√2m(V − E)/h̄2. Estimate the probability of tunnelling across a
10−6 m gap between two slabs of metal of work function 5 eV.
Extra Questions for Enthusiasts (not examinable)
13. The probability density of the positions at which the electrons in a two-slit experimenthit the screen is the square modulus of a complex wave that travels through both slits.Which slit does the electron itself go through? As I tried to explain in lectures, this is not aquestion that quantum mechanics can answer. The point-like “particle” that we associatewith the electron is simply a convenient abstraction or mental picture that we use to helpinterpret the results of certain types of measurement — clicks of an electron counter orflashes of light originating from specific points in space. QM tells us almost nothingabout what “really” produces these experimental results, and does not allow us to ascribea position to the particle abstraction between measurements.
To help make these ideas more concrete, consider the thought experiment illustrated below.
4
d
θ
����������������������������������������
����������������������������������������
x
z
������������������������������������������
������������������������������������������
������������������������������������������
������������������������������������������
When the light bulb is switched off, the electron wave passing through the two slits inter-feres constructively at angles θ satisfying:
d sin θ = nλelectron (n = 0, 1, 2, . . .).
If θ is small, this translates to θ ≈ nλelectron/d.
Now imagine switching on the light bulb (which is shaded to ensure that it only sends lightalong the plane of the barrier containing the slits). If the light is bright enough, most of theelectrons emerging from the slits will scatter one or more photons. By imaging the flashesof light corresponding to the scattered photons, we can find out where they came from andhence measure the positions of the electrons when they are just behind the slits.
What are the results of this experiment? As you might expect, we see flashes of lightoriginating from both slits, but never from both at the same time. Every time we detectan electron, it is either behind one slit or behind the other. However, when we look at thescreen, we find that the two-slit diffraction pattern has disappeared! The measurement ofthe electron position just behind the slits has somehow destroyed the interference pattern.We can have it one way (interference pattern but no measurement of electron path) or theother (measurement of electron path but no interference pattern), but not both.
(i) In order for this experiment to work, the photon wavelength must be less than orapproximately equal to d. Why?
(ii) Assuming that the x component of the momentum of the electron is small (i.e., as-suming that θ is small), show that the impact of a single photon of wavelength λphoton
deflects the electron by an angle ∆θ of order λelectron/λphoton.
5
(iii) Explain why the measurement of the electron position wipes out the electron diffrac-tion pattern.
14. In next year’s QM course, you will prove that different energy eigenfunctions of the samephysical system are “orthogonal” to each other, in the sense that∫ ∞
−∞ψ∗n(x)ψm(x) dx = 0 n 6= m.
Verify this result for the energy eigenfunctions given in Q8. (Since these functions are realthere is no need to worry about the complex conjugation.)
The following result may be used without proof:
sin(nθ) sin(mθ) =1
2[cos((n−m)θ)− cos((n+m)θ)]
15. Show that the function
φ(x) =∞∑n=1
cnψn(x)
is normalised if and only if the (possibly complex) coefficients cn satisfy∑∞n=1 |cn|2 = 1.
(Hint: use the orthogonality relation from Q14 and remember that ψn(x) is normalised.)Hazard a guess at the probability interpretation of |cn|2.
Physical Constants
me ≈ 9.11× 10−31 J ≈ 511 keV/c2
atomic mass unit ≈ 1.66× 10−27 kgh ≈ 6.63× 10−34 Jsh̄ ≈ 1.05× 10−34 Jsc ≈ 3.00× 108 ms−1
e ≈ 1.60× 10−19 C
Numerical Answers
6. 1.26× 10−6 eV.
7. 141 MeV/c2.
9. 610 MeV.
10. (i) 4.67× 10−6 m; 0.13 eV.
12. ∼ e−23,000 ∼ 10−10,000.
6
Quantum Physics Answer Sheet 1
Units and Magnitudes
1. Number of atoms per unit volume is:
n =Mass per unit volume
Mass per atom=
2700
27× 1.66× 10−27= 6.02× 1028 m−3 .
Volume per atom is 1/n = 1.66 × 10−29 m3. If we assume that each atom is a sphere ofradius r, we have
4
3πr3 = 1.66× 10−29
and hence r = 1.58× 10−10 m = 1.58 Å. This is an overestimate because it is impossibleto fill space with spheres (there are always gaps between them). A different estimate maybe obtained by assuming that the spheres are stacked in a simple cubic lattice. Each cubeof side 2r then contains one atom of radius r and the volume per atom is 8r3. This gives
8r3 = 1.66× 10−29
and hence r = 1.28 × 10−10 m = 1.28 Å. It is possible to pack spheres much moreefficiently than in a simple cubic lattice (the packing in Al is actually face-centred cubic —the most efficient regular packing of spheres), and so the true answer must lie somewherebetween these two estimates.
2. The KE of the electrons is 25 keV = 0.025 MeV. Since the KE is the total energy, mγc2,minus the rest energy, mc2, this gives:
0.025 = γmc2 −mc2 =
1√1− v2
c2
− 1
× 0.51 .
Aside: I remember finding these “relativistic” units very confusing the first time I metthem. The statement that the mass m of an electron is 0.51 MeV/c2 is exactly equiv-alent to the statement that mc2 (which is, of course, an energy) is equal to 0.51 MeV.Similarly, if you are told that the momentum p is equal to 1 MeV/c, the implication isthat pc (which also has the dimensions of an energy) is equal to 1 MeV.
A little algebra now gives
1− v2
c2=
1(1 + 0.025
0.51
)2 ,and hence
v
c≈ 0.30 .
1
The relativistic effects (which depend on v2/c2) are small but not negligible. Televisiondesigners presumably need to take them into account.
3. For a very rough estimate, equate kBT (or 3kBT/2 if you insist) to 13.6 eV. This gives:
T =13.6× 1.6× 10−19
1.38× 10−23≈ 160, 000 K .
Travelling Waves
4. To work out the wavelength, note that
ψ(x+
2π
k, t)
= a cos(−k
[x+
2π
k
]− ωt+ φ
)= a cos(−kx− ωt+ φ− 2π) = ψ(x, t) .
Hence, ψ(x, t) changes by one full period as x increases by 2π/k at constant t. In otherwords, the wavelength λ = 2π/k.
Similarly, to work out the time period, note that
ψ(x, t+
2π
ω
)= a cos
(−kx− ω
[t+
2π
ω
]+ φ
)= a cos(−kx− ωt+ φ− 2π) = ψ(x, t) .
