SOLUTION MANUAL Modern Physics with Modern
f f f f f
f Computational Methods: for Scientists and Engineers
f f f f f
f 3rd Edition by Morrison Chapters 1- 15
f f f f f f
,Table of contents f f
1. The Wave-Particle Duality
f f f
2. The Schrödinger Wave Equation
f f f f
3. Operators and Waves
f f f
4. The Hydrogen Atom
f f f
5. Many-Electron Atoms
f f
6. The Emergence of Masers and Lasers
f f f f f f
7. Diatomic Molecules
f f
8. Statistical Physics
f f
9. Electronic Structure of Solids
f f f f
10. Charge Carriers in Semiconductors
f f f f
11. Semiconductor Lasers
f f
12. The Special Theory of Relativity
f f f f f
13. The Relativistic Wave Equations and General Relativity
f f f f f f f
14. Particle Physics
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15. Nuclear Physics
f f
,1
The Wave-Particle Duality - Solutions
f f f f
1. The energy of photons in terms of the wavelength of light is
f f f f f f f f f f f
given by Eq. (1.5). Following Example 1.1 and substituting λ =
f f f f f f f f f f f
200 eV gives:
f f f
hc 1240 eV · nm
= = 6.2 eV
f f f
Ephoton = λ
f f
200 nm f f
2. The energy of the beam each second is:
f f f f f f f
power 100 W
= = 100 J
f
Etotal = time
f f
1s f f
The number of photons comes from the total energy divided by
f f f f f f f f f f
the energy of each photon (see Problem 1). The photon’s energy
f f f f f f f f f f f
must be converted to Joules using the constant 1.602 × 10−19
f f f f f f f f f f f
J/eV , see Example 1.5. The result is:
f f f f f f f f
N =Etotal = 100 J = 1.01 × 1020 f f f
photons E
f f f
pho
ton 9.93 × 10−19 f f
for the number of photons striking the surface each second.
f f f f f f f f f
3.We are given the power of the laser in milliwatts, where 1 mW =
f f f f f f f f f f f f f
10−3 W . The power may be expressed as: 1 W = 1 J/s. Following
f f f f f f f f f f f f f f f
Example 1.1, the energy of a single photon is:
f f f f f f f f f
1240 eV · nm
hc = 1.960 eV
f f f
f f
Ephoton = 632.8 nm f f f
=
λ
f
f
We now convert to SI units (see Example 1.5):
f f f f f f f f
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
f f f f f f f f f f f
Following the same procedure as Problem 2: f f f f f f
1 × 10−3 J/s 15 photons f f f
f
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
f f f f f f f
f
f f f
, 2
4. The maximum kinetic energy of photoelectrons is found
f f f f f f f
f using Eq. (1.6) and the work functions, W, of the metals are
f f f f f f f f f f f
f given in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20
f f f f f f f f f f f
f eV . For part (a), Na has W = 2.28 eV :
f f f f f f f f f f f
(KE)max= 6.20 eV − 2.28 eV = 3.92 eV f f f f f f f f f
Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 eV
f f f f f f f f f f f f f f f
and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .
f f f f f f f f f f f f f f f f f
5.This problem again concerns the photoelectric effect. As in
f f f f f f f f
Problem 4, we use Eq. (1.6):
f f f f f f
hc − f
(KE)max =
Wλ
f
f f
where W is the work function of the material and the term hc/λ
f f f f f f f f f f f f
describes the energy of the incoming photons. Solving for the latter:
f f f f f f f f f f f
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
f f f f f f f f f f f f
f
Solving Eq. (1.5) for the wavelength: f f f f f
1240 eV · nm
λ=
f f f
= 387.5 nm f
3.2
f f
eV f
6. A potential energy of 0.72 eV is needed to stop the flow of electrons.
f f f f f f f f f f f f f
Hence, (KE)max of the photoelectrons can be no more than 0.72 eV.
f f f f f f f f f f f f
Solving Eq. (1.6) for the work function:
f f f f f f f
hc 1240 eV · — 0.72 eV = 1.98 eV
W= (KE)max
f f
λ
f f f f f
nm
f f
=
f
f
460 nm f
7. Reversing the procedure from Problem 6, we start with Eq. (1.6): f f f f f f f f f f
hc 1240 eV ·
(KE)max = − W
f
— 1.98 eV = 3.19 eV
f f
f
nm
f f f f f f
= f
f
λ
240 nm f
Hence, a stopping potential of 3.19 eV prohibits the electrons from
f f f f f f f f f f
reaching the anode.
