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

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

,Table of contents ZX ZX




1. The Wave-Particle Duality
ZX ZX ZX




2. The Schrödinger Wave Equation
ZX ZX ZX ZX




3. Operators and Waves
ZX ZX ZX




4. The Hydrogen Atom
ZX ZX ZX




5. Many-Electron Atoms
ZX ZX




6. The Emergence of Masers and Lasers
ZX ZX ZX ZX ZX ZX




7. Diatomic Molecules
ZX ZX




8. Statistical Physics
ZX ZX




9. Electronic Structure of Solids
ZX ZX ZX ZX




10. Charge Carriers in Semiconductors
ZX ZX ZX ZX




11. Semiconductor Lasers
ZX ZX




12. The Special Theory of Relativity
ZX ZX ZX ZX ZX




13. The Relativistic Wave Equations and General Relativity
ZX ZX ZX ZX ZX ZX ZX




14. Particle Physics
ZX ZX




15. Nuclear Physics
ZX ZX

,1

The Wave-Particle Duality - Solutions
Z X Z X Z X Z X




1. The energy of photons in terms of the wavelength of light is giv
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




en by Eq. (1.5). Following Example 1.1 and substituting λ = 200
ZX ZX ZX ZX ZX Z X ZX ZX ZX ZX ZX Z




eV gives:
X ZX




hc 1240 eV · nm
= = 6.2 eV
Z X Z X ZX




Ephoton = λ
ZX ZX



200 nm ZX ZX




2. The energy of the beam each second is: Z X Z X Z X Z X Z X Z X Z X




power 100 W
= = 100 J
ZX




Etotal = time
ZX ZX



1s ZX ZX




The number of photons comes from the total energy divided by t
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




he energy of each photon (see Problem 1). The photon’s energy m
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




ust be converted to Joules using the constant 1.602 × 10−19 J/eV ,
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




see Example 1.5. The result is:
ZX ZX ZX ZX ZX




N =Etotal = 100 J = 1.01 × 1020 ZX Z X ZX




photons E
ZX ZX ZX




pho
ton 9.93 × 10−19 ZX ZX




for the number of photons striking the surface each second.
Z X Z X Z X Z X Z X Z X Z X Z X Z X




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




0−3 W . The power may be expressed as: 1 W = 1 J/s. Following E
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




xample 1.1, the energy of a single photon is: ZX ZX ZX ZX ZX ZX ZX ZX




1240 eV · nm
hc = 1.960 eV
Z X Z X ZX




Ephoton = 632.8 nm
ZX ZX ZX


Z ZX
Z X



=
λ
X


Z X




We now convert to SI units (see Example 1.5):
ZX ZX ZX ZX ZX ZX ZX Z X




1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




Following the same procedure as Problem 2: ZX ZX ZX ZX ZX ZX




1 × 10−3 J/s 15 photons ZX ZX ZX
Z X



Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
ZX Z X ZX Z X ZX ZX ZX
Z X
ZX ZX Z X

, 2

4.The maximum kinetic energy of photoelectrons is found usin
ZX ZX ZX ZX ZX ZX ZX ZX




g Eq. (1.6) and the work functions, W, of the metals are given in
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV . For p
ZX ZX ZX Z X Z X ZX ZX ZX ZX Z X ZX Z X Z X




art (a), Na has W = 2.28 eV :
Z X Z X Z X Z X Z X ZX Z X ZX




(KE)max = 6.20 eV − 2.28 eV = 3.92 eV ZX ZX ZX ZX ZX ZX Z X ZX ZX




Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 eV
ZX ZX ZX ZX ZX ZX ZX Z X ZX ZX Z X ZX ZX ZX ZX




and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




5.This problem again concerns the photoelectric effect. As in Probl
ZX ZX ZX ZX ZX ZX ZX ZX ZX




em 4, we use Eq. (1.6):
ZX ZX ZX ZX ZX




hc − ZX


(KE)max = ZX




Wλ
ZX




ZX




where W is the work function of the material and the term hc/λ
Z X Z X Z X Z X Z X Z X Z X Z X Z X Z X Z X Z X Z X




describes the energy of the incoming photons. Solving for the latter:
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
ZX ZX ZX Z X ZX ZX Z X ZX ZX Z X ZX ZX


Z X




Solving Eq. (1.5) for the wavelength: ZX ZX ZX ZX ZX




1240 eV · nm
λ=
Z X Z X ZX



= 387.5 nm ZX



3.2 e
ZX ZX


Z X



V
6. A potential energy of 0.72 eV is needed to stop the flow of electrons.
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




Hence, (KE)max of the photoelectrons can be no more than 0.72 eV. S
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




olving Eq. (1.6) for the work function:
ZX ZX ZX ZX ZX ZX




hc 1240 eV · n — 0.72 eV = 1.98 eV
W= —
Z X Z X ZX




λ m
ZX Z X Z X ZX Z X
ZX Z X




(KE)max Z



= X




460 nm ZX




7. Reversing the procedure from Problem 6, we start with Eq. (1.6): ZX ZX ZX Z X ZX ZX ZX Z X ZX ZX




hc 1240 eV · n
(KE)max = − W
Z X




— 1.98 eV = 3.19 eV
Z X Z X ZX
ZX Z X




m
ZX ZX Z X Z X ZX Z X




=
λ
240 nm ZX




Hence, a stopping potential of 3.19 eV prohibits the electrons from r
ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX ZX




eaching the anode. ZX ZX




8. Just at threshold, the kinetic energy of the electron is z
Z X Z X Z X Z X Z X Z X Z X Z X Z X Z X




ero. Setting (KE)max = 0 in Eq. (1.6),
Z X ZX ZX ZX Z X Z X Z X




hc
W= = 1240 eV · n = 3.44 eV Z X Z X ZX




λ0
ZX



m
ZX ZX




360 nm ZX




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