1.2 (a)

Strategy: We are given the wavelength of an electromagnetic wave and asked to calculate

the frequency. Rearranging Equation (1.2) of the text and replacing u with c (the speed of

light) gives:

c

Solution: Because the speed of light is given in meters per second, it is convenient to first

convert wavelength to units of meters. Recall that 1 nm 1 109 m. We write:

456 nm 1 109 m

1 nm 456 109 m 4.56 107 m

Substituting in the wavelength and the speed of light ( 3.00 108 ms1 ), the frequency is:

8 1

7or

4.56 10 m

c

14 1 146.58 10 s 6.58 10 Hz

Check: The answer shows that 6.58 1014 waves pass a fixed point every second. This

very high frequency is in accordance with the very high speed of light.

(b)

Strategy: We are given the frequency of an electromagnetic wave and asked to calculate

the wavelength. Rearranging Equation (1.2) of the text and replacing u with c (the speed of

light) gives:

c

Solution: Substituting in the frequency and the speed of light ( 3.00 108 ms1 ) into the

above equation, the wavelength is:

8 1

9 10.122 m

2.45 10 s

c

The problem asks for the wavelength in units of nanometers. Recall that 1 nm 1 109 m.

0.122 m 1 nm

1 109 m 1.22 108 nm

1.6 (a)

Strategy: We are given the frequency of an electromagnetic wave and asked to calculate

the wavelength. Rearranging Equation (1.2) of the text and replacing u with c (the speed of

light) gives:

c

Solution: Substituting in the frequency and the speed of light (3.00 108 m s-1) into the

above equation, the wavelength is:

8 1

7

14 14.0 10 m

7.5 10 s

24.0 10 nm

Check: The wavelength of 400 nm calculated is in the blue region of the visible spectrum

as expected.

(b)

Strategy: We are given the frequency of an electromagnetic wave and asked to calculate its

energy. Equation (1.3) of the text relates the energy and frequency of an electromagnetic

wave.

E h

Solution: Substituting in the frequency and Planck's constant (6.63 1034 J s) into the

above equation, the energy of a single photon associated with this frequency is:

34 14 1(6.63 10 J s) 7.5 10 sE h 195.0 10 J

Check: We expect the energy of a single photon to be a very small energy as calculated

above, 5.0 1019 J.

1.16 Because the same number of photons are being delivered by both lasers and because both

lasers produce photons with enough energy to eject electrons from the metal, each laser will

eject the same number of electrons. The electrons ejected by the blue laser will have higher

kinetic energy because the photons from the blue laser have higher energy.

1.18 (a) The maximum wavelength is obtained by solving Equation 1.4 when the kinetic energy

of the ejected electrons is equal to zero. In this case, Equation 1.4, together with the relation

c / , gives

34 8 1

9

6.626 10 J s 3.00 10 m s

520 10 m

hcE

193.82 10 J

(b) h (kinetic Energy)

(kinetic Energy) hc

34 8 119

9

6.626 10 3.00 10(kinetic Energy) 3.8227 10

450 10

J s m sJ

m

205.9 10 J

Please remember to carry extra significant figures during the calculation and then round

at the very end to the appropriate number of significant figures.

1.24 We use more accurate values of h and c for this problem.

E hc

=

( 6.6256 1034 J s )( 2.998 108 m s1 )

656.3 109 m 3.027 1019 J

1.28 Equation 1.14: rn 0 h

2n

2

Z me e2

for n = 2:

r2 (8.8542 1012 C2 J1 m1 )(6.626 1034 J s)2 2

2

1 (9.109 1031 kg)(1.602 1019 C)2 2.117x1010 m = 211.7 pm

for n = 3:

r3 (8.8542 1012 C2 J1 m1 )(6.626 1034 J s)2 3

2

1 (9.109 1031 kg)(1.602 1019 C)2 4.764x1010 m = 476.4 pm

1.30 Z 2RH1

n12

1

n22

2 15 ? 1 1

1 3.289832496 10 s ?1

15 ? 153.289832496x10 s 3.289832496x10 Hz

1.36 From the Heisenberg uncertainty principle (Equation 1.22) we have

2

hx p

For the minimum uncertainty we can change the greater than sign to an equal sign.

