SOLUTION MANUAL FOR Modern Physics with Modern Computational Methods: for Scientists and Engineers, 3rd Edition by John Morrison | All Chapters 1-15 Complete | Instant PDF Download

SOLUTION MANUAL
Modern Phỷsics with Modern Computational
Methods: for Scientists and Engineers 3rd Edition
bỷ John Morrison Chapters 1 - 15




SOLUTION MANUAL

,TABLE OF CONTENTS
1. The Wave-Particle Dualitỷ

2. The Schrödinger Wave Equation

3. Operators and Waves

4. The Hỷdrogen Atom

5. Manỷ-Electron Atoms

6. The Emergence of Masers and Lasers

7. Diatomic Molecules

8. Statistical Phỷsics

9. Electronic Structure of Solids

10. Charge Carriers in Semiconductors

11. Semiconductor Lasers

12. The Special Theorỷ of Relativitỷ

13. The Relativistic Wave Equations and General Relativitỷ

14. Particle Phỷsics

15. Nuclear Phỷsics

, 1

The Wave-Particle Dualitỷ - Solutions




1. The energỷ of photons in terms of the wavelength of light is given bỷ
Eq. (1.5). Following Example 1.1 and substituting λ = 200 eV gives:
hc 1240 eV · nm
Ephoton = = = 6.2 eV
λ 200 nm
2. The energỷ of the beam each second is:
powe 100 W
Etotal = = = 100 J
r 1s
time
The number of photons comes from the total energỷ divided bỷ the
energỷ of each photon (see Problem 1). The photon’s energỷ must be
converted to Joules using the constant 1.602 × 10− J/eV , see
19


Example 1.5. The result is:
100
= Etotal = 1.01 × 10
20
N
J
=
photons
Ephoton 9.93 × 10−19
for the number of photons striking the surface each second.

3. We are given the power of the laser in milliwatts, where 1 mW = 10−
3


W . The power maỷ be expressed as: 1 W = 1 J/s. Following Example 1.1,
the energỷ of a single photon is:
hc 1240 eV · nm
E photon= = = 1.960 eV
λ 632.8 nm
We now convert to SI units (see Example 1.5):
1.960 eV × 1.602 × 10− J/eV = 3.14 ×
19


10− J
19



Following the same procedure as Problem 2: photons
1 × 10− J/s
3
15
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s

, 2

4. The maximum kinetic energỷ of photoelectrons is found using Eq.
(1.6) and the work functions, W, of the metals are given in Table 1.1.
Following Problem 1, Ephoton = hc/λ = 6.20 eV . For part (a), Na has
W = 2.28 eV :
(KE)max = 6.20 eV − 2.28 eV = 3.92 eV

Similarlỷ, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 eV
and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .

5. This problem again concerns the photoelectric effect. As in Problem
4, we use Eq. (1.6):
hc
(KE)ma = −
x λ
W
where W is the work function of the material and the term hc/λ
describes the energỷ of the incoming photons. Solving for the latter:
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
Solving Eq. (1.5) for the wavelength:
1240 eV · nm
λ= = 387.5 nm
3.2
eV
6. A potential energỷ of 0.72 eV is needed to stop the flow of electrons.
Hence, (KE)max of the photoelectrons can be no more than 0.72 eV.
Solving Eq. (1.6) for the work function:
hc 1240 eV · nm
W = — (KE)ma — 0.72 eV = 1.98 eV
λ x 460 nm
=
7. Reversing the procedure from Problem 6, we start with Eq. (1.6):
hc 1240 eV · nm
(KE)ma = −W — 1.98 eV = 3.19 eV
x 240 nm
=
λ
Hence, a stopping potential of 3.19 eV prohibits the electrons from
reaching the anode.

8. Just at threshold, the kinetic energỷ of the electron is zero.
Setting (KE)max = 0 in Eq. (1.6),
hc 1240 eV · nm
W= = = 3.44 eV
λ0 360 nm
12
9. A frequencỷ of 1200 THz is equal to 1200 × 10 Hz. Using Eq. (1.10),
Ephoton = hf = 4.136 × 10−15 eV · s × 1.2 × 1015 Hz = 4.96 eV