f f f
8. Just at threshold, the kinetic energy of the electron is
f f f f f f f f f
f zero. Setting (KE)max = 0 in Eq. (1.6),
f f f f f f f
hc
W= f = 1240 eV · = 3.44 eV f f
λ0 nm f
f f
360 nm f
9. A frequency of 1200 THz is equal to 1200 × 1012 Hz. Using Eq. (1.10),
f f f f f f f f f f f f f f
f f f f f
f Computational Methods: for Scientists and Engineers
f f f f f
f 3rd Edition by Morrison Chapters 1- 15
f f f f f f
,Table of contents f f
1. The Wave-Particle Duality
f f f
2. The Schrödinger Wave Equation
f f f f
3. Operators and Waves
f f f
4. The Hydrogen Atom
f f f
5. Many-Electron Atoms
f f
6. The Emergence of Masers and Lasers
f f f f f f
7. Diatomic Molecules
f f
8. Statistical Physics
f f
9. Electronic Structure of Solids
f f f f
10. Charge Carriers in Semiconductors
f f f f
11. Semiconductor Lasers
f f
12. The Special Theory of Relativity
f f f f f
13. The Relativistic Wave Equations and General Relativity
f f f f f f f
14. Particle Physics
f f
15. Nuclear Physics
f f
,1
The Wave-Particle Duality - Solutions
f f f f
1. The energy of photons in terms of the wavelength of light is
f f f f f f f f f f f
given by Eq. (1.5). Following Example 1.1 and substituting λ =
f f f f f f f f f f f
200 eV gives:
f f f
hc 1240 eV · nm
= = 6.2 eV
f f f
Ephoton = λ
f f
200 nm f f
2. The energy of the beam each second is:
f f f f f f f
power 100 W
= = 100 J
f
Etotal = time
f f
1s f f
The number of photons comes from the total energy divided by
f f f f f f f f f f
the energy of each photon (see Problem 1). The photon’s energy
f f f f f f f f f f f
must be converted to Joules using the constant 1.602 × 10−19
f f f f f f f f f f f
J/eV , see Example 1.5. The result is:
f f f f f f f f
N =Etotal = 100 J = 1.01 × 1020 f f f
photons E
f f f
pho
ton 9.93 × 10−19 f f
for the number of photons striking the surface each second.
f f f f f f f f f
3.We are given the power of the laser in milliwatts, where 1 mW =
f f f f f f f f f f f f f
10−3 W . The power may be expressed as: 1 W = 1 J/s. Following
f f f f f f f f f f f f f f f
Example 1.1, the energy of a single photon is:
f f f f f f f f f
1240 eV · nm
hc = 1.960 eV
f f f
f f
Ephoton = 632.8 nm f f f
=
λ
f
f
We now convert to SI units (see Example 1.5):
f f f f f f f f
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
f f f f f f f f f f f
Following the same procedure as Problem 2: f f f f f f
1 × 10−3 J/s 15 photons f f f
f
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
f f f f f f f
f
f f f
, 2
4. The maximum kinetic energy of photoelectrons is found
f f f f f f f
f using Eq. (1.6) and the work functions, W, of the metals are
f f f f f f f f f f f
f given in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20
f f f f f f f f f f f
f eV . For part (a), Na has W = 2.28 eV :
f f f f f f f f f f f
(KE)max= 6.20 eV − 2.28 eV = 3.92 eV f f f f f f f f f
Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 eV
f f f f f f f f f f f f f f f
and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .
f f f f f f f f f f f f f f f f f
5.This problem again concerns the photoelectric effect. As in
f f f f f f f f
Problem 4, we use Eq. (1.6):
f f f f f f
hc − f
(KE)max =
Wλ
f
f f
where W is the work function of the material and the term hc/λ
f f f f f f f f f f f f
describes the energy of the incoming photons. Solving for the latter:
f f f f f f f f f f f
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
f f f f f f f f f f f f
f
Solving Eq. (1.5) for the wavelength: f f f f f
1240 eV · nm
λ=
f f f
= 387.5 nm f
3.2
f f
eV f
6. A potential energy of 0.72 eV is needed to stop the flow of electrons.
f f f f f f f f f f f f f
Hence, (KE)max of the photoelectrons can be no more than 0.72 eV.
f f f f f f f f f f f f
Solving Eq. (1.6) for the work function:
f f f f f f f
hc 1240 eV · — 0.72 eV = 1.98 eV
W= (KE)max
f f
λ
f f f f f
nm
f f
=
f
f
460 nm f
7. Reversing the procedure from Problem 6, we start with Eq. (1.6): f f f f f f f f f f
hc 1240 eV ·
(KE)max = − W
f
— 1.98 eV = 3.19 eV
f f
f
nm
f f f f f f
= f
f
λ
240 nm f
Hence, a stopping potential of 3.19 eV prohibits the electrons from
f f f f f f f f f f
reaching the anode.
f f f
8. Just at threshold, the kinetic energy of the electron is
f f f f f f f f f
f zero. Setting (KE)max = 0 in Eq. (1.6),
f f f f f f f
hc
W= f = 1240 eV · = 3.44 eV f f
λ0 nm f
f f
360 nm f
9. A frequency of 1200 THz is equal to 1200 × 1012 Hz. Using Eq. (1.10),
f f f f f f f f f f f f f f
