SOLUTION MANUAL Modern Physics with Modern
Computational Methods: for Scientists and
Engineers 3rd Edition by Morrison Chapters 1- 15
,Table of contents TV TV
1. The Wave-Particle Duality
TV TV TV
2. The Schrödinger Wave Equation
TV TV TV TV
3. Operators and Waves
TV TV TV
4. The Hydrogen Atom
TV TV TV
5. Many-Electron Atoms
TV TV
6. The Emergence of Masers and Lasers
TV TV TV TV TV TV
7. Diatomic Molecules
TV TV
8. Statistical Physics
TV TV
9. Electronic Structure of Solids
TV TV TV TV
10. Charge Carriers in Semiconductors
TV TV TV TV
11. Semiconductor Lasers
TV TV
12. The Special Theory of Relativity
TV TV TV TV TV
13. The Relativistic Wave Equations and General Relativity
TV TV TV TV TV TV TV
14. Particle Physics
TV TV
15. Nuclear Physics
TV TV
,1
The Wave-Particle Duality - Solutions
T V T V T V T V
1. The energy of photons in terms of the wavelength of light is
TV TV TV TV TV TV TV TV TV TV TV TV
given by Eq. (1.5). Following Example 1.1 and substituting λ
TV TV TV TV TV T V TV TV TV TV
= 200 eV gives:
TV TV TV
hc 1240 eV · nm
= = 6.2 eV
T V T V TV
Ephoton = λ
TV TV
200 nm TV TV
2. The energy of the beam each second is:
T V T V T V T V T V T V T V
power 100 W
= = 100 J
T V
Etotal = time
TV TV
1s TV TV
The number of photons comes from the total energy divided b
TV TV TV TV TV TV TV TV TV TV
y the energy of each photon (see Problem 1). The photon’s ener
TV TV TV TV TV TV TV TV TV TV TV
gy must be converted to Joules using the constant 1.602 × 10−1
TV TV TV TV TV TV TV TV TV TV TV
9 J/eV , see Example 1.5. The result is:
TV TV TV TV TV TV TV TV
N =Etotal = 100 J = 1.01 × 1020 TV T V TV
photons E
TV TV TV
pho
ton 9.93 × 10−19 TV TV
for the number of photons striking the surface each second.
T V T V T V T V T V T V T V T V T V
3. We are given the power of the laser in milliwatts, where 1 mW
TV TV TV TV TV TV TV TV TV TV TV TV TV
= 10−3 W . The power may be expressed as: 1 W = 1 J/s. Follo
TV TV TV TV TV TV TV TV TV TV TV TV TV TV TV
wing Example 1.1, the energy of a single photon is:
TV TV TV TV TV TV TV TV TV
1240 eV · nm
hc = 1.960 eV
T V T V TV
Ephoton = 632.8 nm
TV TV TV
T V T V
=
λ
T V
T V
We now convert to SI units (see Example 1.5):
T V T V T V T V T V T V T V T V
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
TV TV TV TV TV TV T V TV TV TV TV
Following the same procedure as Problem 2: TV TV TV TV TV TV
1 × 10−3 J/s 15 photons TV TV TV
T V
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
TV T V TV T V TV TV TV
T V
TV TV T V
, 2
4. The maximum kinetic energy of photoelectrons is found usi
TV TV TV TV TV TV TV TV
ng Eq. (1.6) and the work functions, W, of the metals are give
TV TV TV TV TV TV TV TV TV TV TV TV
n in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV .
TV TV TV TV TV T V T V TV TV TV TV T V TV
For part (a), Na has W = 2.28 eV :
T V T V T V T V T V T V T V TV T V TV
(KE)max = 6.20 eV − 2.28 eV = 3.92 eV TV TV TV TV TV TV T V TV TV
Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 e
T V TV TV TV TV TV T V T V TV TV T V TV TV TV TV
V
and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .
TV TV TV TV TV TV TV T V TV TV TV TV TV TV TV TV TV
5. This problem again concerns the photoelectric effect. As in Pro
TV TV TV TV TV TV TV TV TV
blem 4, we use Eq. (1.6): TV TV TV TV TV
hc − TV
(KE)max = TV
λ
T V
W TV
where W is the work function of the material and the term hc
T V T V T V T V T V T V T V T V T V T V T V T V
/λ describes the energy of the incoming photons. Solving for the la
T V TV TV TV TV TV TV TV TV TV TV
tter:
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
TV TV TV T V TV T V T V TV T V T V TV T V
T V
Solving Eq. (1.5) for the wavelength: TV T V T V TV T V
1240 eV · nm
λ=
T V T V TV
= 387.5 nm TV
3.2 e
TV TV
T V
V
6. A potential energy of 0.72 eV is needed to stop the flow of electron
TV TV TV TV TV TV TV TV TV TV TV TV TV
s. Hence, (KE)max of the photoelectrons can be no more than 0.72 e
TV TV TV TV TV TV TV TV TV TV TV TV
V. Solving Eq. (1.6) for the work function:
TV TV TV TV TV TV TV
hc 1240 eV · n — 0.72 eV = 1.98 eV
W= —
T V T V TV
λ m
TV T V T V TV T V
TV T V
(KE) max T
= V
460 nm TV
7. Reversing the procedure from Problem 6, we start with Eq. (1.6): T V T V T V T V T V T V T V T V T V T V
hc 1240 eV · n
(KE)max = − W
T V
— 1.98 eV = 3.19 eV
T V T V TV
TV T V
m
T V TV T V T V TV T V
=
λ
240 nm TV
Hence, a stopping potential of 3.19 eV prohibits the electrons fro
TV TV TV TV TV TV TV TV TV TV
m reaching the anode.
TV TV TV
8. Just at threshold, the kinetic energy of the electron is
T V T V T V T V T V T V T V T V T V
T V zero. Setting (KE)max = 0 in Eq. (1.6),
T V TV TV TV T V T V T V
hc 1240 eV · n
W= = = 3.44 eV
T V T V TV
λ0
TV
m
TV T V
360 nm TV
Computational Methods: for Scientists and
Engineers 3rd Edition by Morrison Chapters 1- 15
,Table of contents TV TV
1. The Wave-Particle Duality
TV TV TV
2. The Schrödinger Wave Equation
TV TV TV TV
3. Operators and Waves
TV TV TV
4. The Hydrogen Atom
TV TV TV
5. Many-Electron Atoms
TV TV
6. The Emergence of Masers and Lasers
TV TV TV TV TV TV
7. Diatomic Molecules
TV TV
8. Statistical Physics
TV TV
9. Electronic Structure of Solids
TV TV TV TV
10. Charge Carriers in Semiconductors
TV TV TV TV
11. Semiconductor Lasers
TV TV
12. The Special Theory of Relativity
TV TV TV TV TV
13. The Relativistic Wave Equations and General Relativity
TV TV TV TV TV TV TV
14. Particle Physics
TV TV
15. Nuclear Physics
TV TV
,1
The Wave-Particle Duality - Solutions
T V T V T V T V
1. The energy of photons in terms of the wavelength of light is
TV TV TV TV TV TV TV TV TV TV TV TV
given by Eq. (1.5). Following Example 1.1 and substituting λ
TV TV TV TV TV T V TV TV TV TV
= 200 eV gives:
TV TV TV
hc 1240 eV · nm
= = 6.2 eV
T V T V TV
Ephoton = λ
TV TV
200 nm TV TV
2. The energy of the beam each second is:
T V T V T V T V T V T V T V
power 100 W
= = 100 J
T V
Etotal = time
TV TV
1s TV TV
The number of photons comes from the total energy divided b
TV TV TV TV TV TV TV TV TV TV
y the energy of each photon (see Problem 1). The photon’s ener
TV TV TV TV TV TV TV TV TV TV TV
gy must be converted to Joules using the constant 1.602 × 10−1
TV TV TV TV TV TV TV TV TV TV TV
9 J/eV , see Example 1.5. The result is:
TV TV TV TV TV TV TV TV
N =Etotal = 100 J = 1.01 × 1020 TV T V TV
photons E
TV TV TV
pho
ton 9.93 × 10−19 TV TV
for the number of photons striking the surface each second.
