Fundamental of Physics - Instituto de Física da UFRGS

[Pages:26]Chapter 39

1. Since En L? 2 in Eq. 39-4, we see that if L is doubled, then E1 becomes (2.6 eV)(2)? 2 = 0.65 eV.

2. We first note that since h = 6.626 10?34 J?s and c = 2.998 108 m/s,

c hc h 6.626 1034 J s 2.998 108 m / s

c hc h hc

1240eV nm.

1.602 1019 J / eV 109 m / nm

Using the mc2 value for an electron from Table 37-3 (511 103 eV), Eq. 39-4 can be

rewritten as

cbhg En

n2h2 8mL2

n2 hc 2 8 mc2 L2

.

The energy to be absorbed is therefore

42 12 h2 15hc2

151240eV nm2

E E4 E1

8me L2

8 mec2

L2 8 511103eV

0.250nm2 90.3eV.

3. We can use the mc2 value for an electron from Table 37-3 (511 103 eV) and hc = 1240 eV ? nm by writing Eq. 39-4 as

b g n2h2 n2 hc 2

c h En 8mL2

8 mc2

. L2

For n = 3, we set this expression equal to 4.7 eV and solve for L:

nbhcg

b g 3 1240eV nm

c h c hb g L

0.85nm.

8 mc2 En 8 511103 eV 4.7 eV

4. With m = mp = 1.67 10? 27 kg, we obtain

E1

h2 8mL2

n2

6.631034 J.s 2

8(1.67

1027

kg)

100 1012

m

2

12

3.291021 J 0.0206 eV.

1523

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CHAPTER 39

Alternatively, we can use the mc2 value for a proton from Table 37-3 (938 106 eV) and hc = 1240 eV ? nm by writing Eq. 39-4 as

b g n2h2

n2 hc 2

d i En

8mL2

8 mpc2

. L2

This alternative approach is perhaps easier to plug into, but it is recommended that both approaches be tried to find which is most convenient.

5. To estimate the energy, we use Eq. 39-4, with n = 1, L equal to the atomic diameter, and m equal to the mass of an electron:

E

n2

h2 8mL2

8

12 6.631034 J s 2

9.111031 kg 1.41014 m

2

3.07 1010 J=1920MeV 1.9 GeV.

6. (a) The ground-state energy is

E1

h2 8me L2

n2

6.631034 J s 2

8(9.111031

kg)

200 1012 m

2

12

1.511018 J

9.42eV.

(b) With mp = 1.67 10? 27 kg, we obtain

E1

h2 8mp L2

n2

6.631034 J s 2

8(1.67

1027

kg)

200 1012 m

2

12

8.2251022 J

5.13103 eV.

7. According to Eq. 39-4 En L? 2. As a consequence, the new energy level E'n satisfies

FGIJ FGIJ En

L 2

L2 1 ,

HK HK En L

L 2

which gives L 2L. Thus, the ratio is L / L 2 1.41.

8. Let the quantum numbers of the pair in question be n and n + 1, respectively. Then

Letting

En+1 ? En = E1 (n + 1)2 ? E1n2 = (2n + 1)E1.

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b g b g c h En1 En 2n 1 E1 3 E4 E3 3 42 E1 32 E1 21E1,

we get 2n + 1 = 21, or n = 10. Thus, (a) the higher quantum number is n + 1 = 10 + 1 = 11, and (b) the lower quantum number is n = 10. (c) Now letting

b g b g c h En1 En 2n 1 E1 2 E4 E3 2 42 E1 32 E1 14E1,

we get 2n + 1 = 14, which does not have an integer-valued solution. So it is impossible to find the pair of energy levels that fits the requirement.

9. Let the quantum numbers of the pair in question be n and n + 1, respectively. We note

that

b g b g En1 En

n 1 2h2 8mL2

n2h2 8mL2

2n 1 h2 8mL2

Therefore, En+1 ? En = (2n + 1)E1. Now

b g En1 En E5 52 E1 25E1 2n 1 E1,

which leads to 2n + 1 = 25, or n = 12. Thus,

(a) The higher quantum number is n+1 = 12+1 = 13.

(b) The lower quantum number is n = 12.