Hence, ψ(x, t) changes by one full period as t increases by 2π/ω at constant x. In otherwords, the time period T = 2π/ω. The frequency ν = 1/T is therefore given by ν =ω/2π.
To work out the velocity of the crests, consider the crest at the point xcrest where theargument of the cosine function is zero:
−kxcrest − ωt+ φ = 0 .
This equation rearranges to
xcrest =φ
k− ω
kt .
Comparing with the equation xcrest = x0 + vt that describes uniform motion at velocity v,we see that the wave crest is at position x0 = φ/k at time t = 0 and is moving at velocity−ω/k.
5. The phase velocity is
v =ω
k=
√g
k=
√gλ
2π≈√
9.8× 10
2π= 3.96 ms−1 .
2
Complex Representation of Waves, Interference and Diffraction
6. Since eiθ = cos θ + i sin θ, it follows that cos θ = Re(eiθ) and sin θ = Re(−ieiθ). Hence
cos θ + sin θ = Re(eiθ − ieiθ
)= Re
((1− i)eiθ
).
The prefactor 1− i may be rewritten in the form reiφ, where:
r =√
[Re(1− i)]2 + [Im(1− i)]2 =√
12 + (−1)2 =√
2 ,
φ = tan−1
(Im(1− i)Re(1− i)
)= tan−1(−1) = −π
4.
(An alternative way to find r and φ is to draw the complex number 1 − i in the Arganddiagram and note its length and argument.)
Hence
cos θ + sin θ = Re(√
2e−iπ/4eiθ)
=√
2Re(ei(θ−π/4)
)=√
2 cos(θ − π/4) ,
as required.
7. (i) Using real arithmetic only:
ψ(x, t) = a cos(kx− ωt) + a cos(−kx− ωt)= a(cos kx cosωt+ sin kx sinωt) + a(cos kx cosωt− sin kx sinωt)
= 2a cos kx cosωt ,
where the first step used the trigonometic identity
cos(θ + φ) = cos θ cosφ− sin θ sinφ,
and the information that cos θ is an even function [cos(−θ) = cos θ] while sin θ is anodd function [sin(−θ) = − sin θ].
(ii) Using complex arithmetic:
ψ(x, t) = aei(kx−ωt) + aei(−kx−ωt) .
(Actually, of course, ψ(x, t) is the real part of the expression on the RHS. From nowon we shall take this as understood.) Hence:
ψ(x, t) = ae−iωt(eikx + e−ikx
)= 2ae−iωt cos kx .
This step used the identities
eiθ = cos θ + i sin θ and e−iθ = cos θ − i sin θ ,
3
from which it follows that
eiθ + e−iθ = 2 cos θ (and eiθ − e−iθ = 2i sin θ) .
Taking the real part of ψ(x, t) now gives the real wavefunction
ψ(x, t) = 2a cos kx cosωt ,
exactly as in part (i).
The amplitude atotal of ψ(x, t) depends on position,
atotal(x) = 2a cos kx ,
and hence so does the intensity,
Itotal(x) = a2total(x) = 4a2 cos2 kx .
The position average of the intensity is 4a2 times the position average of cos2 kx. Aslong as you average over a whole number of half periods, the average value of cos2 θ (orsin2 θ) is equal to 1/2 (this is easy to prove by starting from the trigonometic identitycos2 θ = (1 + cos(2θ))/2 and noting that cos(2θ) averages to zero). Hence, the averageintensity is 2a2.
This is expected because the average intensity is proportional to the average energy per unitvolume. Assuming that energy is conserved, the average energy density of the standingwave must equal the sum of the average energy densities of the two travelling waves.Both travelling waves have intensity a2 (independent of position), and hence the averageintensity of the standing wave must be 2a2.
8. (i) The wave emerging from the segment ∆y at height y is
Aei(k[ζ−(−y sin θ)]−ωt)∆y = Aei(kζ−ωt+ky sin θ)∆y .
(ii) The total wave emerging in the ζ direction is the sum of the waves emerging from allthe little segments:
ψ(ζ, t) =∑
segments
Aei(kζ−ωt+ky sin θ)∆y ,
where, for each segment (or, equivalently, for each term in the sum), y is the positionof the centre of that segment. In the limit as ∆y → 0, the sum turns into the integral:
ψ(ζ, t) =∫ d/2
−d/2Aei(kζ−ωt+ky sin θ)dy .
Since ei(kζ−ωt+ky sin θ) = ei(kζ−ωt)eiky sin θ, this is equivalent to
ψ(ζ, t) = Aei(kζ−ωt)∫ d/2
−d/2eiky sin θdy ,
as required.
4
If you are confused about the relationship between the sum and the integral, con-sider the diagram below. The integral of f(y) from a to b is the sum of the areas ofthe segments of width ∆y.
∆ y
f(y)
ya b
Since the area of each segment is approximately f(y)∆y, it follows that∫ b
af(y)dy ≈
∑segments
f(y)∆y .
In fact, the integral is normally defined as the limit of this sum as ∆y tends to zero.
(iii) The integral ∫ d/2
−d/2eiky sin θdy
may be written as ∫ d/2
−d/2eαydy ,
where α = ik sin θ. This is easy to integrate:
∫ d/2
−d/2eαydy =
[eαy
α
]d/2−d/2
=1
α
(eαd/2 − e−αd/2
)
=e
12ikd sin θ − e− 1
2ikd sin θ
ik sin θ=
d sin(kd sin θ
2
)kd sin θ
2
,
where the last step used the identity eiφ − e−iφ = 2i sinφ discussed in the answer toquestion 7.The wavefunction ψ(ζ, t) is therefore given by:
ψ(ζ, t) = Adsin
(kd sin θ
2
)kd sin θ
2
ei(kζ−ωt) .
5
The intensity I is the square modulus of the complex amplitude (which is everythingin front of the ei(kζ−ωt) factor). Hence
I =|A|2d2 sin2
(kd sin θ
2
)(kd sin θ
2
)2 ,
as required.
(iv) The diffraction pattern looks like this (note that the horizontal axis shows values ofkd sin θ
2πinstead of values of kd sin θ
2):
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-4 -3 -2 -1 0 1 2 3 4
Inte
nsity / (
|A
|^2 d
^2 )
k d sin(theta) / ( 2 pi )
Single Slit Diffraction Pattern
The first zero occurs where kd sin θ2π
= 1 and hence where
k =2π
λ=
2π
d sin θ.
