Solution Manual for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by John Morrison, All 1- 15 Chapters Covered, Verified Latest Version.

SOLUTION MANUAL Modern Physics with Modern
Computational Methods: for Scientists and
Engineers 3rd Edition by Morrison Chapters 1- 15

,Table of contents
1. The Wave-Particle Duality

2. The Schrödinger Wave Equation

3. Operators and Waves

4. The Hydrogen Atom

5. Many-Electron Atoms

6. The Emergence of Masers and Lasers

7. Diatomic Molecules

8. Statistical Physics

9. Electronic Structure of Solids

10. Charge Carriers in Semiconductors

11. Semiconductor Lasers

12. The Special Theory of Relativity

13. The Relativistic Wave Equations and General Relativity

14. Particle Physics

15. Nuclear Physics

,1

The Wave-Particle Duality - Solutions




1. The energy of photons in terms of the wavelength of light is
given by Eq. (1.5). Following Eẍample 1.1 and substituting λ =
200 eV gives:
hc 1240 eV · nm
= = 6.2 eV
Ephoton = λ 200 nm
2. The energy of the beam each second is:
power 100 W
= = 100 J
Etotal = time 1s
The number of photons comes from the total energy divided by
the energy of each photon (see Problem 1). The photon’s energy
must be converted to Joules using the constant 1.602 × 10−19
J/eV , see Eẍample 1.5. The result is:
N =Etotal = 100 J = 1.01 × 1020
photons E
phot
on 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 may be eẍpressed as: 1 W = 1
J/s. Following Eẍample 1.1, the energy of a single photon is:
1240 eV · nm
hc = 1.960 eV
Ephoton = 632.8 nm
=
λ
We now convert to SI units (see Eẍample 1.5):
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
Following the same procedure as Problem 2:
1 × 10−3 J/s 15 photons
Rate of emission = −19 = 3.19 × 10
3.14 × 10 J/photon s

, 2

4. The maẍimum kinetic energy 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)maẍ = 6.20 eV − 2.28 eV = 3.92 eV
Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)maẍ = 2.12
eV
and for Ag metal in part (c), W = 4.73 eV , giving (KE)maẍ = 1.47 eV .

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

8. Just at threshold, the kinetic energy of the electron is
zero. Setting (KE)maẍ = 0 in Eq. (1.6),
hc
W = = 1240 eV · = 3.44 eV
λ0 nm

360 nm
9. A frequency of 1200 THz is equal to 1200 × 1012 Hz. Using Eq. (1.10),