(c) Now let

b g En1 En E6 62 E1 36E1 2n 1 E1,

which gives 2n + 1 = 36, or n = 17.5. This is not an integer, so it is impossible to find the pair that fits the requirement.

10. The energy levels are given by En = n2h2/8mL2, where h is the Planck constant, m is

the mass of an electron, and L is the width of the well. The frequency of the light that will

excite the electron from the state with quantum number ni to the state with quantum

number nf is

f

E h

h 8mL2

n2f ni2

and the wavelength of the light is

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CHAPTER 39

d i c 8mL2c . f h n2f ni2

We evaluate this expression for ni = 1 and nf = 2, 3, 4, and 5, in turn. We use h = 6.626 10? 34 J ? s, m = 9.109 10? 31kg, and L = 250 10? 12 m, and obtain the following results:

(a) 6.87 10? 8 m for nf = 2, (the longest wavelength).

(b) 2.58 10? 8 m for nf = 3, (the second longest wavelength).

(c) 1.37 10? 8 m for nf = 4, (the third longest wavelength).

11. We can use the mc2 value for an electron from Table 37-3 (511 103 eV) and hc = 1240 eV ? nm by rewriting Eq. 39-4 as

cbhg En

n2h2 8mL2

n2 hc 2

8 mc2 L2

.

(a) The first excited state is characterized by n = 2, and the third by n' = 4. Thus,

hc 2

E 8 mc2 L2

n2 n2

1240eV nm2

8 511103 eV

0.250nm2

42 22 6.02eV 16 4

72.2eV .

Now that the electron is in the n' = 4 level, it can "drop" to a lower level (n'') in a variety of ways. Each of these drops is presumed to cause a photon to be emitted of wavelength

c h

hc

8 mc2 L2

.

c h En En hc n2 n2

For example, for the transition n' = 4 to n'' = 3, the photon emitted would have

wavelength

c hb g 8 511103 eV 0.250 nm 2

b gc h

29.4 nm,

1240eV nm 42 32

and once it is then in level n'' = 3 it might fall to level n''' = 2 emitting another photon. Calculating in this way all the possible photons emitted during the de-excitation of this system, we obtain the following results:

(b) The shortest wavelength that can be emitted is l 41 13.7nm.

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(c) The second shortest wavelength that can be emitted is l 42 17.2nm. (d) The longest wavelength that can be emitted is l 21 68.7 nm. (e) The second longest wavelength that can be emitted is l 32 41.2 nm. (f) The possible transitions are shown next. The energy levels are not drawn to scale.

(g) A wavelength of 29.4 nm corresponds to 4 3 transition. Thus, it could make either the 3 1 transition or the pair of transitions: 3 2 and 2 1 . The longest wavelength that can be emitted is l 21 68.7 nm.

(h) The shortest wavelength that can next be emitted is l 31 25.8nm.

12. The frequency of the light that will excite the electron from the state with quantum number ni to the state with quantum number nf is

f E h h 8mL2

n2f ni2

and the wavelength of the light is

d i c 8mL2c . f h n2f ni2

The width of the well is

L

hc(n2f ni2 ) 8mc2

The longest wavelength shown in Figure 39-28 is 80.78 nm which corresponds to a jump from ni 2 to nf 3 . Thus, the width of the well is

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CHAPTER 39

L hc(n2f ni2 ) (80.78 nm)(1240eV nm)(32 22 ) 0.350nm 350 pm.

8mc2

8(511103 eV)

z 13. The probability that the electron is found in any interval is given by P 2 dx,

where the integral is over the interval. If the interval width x is small, the probability can be approximated by P = ||2 x, where the wave function is evaluated for the center

of the interval, say. For an electron trapped in an infinite well of width L, the ground state

probability density is

FG IJ 2 2 sin2 x ,

H K L

L

so

P FGH2LxIJKsin2 FGHLxIJK.

(a) We take L = 100 pm, x = 25 pm, and x = 5.0 pm. Then,

Lb gO Lb gO 2 5.0pm M P M P P

sin2 25pm

0.050.

N Q N Q 100pm

100 pm

(b) We take L = 100 pm, x = 50 pm, and x = 5.0 pm. Then,

Lb gO Lb gO 2 5.0pm M P M P P

sin2 50 pm

0.10.