This implies thatλ = d sin θ .
The difference in the lengths of the paths emerging from the top and bottom of theslit is one wavelength.
6
Quantum Physics Answer Sheet 2
Photons
1. The total energy entering each eye per second is the energy striking a unit area per secondtimes the area of the pupil:
energy entering eye per second = 1.4 × 10−10 × π(0.0035)2
≈ 5.39 × 10−15 J .
Average number of photons entering eye per second is
energy entering eye per secondenergy per photon
=5.39 × 10−15
hc/λ
=5.39 × 10−15 × 500 × 10−9
6.63 × 10−34 × 3.00 × 108
≈ 13, 500 .
The average number of photons inside eye at any one time isnumber entering per second × length of eye
distance a photon travels per second
=13500 × 0.04
3.00 × 108≈ 1.8 × 10−6 .
The actual number of photons in the eye is almost always zero.
Since light always arrives as individual photons, all light detectors must be capable ofdetecting individual photons. A more interesting question is whether the arrival of a singlephoton is sufficient to trigger one of the detectors (rods and cones) in the retina, or whetherit is necessary to bombard that detector with many photons in close succession. Given thatthe eye takes much less than 1 s to process a new image, it is reasonable to assume thatthe “memory” of the detectors is less than, say, 0.2 s. Any effects caused by photons thatarrived more than 0.2 s ago can therefore be ignored. Within 0.2 s, only 2,700 photonsenter the eye, all of which are focused onto the small area of the retina where the imageis formed. Are there more than 2,700 detectors in this area? I have no idea, but I doubt it.In other words, the ability of the eye to see the star provides no convincing evidence thatthe detectors in the retina are triggered by single photons — it may be necessary to hit thesame detector with several photons in quick succession.
2. The electron energy is 30 keV and so the energy hν = hc/λ of the X-ray photons producedmust be less than or equal to 30 keV. The minimum photon wavelength is therefore
λ =hc
30 keV=
6.63 × 10−34 × 3.00 × 108
30 × 103 × 1.6 × 10−19= 4.14 × 10−11 m.
1
3. (i) Light of wavelength greater than λmax = 310 nm is incapable of producing a current.Hence the work function W is given by:
W =hc
λmax
=6.63 × 10−34 × 3.00 × 108
310 × 10−9≈ 6.42 × 10−19 J .
W =6.42 × 10−19
1.60 × 10−19≈ 4.00 eV .
(ii) The energy of a photon of wavelength 200 nm is
E =hc
λ=
6.63 × 10−34 × 3.00 × 108
200 × 10−9≈ 9.95 × 10−19 J ,
or
E =9.95 × 10−19
1.60 × 10−19≈ 6.22 eV .
(iii) The stopping potential V0 at 200 nm is given by Einstein’s equation, W + eV0 = E.Hence
V0 = 6.22 − 4.00 = 2.22 V .
The maximum KE of the emitted electrons is 2.22 eV.
4. (i) The data look like this:
ν15
(10 Hz)
eV0(eV)
0.5 0.750.25 1.0 1.25
2.0
3.0
1.0
0.0
−1.0
−2.0
Since the y intercept is −W , the work function W is 2.0 eV.
(ii) The slope of the line is
(3.0 − 1.0) eV
(1.25 − 0.75) × 1015 Hz=
2 × 1.60 × 10−19 J
0.5 × 1015 s−1= 6.4 × 10−34 Js .
Assuming that the experimental errors were reflected in the precision with whichthe measured values were quoted, this is consistent with the accepted value of 6.63×10−34 Js. (The fact that the three data points lie on a perfect straight line is suspicious,though.)
2
5. (i) The wavelength of the incident photons is
λ =hc
E≈ 6.63 × 10−34 × 3.00 × 108
20 × 103 × 1.60 × 10−19≈ 6.22 × 10−11 m .
The change in wavelength is given by the Compton formula with θ = 60o:
λ′ − λ =h
mc(1 − cos θ) ≈ 6.63 × 10−34 × 0.5
9.11 × 10−31 × 3.00 × 108≈ 1.21 × 10−12 m .
Combining the values of λ and λ′− λ gives the wavelength of the scattered photons:
λ′ ≈ 6.34 × 10−11 m .
(ii) The energy lost by a photon as it scatters is:
E − E ′ = hc(
1
λ− 1
λ′
)= 6.63 × 10−34 × 3.00 × 108
(1
6.22 × 10−11− 1
6.34 × 10−11
)≈ 6.05 × 10−17 J
≈ 378 eV .
Warning: this answer is the difference of two much larger numbers (the incoming and outgo-ing photon energies) and is subject to considerable rounding error. For example, if you storeintermediate values such as λ and λ′ to full calculator precision, the final result changes byseveral eV. Short of using more accurate values for the fundamental constants, there is littlethat can be done about this.
All this energy is transferred to the electron as recoil energy. The work function ofa typical solid is only 5 or 10 eV, so some of the recoiling electrons will certainlyescape from the metal.
The largest change in wavelength would be obtained when θ = 1800, in which case λ′ −λ = 2h/mc ≈ 4.85×10−12 m. The maximum possible wavelength of the scattered photon(assuming only one scattering) is 6.22 × 10−11 + 4.85 × 10−12 ≈ 6.71 × 10−11 m.
6. (i) The initial photon wavelength is
λinit =hc
E≈ 6.63 × 10−34 × 3.00 × 108
106 × 1.60 × 10−19≈ 1.24 × 10−12 m .
The final photon wavelength after 1026 Compton scattering events is 500 nm. Ifwe assume that each scattering event increases the photon wavelength by the sameamount ∆λ, we obtain
1026∆λ ≈ (500 × 10−9 − 1.24 × 10−12) m ,
and hence∆λ ≈ 5 × 10−33 m .
3
(ii) The Compton formula says that
∆λ =h
mc(1 − cos θ) .
Since ∆λ (≈ 5 × 10−33 m) � h/mc (≈ 2.43 × 10−12 m), the average scatteringangle θ must be very small (so that cos θ is very close to 1). We can therefore makethe approximation cos θ ≈ 1 − θ2/2 to obtain ∆λ ≈ hθ2/2mc, and hence
θ ≈√
2mc∆λ
h
≈√
2 × 9.11 × 10−31 × 3.00 × 108 × 5 × 10−33
6.63 × 10−34
≈ 6.42 × 10−11 radians
≈ 3.68 × 10−9 degrees .