34

? 4 ?

? 1

1.055 10 J s1 10 kg m s

2 2 (4 10 m)

hp

x

Because p = m u, the uncertainty in the velocity is given by

u p

m

1 1024 kg m s1

9.1095 1031 kg 1 106 m s1

The minimum uncertainty in the velocity is 1 106 m s1.

1.37 Below are the probability distributions for the first three energy levels for a one-dimensional

particle in a box, as shown in Figure 1.24. The average value for x for all of the states for a

one-dimensional particle in a box is L/2 (the middle of the box). This is true even for those

states (like the second state) that have no probability of being exactly in the middle of the box.

1.38 The difference between the energy levels n and (n + 1) for a one-dimensional particle in a box

is given by Equation 1.31:

Enn1 h

2

8mL2

(2n 1) .

All we have to do is to plug the numbers into the equation remembering that 1 =10-10 m.

Enn1 (6.626 1034 J s)2[2(2) 1]

8(9.109 1031 kg)[5.00 1010 m]2 1.20 1018 J

Energy is equal to Plancks constant times frequency. We can rearrange this equation to

solve for frequency.

E

h

1.20 1018 J

6.626 1034 J s 1.811015 s1 1.81 1015 Hz

1.39 From Equation 1.15 we can derive an equation for the change in energy going from the n=1 to

n=2 states for a hydrogen atom:

E 3Z

2e

4me

32h20

2.

From Equation 1.31 we can derive an equation for the change in energy going from the n=1 to

n=2 states for a particle in a box:

E 3h

2

8mL2

.

We can combine the two equations and solve for L.

3Z

2e

4me

32h20

2

3h2

8mL2

L 4h

402

Z2e

4me

2

2h20

Ze2me

2(6.626 1034 J s)2 (8.8542 1012 C2 J1 m1 )

1 (1.602 1019 C)2 (9.109 1031 kg) 3.326 1010 m

The radius of the ground state of the hydrogen atom (the Bohr radius) is 5.29 10-11 m (see

Example 1.4). This leads to a diameter of 1.06 10-10 m, which is only a third the size of L that we

calculated. So, it seems like a crude estimate.

1.40

The average value for x will be greater than L/2. The larger the electric field, the larger the

average value of x.

1.42 (a) 2p: n 2, l 1, ml 1, 0, or 1

(b) 6s: n 6, l 0, ml 0 (only allowed value)

(c) 5d: n 5, l 2, ml 2, 1, 0, 1, or 2

An orbital in a subshell can have any of the allowed values of the magnetic quantum number

for that subshell. All the orbitals in a subshell have exactly the same energy.

1.46 A 2s orbital is larger than a 1s orbital. Both have the same spherical shape. The 1s orbital

is lower in energy than the 2s. The 1s orbital does not have a node while the 2s orbital does.

1.50 (a) False. n 2 is the first excited state.

(b) False. In the n 4 state, the electron is (on average) further from the nucleus and

hence easier to remove.

(c) True.

(d) False. The n 4 to n 1 transition is a higher energy transition, which corresponds to

a shorter wavelength.

(e) True.

1.70 The kinetic energy acquired by the electrons is equal to the voltage times the charge on the

electron. After calculating the kinetic energy, we can calculate the velocity of the electrons

(KE 1/2mu2). Finally, we can calculate the wavelength associated with the electrons using

the de Broglie equation.

KE (5.00 103 V) 1.602 1019 J

1 V 8.01 1016 J

We can now calculate the velocity of the electrons.