T V T V T V T V T V T V T V T V T V
3. We are given the power of the laser in milliwatts, where 1 mW
TV TV TV TV TV TV TV TV TV TV TV TV TV
= 10−3 W . The power may be expressed as: 1 W = 1 J/s. Follo
TV TV TV TV TV TV TV TV TV TV TV TV TV TV TV
wing Example 1.1, the energy of a single photon is:
TV TV TV TV TV TV TV TV TV
1240 eV · nm
hc = 1.960 eV
T V T V TV
Ephoton = 632.8 nm
TV TV TV
T V T V
=
λ
T V
T V
We now convert to SI units (see Example 1.5):
T V T V T V T V T V T V T V T V
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
TV TV TV TV TV TV T V TV TV TV TV
Following the same procedure as Problem 2: TV TV TV TV TV TV
1 × 10−3 J/s 15 photons TV TV TV
T V
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
TV T V TV T V TV TV TV
T V
TV TV T V
, 2
4. The maximum kinetic energy of photoelectrons is found usi
TV TV TV TV TV TV TV TV
ng Eq. (1.6) and the work functions, W, of the metals are give
TV TV TV TV TV TV TV TV TV TV TV TV
n in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV .
TV TV TV TV TV T V T V TV TV TV TV T V TV
For part (a), Na has W = 2.28 eV :
T V T V T V T V T V T V T V TV T V TV
(KE)max = 6.20 eV − 2.28 eV = 3.92 eV TV TV TV TV TV TV T V TV TV
Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 e
T V TV TV TV TV TV T V T V TV TV T V TV TV TV TV
V
and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .
TV TV TV TV TV TV TV T V TV TV TV TV TV TV TV TV TV
5. This problem again concerns the photoelectric effect. As in Pro
TV TV TV TV TV TV TV TV TV
blem 4, we use Eq. (1.6): TV TV TV TV TV
hc − TV
(KE)max = TV
λ
T V
W TV
where W is the work function of the material and the term hc
T V T V T V T V T V T V T V T V T V T V T V T V
/λ describes the energy of the incoming photons. Solving for the la
T V TV TV TV TV TV TV TV TV TV TV
tter:
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
TV TV TV T V TV T V T V TV T V T V TV T V
T V
Solving Eq. (1.5) for the wavelength: TV T V T V TV T V
1240 eV · nm
λ=
T V T V TV
= 387.5 nm TV
3.2 e
TV TV
T V
V
6. A potential energy of 0.72 eV is needed to stop the flow of electron
TV TV TV TV TV TV TV TV TV TV TV TV TV
s. Hence, (KE)max of the photoelectrons can be no more than 0.72 e
TV TV TV TV TV TV TV TV TV TV TV TV
V. Solving Eq. (1.6) for the work function:
TV TV TV TV TV TV TV
hc 1240 eV · n — 0.72 eV = 1.98 eV
W= —
T V T V TV
λ m
TV T V T V TV T V
TV T V
(KE) max T
= V
460 nm TV
7. Reversing the procedure from Problem 6, we start with Eq. (1.6): T V T V T V T V T V T V T V T V T V T V
hc 1240 eV · n
(KE)max = − W
T V
— 1.98 eV = 3.19 eV
T V T V TV
TV T V
m
T V TV T V T V TV T V
=
λ
240 nm TV
Hence, a stopping potential of 3.19 eV prohibits the electrons fro
TV TV TV TV TV TV TV TV TV TV
m reaching the anode.
TV TV TV
8. Just at threshold, the kinetic energy of the electron is
T V T V T V T V T V T V T V T V T V
T V zero. Setting (KE)max = 0 in Eq. (1.6),
T V TV TV TV T V T V T V
hc 1240 eV · n
W= = = 3.44 eV
T V T V TV
λ0
TV
m
TV T V
360 nm TV