(iii) In 106 years, a photon travels a distance:
d = ct = 3.00 × 108 × 60 × 60 × 24 × 365 × 106
≈ 9.46 × 1021 m .
During this time, it scatters 1026 times. Hence, the average distance travelled by aphoton between scattering events is 9.46 × 1021/1026 ≈ 9.46 × 10−5 m or roughly0.1 mm.
4
Quantum Physics Answer Sheet 3
De Broglie Waves
1. In order to use neutron diffraction to study atomic positions and atomic-scale magneticfields, the neutron de Broglie wavelength must be comparable to the size of an atom:
λ ≈ 10−10 m .
The kinetic energy of the neutron is thus:
1
2mv2 =
p2
2m=
h2
2mλ2≈ (6.63× 10−34)2
2× 1.67× 10−27 × 10−20
≈ 1.32× 10−20 J ≈ 0.083 eV .
This is the same as the average energy 3kBT/2 of a neutron in thermal equilibrium attemperature
T ≈ 2× 1.32× 10−20
3× 1.38× 10−23≈ 640 K .
2. The de Broglie wavelength of a 100 eV electron is given by:
100 eV =p2
2m=
h2
2mλ2,
and hence
λ =6.63× 10−34
√2× 9.11× 10−31 × 100× 1.60× 10−19
≈ 1.23× 10−10 m .
w/2
dθ
l
1
From Q8 of Problem Sheet 1, the first zero in the diffraction pattern from a slit of width doccurs where sin θ = λ/d. Hence,
θ = sin−1(λ/d) ≈ sin−1(1.23× 10−10/10−6) ≈ 1.23× 10−4 radians .
The width w of the central diffraction peak is 2l tan θ, where l = 1 m is the distance fromthe slit to the screen. Hence,
w = 2× 1× tan θ ≈ 2θ ≈ 2.46× 10−4 m .
3. (i) When the electron and positron annihilate, both momentum and energy must be con-served. Since the electron and positron are initially at rest, the total momentum isequal to zero and must remain equal to zero. This explains why the two photons haveequal and opposite momentum. Since the energy and (magnitude of) momentum ofa photon are related via E = pc, photons with the same (magnitude of) momentummust also have the same energy.
(ii) The photon energy can be obtained by conservation of energy:
2mc2 = 2× 511 keV = 2Ephoton ,
and hence Ephoton ≈ 511 keV. The photon momentum is
pphoton = Ephoton/c ≈ 511 keV/c
≈ 511× 103 × 1.6× 10−19
3.00× 108≈ 2.73× 10−22 kg ms−1 ,
and the photon wavelength is
λ =h
p≈ 6.63× 10−34
2.73× 10−22≈ 2.43× 10−12 m .
4. A particle of mass m and momentum p has kinetic energy p2/2m. If the kinetic energy isequal to 3kBT/2:
p2
2m=
3kBT
2,
thenp =
√3mkBT .
Combining this result with de Broglie’s equation, p = h/λ, gives:
λ =h
p=
h√3mkBT
,
as required.
Avogadro’s number of He atoms occupy a volume of 27.6× 10−6 m3. Hence, the volumeper atom d3 is
27.6× 10−6
6.02× 1023≈ 4.58× 10−29 m3 .
2
Taking the cube root, we obtain d ≈ 3.58× 10−10 m.
To find the temperature T at which λ = d, we have to solve the equation
h√3mkBT
= d .
Hence
T =1
3mkB
(h
d
)2
≈ 1
3× 4× 1.66× 10−27 × 1.38× 10−23
(6.63× 10−34
3.58× 10−10
)2
≈ 12.5 K .
When the temperature is comparable to or smaller than this value, the de Broglie wave-length of the He atoms will be the same as or greater than the interparticle spacing, andthe wave-like properties of the atoms will be important.
5. The figure below shows that in order for the de Broglie wave of wavelength λ to “fit in” tothe box, the box side d must be an integer multiple of λ/2: d = nλ/2, where n = 1, 2, . . .
λ/2d = 2λ/2 d =
d
The maximum possible de Broglie wavelength is therefore 2d. The smallest possible mo-mentum is
pmin =h
λmax
=h
2d≈ 6.63× 10−34
2× 3.58× 10−10≈ 9.26× 10−25 kg ms−1 .
The smallest possible KE is
KEmin =p2
min
2m≈ (9.26× 10−25)2
2× 4× 1.66× 10−27≈ 6.46× 10−23 J .
The thermal KE of 3kBT/2 would equal KEmin when
T =2 KEmin
3kB≈ 2× 6.46× 10−23
3× 1.38× 10−23≈ 3.12 K .
3
The Bohr Atom
6. The Rydberg formula is1
λ= RH
(1
n2f
− 1
n2i
),
where RH = 1.097× 107 m−1. In this case, nf = 2 and λ = c/ν = (3.00× 108)/(7.316×1014) = 4.10× 10−7 m.
Rearranging the Rydberg formula gives:
n2i =
(1
n2f
− 1
RHλ
)−1
≈(
1
4− 1
1.097× 107 × 4.10× 10−7
)−1
≈ 36.1 .
Hence, the initial energy level was the n = 6 level.
7. The Bohr orbit of the electron must still contain a whole number of de Broglie wave-lengths. The angular momentum L = mvr must therefore be quantised just as in a hydro-gen atom:
L = nh̄ , n = 1, 2, 3, . . . .
However, since the Coulomb attraction between the orbiting electron and the nucleus islarger by a factor Z than in a hydrogen atom, the equation linking centripetal force andcentripetal acceleration becomes:
Ze2
4πε0r2=
mv2
r=
(mvr)2
mr3.
Replacing mvr by nh̄ and rearranging gives the following formula,
rn =4πε0(nh̄)2
Zme2,
for the radius of the nth Bohr orbit.
The total energy of this orbit is
En = KEn + PEn
=1
2mv2
n −Ze2
4πε0rn
=1
2
Ze2
4πε0rn− Ze2
4πε0rn
(since
mv2n
rn=
Ze2
4πε0r2n
)
= − Z2me4
2(4πε0h̄)2n2(substituting for rn)
≈ −Z2 × 13.6 eV
n2.
4
8. (i) The shortest wavelength (highest energy) photon that a hydrogen atom can emit with-out ending up in the ground-state (nf = 1) energy level is produced when the atomdecays from an initial state with very large ni to the nf = 2 final state. The wave-length of the photon emitted in this transition satisfies the equation
1
λ= RH
(1
4− 1
n2i
)≈ RH
4,
and henceλ ≈ 4
RH
≈ 3.65× 10−7 m .