KE

1

2mu

2

8.01 1016 J

1

2(9.1094 1031 kg)u2

u 4.19 107 m s-1

Finally, we can calculate the wavelength associated with the electrons using the de Broglie

equation.

h

mu

(6.63 1034 J s)

(9.1094 1031 kg)(4.19 107 m s1) 1.74 1011 m 17.4 pm

1.72 (a) The average kinetic energy of 1 mole of an ideal gas is 3/2RT. Converting to the

average kinetic energy per atom, we have:

3

2(8.314 J mol

1K

1)(298 K)

1 mol

6.022 1023 atoms 6.171 1021 J atom1

(b) To calculate the energy difference between the n 1 and n 2 levels in the hydrogen

atom, we modify Equation 1.5 of the text.

E RH

1

ni

2

1

nf

2

(2.180 10

18J)

1

12

1

2

E 1.635 1018 J

(c) For a collision to excite an electron in a hydrogen atom from the n 1 to n 2 level,

KE E

3

2RT

NA

1.635 1018 J

T 2

3

(1.635 1018 J)(6.022 1023 mol1)

(8.314 J mol1

K1

) 7.90 104 K

This is an extremely high temperature. Other means of exciting H atoms must be used to

be practical.

1.75 Below are the solutions for the particle in a one dimensional box (Equation 1.35).

n(x)

2

L

1/2

sinn x

L

n 1, 2,3,....

The equation below yields the average value of x, for a particle in a one dimensional box.

f x (x)2

dx0

L

x2

L

sin

2 n x

L

dx

0

L

2

L

xsin

2 n x

L

dx

0

L

We can use the following relation obtained from a table of integrals:

xsin2

ax dx x

2

4

x sin(2ax)

4a

cos 2ax 8a

2

f x (x)2

dx0

L

2

L

xsin

2 n x

L

dx

0

L

2

L

x

2

4

x sin2n x

L

4n

L

cos2n x

L

8n

L

2

We now evaluate this for x equal to L and 0.

f 2

L

L

2

4

1

8n

L

2

2

L

1

8n

L

2

L

2

The result above shows us that for all states for a particle in a box, the average value of x is

L/2. This makes sense, because the squared wavefunctions are all symmetric about x = L/2.

The equation below yields the average value of x2, for a particle in a one dimensional box.

f x2 (x)2

dx0

L

x2 2

L

sin

2 n x

L

dx

0

L

2

L

x

2sin

2 n x

L

dx

0

L

We can use the following relation obtained from a table of integrals:

x2sin

2ax dx

x3

6

x cos(2ax)

4a2

2a

2x

2 1 sin 2ax 8a

3

f x2 (x)2

dx0

L

2

L

x

2sin

2 n x

L

dx

0

L

2

L

x

3

6

x cos2n x

L

4n

L

2

2n

L

2

x2 1

sin2n x

L

8n

L

3

We now evaluate this for x equal to L and 0.

f 2

L

L

3

6

L

4n

L

2

L21

3

1

2 n 2

for n = 1 the average value of x2 is 0.283 L2

for n = 2 the average value of x2 is 0.308 L2

for n = 3 the average value of x2 is 0.328 L2

1.76 The average value of r using Equation 1.41:

r rP r dr0

r3

R(r)2

dr0

r r 3 R10

(r )2

dr0

r3

2Z

a0

3/2

e

Zr

ao

2

dr0

4Z

a0

3

r3e

2Zr

ao dr

0

From an integral table:

xne ax

dx0

n!

an1

To use the above integral to solve our problem, n = 3, x = r, and

a 2Z

ao

,

r 4Z

a0

3

r3e

2Zr

ao dr

0

4Z

a0

3

6ao

4

16Z4

3ao

2Z

3

2a

o.

Hence, the average value for r for a 1s orbital is

3

2a

o. Remember that Z is the nuclear charge

and is equal to 1 for a hydrogen atom.

To calculate the most probable value of r, we set the derivative of the radial probability

function to zero and solve for r.