N Q N Q 100pm

100 pm

(c) We take L = 100 pm, x = 90 pm, and x = 5.0 pm. Then,

Lb gO Lb gO 2 5.0pm M P M P P

sin2 90 pm

0.0095.

N Q N Q 100pm

100 pm

14. We follow Sample Problem 39-3 in the presentation of this solution. The integration result quoted below is discussed in a little more detail in that Sample Problem. We note that the arguments of the sine functions used below are in radians.

(a) The probability of detecting the particle in the region 0 x L / 4 is

2 L

L

/ 4 sin2

0

y

dy

2

y 2

sin 2 4

y

/4 0

0.091.

(b) As expected from symmetry,

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2 L

L

sin2

/4

y

dy

2

y 2

sin 2 4

y

/4

0.091.

(c) For the region L / 4 x 3L / 4 , we obtain

2 L

L

3 / 4 sin2

/4

y

dy

2

y 2

sin 2 4

y

3 / 4 /4

0.82

which we could also have gotten by subtracting the results of part (a) and (b) from 1; that is, 1 ? 2(0.091) = 0.82.

15. The position of maximum probability density corresponds to the center of the well: x L / 2 (200 pm) / 2 100 pm.

(a) The probability of detection at x is given by Eq. 39-11:

2

p(

x)

2 n

(

x)dx

2 L

sin

n L

x

dx

2 L

sin 2

n L

x

dx

For n 3 , L 200 pm and dx 2.00 pm (width of the probe), the probability of detection at x L / 2 100 pm is

p(x

L

/

2)

2 L

sin 2

3 L

L 2

dx

2 L

sin 2

3 2

dx

2 L

dx

2 200 pm

2.00

pm

0.020

.

(b) With N 1000 independent insertions, the number of times we expect the electron to be detected is

n Np (1000)(0.020) 20 .

16. From Eq. 39-11, the condition of zero probability density is given by

sin

n L

x

0

n x m L

where m is an integer. The fact that x 0.300L and x 0.400L have zero probability density implies

sin 0.300n sin 0.400n 0

which can be satisfied for n 10m , where m 1, 2,... However, since the probability

density is non-zero between x 0.300L and x 0.400L , we conclude that the electron is in the n 10 state. The change of energy after making a transition to n 9 is then equal to

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CHAPTER 39

|

E

|

h2 8mL2

6.631034 J s 2

n2 n2 8 9.111031 kg

2.00 1010 m 2

102 92

2.861017 J

17. According to Fig. 39-9, the electron's initial energy is 106 eV. After the additional energy is absorbed, the total energy of the electron is 106 eV + 400 eV = 506 eV. Since it is in the region x > L, its potential energy is 450 eV (see Section 39-5), so its kinetic energy must be 506 eV ? 450 eV = 56 eV.

18. From Fig. 39-9, we see that the sum of the kinetic and potential energies in that particular finite well is 233 eV. The potential energy is zero in the region 0 < x < L. If the kinetic energy of the electron is detected while it is in that region (which is the only region where this is likely to happen), we should find K = 233 eV.

19. Schr?dinger's equation for the region x > L is

d 2 dx2

82m h2

E U0

0.

If = De2kx, then d 2/dx2 = 4k2De2kx = 4k2 and

d 2 dx2

82m h2

E U0

4k

2

82m h2

E U0

.

This is zero provided

b g

k h

2m U0 E .

The proposed function satisfies Schr?dinger's equation provided k has this value. Since U0 is greater than E in the region x > L, the quantity under the radical is positive. This means k is real. If k is positive, however, the proposed function is physically unrealistic. It increases exponentially with x and becomes large without bound. The integral of the probability density over the entire x-axis must be unity. This is impossible if is the proposed function.

20. The smallest energy a photon can have corresponds to a transition from the nonquantized region to E3. Since the energy difference between E3 and E4 is

E E4 E3 9.0 eV 4.0 eV 5.0 eV ,

the energy of the photon is Ephoton K E 2.00 eV 5.00 eV 7.00 eV .

21. Using E hc / (1240eV nm)/ , the energies associated with a , b and c are

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