(ii) For the H atom, the normal Rydberg formula applies:
1
λ= RH
(1
n2f
− 1
n2i
),
with λ = 121.5 nm. Hence1
n2f
− 1
n2i
=1
λRH
≈ 1
121.5× 10−9 × 1.097× 107≈ 0.75 .
The only solution of this equation with ni and nf integers is ni = 2 and nf = 1. Inother words, the transition is from the first excited state to the ground state.For the He+ ion, the energies of the states are Z2 = 22 = 4 times what they were inthe H atom (see Q8). Hence, the Rydberg formula becomes
1
λ= 4RH
(1
n2f
− 1
n2i
),
and the equation for nf and ni is1
n2f
− 1
n2i
≈ 0.75
4=
3
16.
This equation can be solved by doubling the values of ni and nf obtained for the Hatom. In other words, the He+ transition is from the ni = 4 level to the nf = 2 level.
9. According to the Bohr model, the binding energy and radius of an H atom in its groundstate are
Ebinding =me4
2(4πε0h̄)2≈ 13.6 eV ,
r =4πε0h̄
2
me2≈ 0.53 Å .
In the case of a positronium atom, the electron mass m has to be replaced by the reducedmass mreduced = (1/m+ 1/m)−1 = 0.5m. Hence
Ebinding =0.5me4
2(4πε0h̄)2≈ 6.8 eV ,
r =4πε0h̄
2
0.5me2≈ 1.06 Å .
5
10. (i) The potential in this example is peculiar, but a Bohr orbit of radius r still has length2πr, and the de Broglie wavelength of the particle must still “fit in” to this length:
2πr = nλ (n any integer > 0) .
Since p = mv = h/λ, this condition translates to
mvr = mvnλ
2π= p
nλ
2π=nh
2π= nh̄ .
In other words, the angular momentum must still be a multiple of h̄.The next step in deriving the Bohr model of the hydrogen atom is to write downNewton’s second law: force = mass× centripetal acceleration. In this case, the forceis
F = −dVdr
= −C ,
where the−ve sign shows that the force acts towards the origin (in the−r direction).Hence, Newton’s second law reads:
C =mv2
r=
(mvr)2
mr3.
Using the angular momentum quantisation conditions, mvr = nh̄, then gives
C =h̄2n2
mr3n
,
or
rn =
(h̄2n2
mC
)1/3
,
as required.
(ii) The energy En of the nth orbit is:
En = KE + PE =1
2mv2
n + Crn
=(mvnrn)2
2mr2n
+ Crn =h̄2n2
2mr2n
+ Crn
=h̄2n2
2m(h̄2n2
mC
)2/3+ C
(h̄2n2
mC
)1/3
=C
2
(h̄2n2
mC
)1/3
+ C
(h̄2n2
mC
)1/3
=3
2
(C2h̄2n2
m
)1/3
,
as required.
6
Quantum Physics Answer Sheet 4
Working with Wavefunctions
1. (i) The probability density f(x, t) is proportional to |ψ(x, t)|2. Hence, the probabilityf(x, t)dx that the particle is found between x and x+ dx at time t is proportional to
|ψ(x, t)|2 dx =
{cos2(πx/d)e−i(h̄π
2/2md2)tei(h̄π2/2md2)t dx |x| < d/2
0 otherwise
=
{cos2(πx/d) dx |x| < d/20 otherwise
which is independent of time. The particle is most likely to be found at x = 0, wheref(x) is largest.
(ii) In order to normalise ψ(x, t) we have to evaluate
N =∫ ∞−∞|ψ(x, t)|2 dx =
∫ d/2
−d/2cos2(πx/d) dx =
d
2.
(The integration is easy because cos2 θ repeats every π radians and averages to 1/2over any whole number of repeats.)Hence, the normalised ground-state wavefunction is:
ψ(x, t) =
√
2
dcos(πx/d)e−i(h̄π
2/2md2)t |x| < d/2 ,
0 otherwise .
(iii) 〈x〉 =∫ d/2
−d/2x
2
dcos2
(πx
d
)dx = 0 (integrand is odd).
〈x2〉 =∫ d/2
−d/2x2 2
dcos2
(πx
d
)dx
=d3
π3
2
d
∫ π/2
−π/2θ2 cos2 θ dθ (where θ = πx/d)
=(
1
12− 1
2π2
)d2 (using integral given in question).
Hence
∆x =√〈(x− 〈x〉)2〉 =
√〈x2〉 = d
√1
12− 1
2π2≈ 0.181 d .
1
2. (i) The electron probability density f(r, t) is the square modulus of the normalisedwavefunction and is proportional to the square modulus of the unnormalised wave-function. Hence
f(r, t) ∝(e−r/a0e−iE0t/h̄
) (e−r/a0e−iE0t/h̄
)∗= e−2r/a0 ,
which is independent of time. The probability density is largest at r = 0 (when theelectron is right on top of the nucleus).
(ii) The normalisation condition in three dimensions is∫all space
|ψ(r, t)|2dV = 1 .
To verify that the wavefunction given in the question is normalised, we have to showthat the normalisation integral,
N =∫
all space
1
πa30
e−2r/a0 dV ,
is equal to 1. Since the integrand only depends on the length r of the vector r,the integral can be evaluated by summing contributions from spherical shells. Thecontribution from the shell with inner and outer radii r and r + dr is the volume ofthat shell, 4πr2dr, times the value of the integrand for that shell. Hence
N =∫ ∞r=0
1
πa30
e−2r/a04πr2 dr
=4π
πa30
(a0
2
)3 ∫ ∞ξ=0
e−ξξ2 dξ (where ξ = 2r/a0)
=1
22! = 1 (using integral given in question)
as required.
(iii) Let p(r)dr be the probability that the electron is found in the spherical shell withinner and outer radii r and r + dr. Then
p(r) dr = |ψ(r, t)|2 4πr2dr =1
πa30
e−2r/a04πr2dr =4
a30
r2e−2r/a0 dr .
The most probable distance of the electron from the nucleus is the value of r forwhich p(r) is maximised. To find the maximum of p(r), we set dp/dr equal to zero:
dp
dr=
4
a30
(2r − 2
a0
r2)e−2r/a0 = 0 .