P r r 2 R r 2

r 2 2Z

a0

3/2

e

Zr

ao

2

4Z

a0

2

r2e

2Zr

ao

d

drP r 4

Z

a0

2

r2e

2Zr

ao 4

Z

a0

2

2re

2Zr

ao

2Z

ao

r2e

2Zr

ao

8

Z

a0

2

r Z

ao

r2

e

2Zr

ao

When r

ao

Z a

o the derivative is zero, which corresponds to the most likely value of r.

This is consistent with ao being the Bohr radius.

For the 2s orbital

The average value of r:

r rP r dr0

r3

R(r)2

dr0

r r 3 R10

(r )2

dr0

r3 2

4

Z

a0

3/2

2 Zr

ao

e

Zr

2ao

2

dr0

r 1

8

Z

a0

3

4 r3e

Zr

ao dr

4Z

ao

r4e

Zr

ao dr

Z2

ao

2r

5e

Zr

ao dr

Using the same integral solution as above, we can solve for the three integrals in the square

brackets

4 r3e

Zr

ao dr

0

46a

o

4

Z4

24ao

4

Z4

4Z

ao

r4e

Zr

ao dr

4Z

ao

24ao

5

Z5

96ao

4

Z4

Z2

ao

2r

5e

Zr

ao dr

Z2

ao

2

120ao

6

Z6

120ao

4

Z4

When we plug in the results for the integrals into the above equation we get that the average

distance from the nucleus for a 2s orbital is six times the Bohr radius. Notice that this is four

times the average distance from the nucleus for a 1s orbital.

r 1

8

Z

a0

3

24ao

4

Z4

96a

o

4

Z4

120a

o

4

Z4

1

8

Z

a0

3

48ao

4

Z4

6

ao

Z 6a

o

To calculate the most probable value of r, we can set the derivative of the radial probability

function to zero and solve for r.

P r r 2 R r 2

r 22

4

Z

a0

3/2

2 Zr

ao

e

Zr

2ao

2

1

8

Z

a0

3

r2

4 4Zr

ao

Z

2r

2

ao

2

e

Zr

ao

P r 1

8

Z

a0

3

4r2e

Zr

ao

4Zr3

ao

e

Zr

ao

Z2r

4

ao

2e

Zr

ao

d

drP r

1

8

Z

a0

2

d

dr4r

2e

Zr

ao

4Zr3

ao

e

Zr

ao

Z2r

4

ao

2e

Zr

ao

d

drP r

1

8

Z

a0

2

4 2r Zr

2

ao

4Z

ao

3r2

Zr3

ao

Z2

ao

24r

3 Zr

4

ao

e

Zr

ao

d

drP r

1

8

Z

a0

2

8r 16Zr

2

ao

8Z

2r

3

ao

2

Z3r

4

ao

3

e

Zr

ao

d

drP r

1

8

Z

a0

2

rZ3

ao

3

8ao

3

Z3

16ao

2r

Z2

a

or

2

Z r 3

e

Zr

ao

The above polynomial can be solved computationally to determine three roots. The largest

root, which corresponds to the most probable value is r

3 5 aoZ

3 5 ao .

1.77

2

L

1/2

sinn x

L

2

L

1/2

sinm x

L

0

L

dx 2

Lsin

n x

L

sin

m x

L

0

L

dx

Using a table of integrals, it can be shown that the last integral listed above is zero if n does

not equal m, and is L/2 if n equals m. Hence, the wavefunctions for a one-dimensional

particle in a box are orthogonal.

1.78 The five lowest energy levels for a two-dimensional particle in a box correspond to the

following sets of quantum numbers: (nx=1, ny=1), (nx=1, ny=2), (nx=2, ny=1), (nx=2, ny=2),

(nx=3, ny=1) , (nx=1, ny=3) , (nx=3, ny=2) , (nx=2, ny=3). These wavefunctions are plotted

below with the lowest energy wavefunction at the bottom. Notice that the states (nx=1,

ny=2) and (nx=2, ny=1) have the same energy as do the states (nx=1, ny=3) and (nx=3, ny=1)

and the states (nx=3, ny=2) and (nx=2, ny=3).