The only solution to this equation is r = r2/a0 and hence r = a0. The most probabledistance of the electron from the nucleus is therefore a0. (Since p(r) is a non-negativefunction which is zero at r = 0, tends to zero as r → ∞, and has only a single
2
stationary point in between, there is no need to check the type of that stationarypoint. It can only be a maximum. Why?)In part (i) we found that the probability density is maximised at the origin. Here wefound that the most likely distance of the electron from the origin is a0. These resultsare not inconsistent. The probability density f(r) is defined as follows:
prob. electron is found in volume element dV at r = f(r) dV .
In other words, f(r) is the probability per unit volume. The probability of finding theelectron in the spherical shell with inner and outer radii r and r + dr is the productof the probability density, f(r), and the volume 4πr2dr. Although the probabilitydensity peaks at the origin and decreases as r increases, the volume of a shell offixed thickness dr increases with r. In fact, for r < a0, the shell volume increasesfaster than the probability density decreases. The probability of finding the electronwithin a spherical shell of thickness dr therefore increases as r increases from 0 toa0.
(iv) The mean distance from the nucleus is
〈r〉 =∫
all spacer |ψ(r, t)|2 dV =
∫ ∞r=0
re−2r/a0
πa30
4πr2dr
=4
a30
(a0
2
)4 ∫ ∞0
ξ3e−ξ dξ (where ξ = 2r/a0)
=a0
43! (using integral given in question)
=3a0
2.
The mean square distance from the nucleus is
〈r2〉 =∫
all spacer2 |ψ(r, t)|2 dV =
∫ ∞r=0
r2 e−2r/a0
πa30
4πr2dr
=4
a30
(a0
2
)5 ∫ ∞0
ξ4e−ξ dξ (where ξ = 2r/a0)
=a2
0
84! (using integral given in question)
= 3a20 .
Hence, the root mean square distance (the square root of the mean square distance)of the electron from the nucleus is
√3 a0.
(v) The probability of finding the electron within a sphere of radius 10−15 m is givenexactly by the following integral:
∫ 10−15 m
r=0|ψ(r, t)|24πr2dr .
3
However, since 10−15 m is such a small radius, the probability density |ψ(r, t)|2 isalmost constant throughout the region of integration. A very good approximationto the probability may therefore be obtained by multiplying the probability densityat the origin, |ψ(0, t)|2 = 1/πa3
0, by the volume of the tiny sphere, 43π(10−15)3 m3.
Using this method, the estimate of the probability of finding the electron on top ofthe nucleus is
43π (10−15)
3
πa30
=4
3
10−45
(0.53)3 × 10−30≈ 9× 10−15 .
Momentum Measurements
3. Sinceeiθ = cos θ + i sin θ and e−iθ = cos θ − i sin θ ,
it follows thatsin(kx) =
1
2i
(eikx − e−ikx
).
Hence, in the region x < 0, we have:
ψ(x, t) =−i2eikx−iωt +
i
2e−ikx−iωt .
The first term is a right-going travelling wave and the second term is a left-going travellingwave.
In lectures, we considered a general superposition of travelling waves,
ψ(x, t) =∑n
A(kn)ei(knx−ωt) ,
and argued that |A(kn)|2 was proportional to the probability of obtaining the result h̄knin a measurement of momentum. The wavefunction in this question is a simple two-termexample of the general form considered in lectures. The wavevectors of the two terms arek and −k and their complex amplitudes are A(k) = −i/2 and A(−k) = i/2. Hence, thetwo possible results of a momentum measurement are h̄k and −h̄k. Since the two termshave the same intensity, |A(k)|2 = |A(−k)|2, the probabilities of measuring h̄k and −h̄kare both 1/2.
Physically, the wavefunction represents a particle in a very (infinitely) spread out wavepacketof momentum h̄k, which is being reflecting elastically from a potential barrier at x = 0.Since the wavepacket is so spread out, it seems reasonable that the probability of mea-suring the incident momentum h̄k is equal to the probability of measuring the reflectedmomentum −h̄k.
4
The Uncertainty Principle
4. The size of a nucleus is about 10−15 m and so the position uncertainty ∆x of an electroncontained in a nucleus is also about 10−15 m. According to the uncertainty principle, themomentum uncertainty of such an electron satisfies
∆p ≥ h̄
2∆x≈ 5.25× 10−20 kg m s−1 .
The confined electron is not going anywhere on average, so its average momentum 〈p〉must be zero. Hence
(∆p)2 = 〈(p− 〈p〉)2〉 = 〈p2〉 .
Since c∆p = 1.575 × 10−11 J ≈ 98 MeV � mc2 = 511 keV, the kinetic energy ofconfinement is sufficient to make the electron highly relativistic. The kinetic energy of theelectron is therefore given by the relativisitic formula:
KE =√p2c2 +m2c4 −mc2 ≈ |p|c .
To obtain an order of magnitude estimate of this quantity, we approximate |p| by ∆p =√〈p2〉 to obtain
KE ∼ 98 MeV.
This argument shows that the KE of an electron confined within a nucleus is of order100 MeV. Since the kinetic energy of the electron emitted is between 1 and 10 MeV, itseems more likely that the electron was created during the radioactive decay process thanthat it was confined within the nucleus all along.
5. (i) Starting from the definition of the rms momentum,
(∆p)2 = 〈(p− 〈p〉)2〉 ,
and using the result 〈p〉 = 0 given in the question, we obtain (∆p)2 = 〈p2〉. Similarly,we can show that (∆x)2 = 〈x2〉. The expression for the total energy,
〈E〉 =〈p2〉2m
+1
2s〈x2〉 ,
may therefore be rewritten as
〈E〉 =(∆p)2
2m+
1
2s(∆x)2 =
(∆p)2
2m+
1
2mω2(∆x)2 ,
where the last step followed because ω =√s/m.
(ii) The uncertainty principle states that ∆x∆p ≥ h̄/2. If we use this to eliminate ∆pfrom the expression for 〈E〉 we obtain
〈E〉 ≥ 1
2m
h̄2
4(∆x)2+
1
2mω2(∆x)2 =
h̄2
8m(∆x)2+
1
2mω2(∆x)2 .
5
The value of ∆x that makes the right-hand side as small as possible (and henceimposes the weakest possible condition on 〈E〉) may be found by differentiatingwith respect to ∆x and setting the result equal to zero:
d
d(∆x)
[h̄2
8m(∆x)2+
1
2mω2(∆x)2
]=
−h̄2
4m(∆x)3+mω2∆x = 0 .
Solving this equation gives ∆x =√h̄/2mω. Substituting back into the inequality
for 〈E〉 gives
〈E〉 ≥ h̄2
8m(h̄/2mω)+
1
2mω2
(h̄
2mω
)=
1
4h̄ω +
1
4h̄ω ,
and hence〈E〉 ≥ 1
2h̄ω ,
as required.
6. The energy-time uncertainty principle,
∆E∆t ≥ h̄
2,
relates the lifetime ∆t of a state to its energy uncertainty ∆E. For ∆t = 2.6× 10−10 s weobtain
∆E ≥ 1.05× 10−34
2× 2.6× 10−10≈ 2.02× 10−25 J ≈ 1.26× 10−6 eV.
Hence, the minimum energy uncertainty of the emitted photons is 1.26× 10−6 eV.
7. In order for a virtual particle of energy mc2 to pop out of the vacuum, the energy uncer-tainty ∆E must be (at least) mc2. According to the energy-time uncertainty principle inthe form ∆E∆t ∼ h̄, the lifetime of such a particle is ∆t ∼ h̄/∆E ∼ h̄/mc2. The speedof the particle must be less than the speed of light, and hence the distance it travels duringits lifetime must be less than c∆t ∼ h̄/mc.Equating this distance to the range of the nuclear force gives
h̄
mc≈ 1.4× 10−15 m ,
and hence
mc2 ≈ h̄c
1.4× 10−15=
1.05× 10−34 × 3.00× 108
1.4× 10−15= 2.25× 10−11 J .
In eV, this becomes
mc2 ≈ 2.25× 10−11
1.60× 10−19≈ 141× 106 eV = 141 MeV ,
or, equivalently, m ≈ 141 MeV/c2.
6
The Schrödinger Equation
8. The time-independent Schrödinger equation is
− h̄2
2m
d2ψ(x)
dx2+ V (x)ψ(x) = Eψ(x) .
If the wavefunction
ψn(x) =
{ √2d
sin(nπxd
)0 < x < d
0 otherwise
is a normalised energy eigenfunction, the following conditions hold:
(a) ψn(x) satisfies the boundary conditions: ψn(0) = ψn(d) = 0.
(b) ψn(x) satisfies the time-independent Schrödinger equation for 0 < x < d.
(c) ψn(x) is normalised.
Consider these conditions one by one:
(a) By inspection, ψn(x) satisfies the boundary conditions for n = 1, 2, . . ..
(b) Substitute ψn(x) into the left-hand side of the Schrödinger equation for 0 < x < d:
− h̄2
2m
d2ψn(x)
dx2+ V (x)ψn(x) = − h̄2
2m
d2ψn(x)
dx2[V (x) = 0 for 0 < x < d]
= − h̄2
2m
√2
d
d2
dx2sin
(nπx
d
)
=h̄2
2m
(nπ
d
)2√
2
dsin
(nπx
d
)
= Enψn(x) , where En =h̄2n2π2
2md2.
Hence, ψn(x) satisfies the time-independent Schrödinger equation in the region 0 <x < d. The corresponding energy eigenvalue is En = h̄2n2π2/2md2.
(c) If ψn(x) is normalised, the integral
N =∫ d
0|ψn(x)|2 dx
must be equal to 1. Check this by evaluating the integral:
N =2
d
∫ d
0sin2
(nπx
d
)dx =
2
d× d
2= 1 .
Hence, ψn(x) is normalised.
7
9. Using the result of Q8, the energy eigenvalues are:
En =h̄2n2π2
2md2n = 1, 2, . . . .
The energy emitted as the nucleon falls from the n = 2 level to the n = 1 level is
E2 − E1 =3h̄2π2
2md2≈ 3× (1.05× 10−34)2 × π2
2× 1.67× 10−27 × (10−15)2
≈ 9.8× 10−11 J ≈ 610 MeV .
This is a sensible number. The energy released in fission is about 200 MeV per nucleus.
10. The classical angular frequency of vibration is ωvib =√s/mreduced, where
mreduced =(
1
mC
+1
mO
)−1
=(
1
12+
1
16
)−1
≈ 6.86 atomic mass units.
Hence
ωvib =
√1857
6.86× 1.66× 10−27≈ 4.04× 1014 Radians s−1 .
(i) The energy of the photon emitted when a CO molecule makes a transition betweenneighbouring vibrational states is the difference between the energies of those states.The energy levels of a QM simple harmonic oscillator are (n + 1
2)h̄ωvib, where n =
0, 1, 2, . . .. Hence, the energy E of the emitted photon is h̄ωvib. Since the energyE and angular frequency ω of the photon are related via E = h̄ω, it follows thatω = ωvib. The frequency of the emitted photon is therefore the same as the classicalvibrational frequency of the molecule, which makes sense. The photon wavelengthis
λ =c
ν=
c
νvib
=2πc
ωvib
≈ 4.67× 10−6 m .
(ii) The vibrational zero-point energy of a CO molecule is
1
2h̄ωvib ≈ 2.12× 10−20 J ≈ 0.13 eV .
11. Substitute the trial solution ψ(x) = e−αx2 into the left-hand side of the Schrödinger equa-
tion:
− h̄2
2m
d2ψ(x)
dx2+
1
2sx2ψ(x) = − h̄2
2m
d2
dx2e−αx
2
+1
2sx2e−αx
2
= − h̄2
2m
d
dx
(−2αxe−αx
2)
+1
2sx2e−αx
2
=h̄2α
m
d
dx
(xe−αx
2)
+1
2sx2e−αx
2
8
=h̄2α
m
(e−αx
2 − 2αx2e−αx2)
+1
2sx2e−αx
2
=h̄2α
me−αx
2
+
(1
2s− 2h̄2α2
m
)x2e−αx
2
.
If ψ(x) is to be a solution of the Schrödinger equation, this must equal a constant, E,times e−αx2 . The value of α must therefore be chosen to ensure that the coefficient of thex2e−αx
2 term is zero. Hence
α2 =ms
4h̄2 ⇒ α =
√sm
2h̄.
If this condition is met, ψ(x) is an energy eigenfunction with energy eigenvalue
E =h̄2α
m=
h̄2
m
√sm
2h̄=
1
2h̄
√s
m=
1
2h̄ω ,
where ω =√s/m is the classical angular frequency of the oscillator.
Q5 showed that the energy of a quantum mechanical simple harmonic oscillator must be≥ 1
2h̄ω, and so the eigenfunction found here must be the (unnormalised) ground state.
12. Inside the barrier, the Schrödinger equation is
− h̄2
2m
d2ψ(x)
dx2+ V ψ(x) = Eψ(x) ,
with V > E. A simple rearrangement gives
d2ψ(x)
dx2=
2m(V − E)
h̄2 ψ(x) = γ2ψ(x) ,
where
γ =
√2m(V − E)
h̄2
is real because V − E is greater than zero. By inspection, the two independent solutionsof this second-order differential equation are eγx and e−γx, and so the general solution is
ψ(x) = Ae−γx +Beγx ,
where A and B are arbitrary constants to be determined from the boundary conditions.
Assume now that the barrier stretches from x = 0 to x = a and that the particles areincident from the left. We discard the eγx solution on physical grounds: it would makelittle sense if the wavefunction, and hence the probability density, increased with distanceinto the barrier. (In fact, as you will find out next year, the exact wavefunction inside abarrier of finite width does include a small amount of the eγx solution. However, as longas the barrier is wide and the tunnelling probability low, the value of B is so small thatthis contribution can be ignored.) Inside the barrier, then, the wavefunction ψ(x) is equalto Ae−γx, as shown in the figure.
9
x=ax=0
Metal
ikx
Gap Metal
eikx t eA
xe
−γ
It follows that|ψ(x = a)|2
|ψ(x = 0)|2≈
(e−γa
)2= e−2γa .
Since |ψ(x)|2 is proportional to the probability density at position x, this equation showsthat the probability density at the right-hand edge of the barrier is e−2γa (which is normallya very small number) times the probability density at the left-hand edge. The tunnellingprobability is therefore equal to e−2γa.
The work function W is the minimum energy required to remove an electron from a pieceof metal. If additional energy W is supplied to a metallic electron with energy E, thatelectron is only just able to escape from the surface. This implies that the potential energyV of the electron outside the metal must be E +W . If no extra energy is supplied (so thatthe electron from the metal still has energy E), the Schrödinger equation in the gap regiontakes the form,
− h̄2
2m
d2ψ(x)
dx2+ (E +W )ψ(x) = Eψ(x) ,
and hence γ =√
2m(E +W − E)/h̄2 =√
2mW/h̄2.
In the case when a = 10−6 m and W = 5 eV, we obtain
γ =
√2× 9.11× 10−31 × 5× 1.60× 10−19
1.05× 10−34≈ 1.15× 1010 m−1 .
The tunnelling probability is therefore
e−2γa ≈ e−23,000 = 10−23,000×log10 e ≈ 10−10,000 .
This is very small! Electron tunnelling is important in scanning tunnelling microscopes,where the gap is comparable to the size of an atom (10−10 m), but not for the much largergap considered here.
Extra Questions for Enthusiasts
13. (i) In order to tell which slit the electron went through, the optical instrument that detectsthe scattered photons must be able to resolve features of size d. Since the resolutioncannot be much better than the wavelength of the light used, λphoton must be less thanor approximately equal to d.
10
(ii) The figure shows a photon moving in the x direction scattering from an electronmoving in the z direction.
x
hν
hν
e
e-
- z
The xmomentum transferred to the electron may take any value from 0 (if the photonis not deflected at all) to 2h/λphoton (if the photon scatters back towards its source),but the typical value is around h/λphoton. The initial z momentum of the electron ispz = h/λelectron. The value of pz changes slightly when the photon scatters, but ifthe change in electron direction is small the change in pz will also be small. Hence,even after the photon has been scattered, pz ≈ h/λelectron.After the scattering event, the electron’s momentum vector looks something like this:
∆θ
p
λelectron~ h/z
p
p ~ h/λphotonx
If the scattering angle ∆θ is small, so that tan(∆θ) ≈ ∆θ, it follows that
∆θ ≈ tan(∆θ) =pxpz∼ λelectron
λphoton
.
(iii) The angular spacing between adjacent diffraction maxima or minima is λelectron/d.Since λphoton ≤ d (see part (i)), the angular spacing must be ≤ λelectron/λphoton. Butthis is exactly equal to ∆θ, the uncertainty in the direction of the electron caused byscattering the photon (part (ii)). Hence, the scattering of the photon is sufficient tosmear out the diffraction pattern completely.
14. The integral we have to evaluate is∫ ∞−∞
ψ∗n(x)ψm(x) dx =2
d
∫ d
0sin
(nπx
d
)sin
(mπx
d
)dx
=2
d
d
π
∫ π
0sin(nθ) sin(mθ) dθ (where θ = πx/d)
=1
π
∫ π
0[cos((n−m)θ)− cos((n+m)θ)] dθ ,
11
where the last step used the expression for sin(nθ) sin(mθ) given in the question. For anynon-zero integer j, the integral
∫ π
0cos(jθ) dθ =
[1
jsin(jθ)
]π0
= 0 .
Hence, if n and m are unequal positive integers,∫ ∞−∞
ψ∗n(x)ψm(x) dx = 0 .
If n = m, so that cos((n−m)θ) = 1, we obtain∫ ∞−∞
ψ∗n(x)ψm(x) dx =∫ ∞−∞
ψ∗n(x)ψn(x) dx
=1
π
∫ π
0[1− cos(2nθ)]dθ =
1
π
∫ π
0dθ = 1 ,
demonstrating that ψn(x) is normalised.
15. If φ(x) =∑∞n=1 cnψn(x) is normalised, then∫ ∞
−∞φ∗(x)φ(x) dx = 1 .
Substituting the expansion of φ(x) into the normalisation integral gives∫ ∞−∞
φ∗(x)φ(x) dx =∫ ∞−∞
∞∑n=1
c∗nψ∗n(x)
∞∑m=1
cmψm(x) dx
=∞∑n=1
∞∑m=1
c∗ncm
∫ ∞−∞
ψ∗n(x)ψm(x) dx .
The integral in this expression is equal to 1 if n = m and to 0 otherwise. Hence, all termswith n 6= m vanish and only the n = m terms are left:∫ ∞
−∞φ∗(x)φ(x) dx =
∞∑n=1
c∗ncn
∫ ∞−∞
ψ∗n(x)ψn(x) dx =∞∑n=1
c∗ncn .
This shows that φ(x) is normalised if and only if
∞∑n=1
c∗ncn = 1 .
Since c∗ncn is greater than or equal to zero, and since the sum of c∗ncn over all n gives 1,it is reasonable (and right) to guess that c∗ncn is the probability that a measurement of theenergy gives the result En (where En is the energy eigenvalue corresponding to the energyeigenfunction ψn(x)).
12