Solution Manual for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by John Morrison, ISBN: 9780128177907, 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 FGH FGH




1. The Wave-Particle Duality
FGH FGH FGH




2. The Schrödinger Wave Equation
FGH FGH FGH FGH




3. Operators and Waves
FGH FGH FGH




4. The Hydrogen Atom
FGH FGH FGH




5. Many-Electron Atoms
FGH FGH




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




7. Diatomic Molecules
FGH FGH




8. Statistical Physics
FGH FGH




9. Electronic Structure of Solids
FGH FGH FGH FGH




10. Charge Carriers in Semiconductors
FGH FGH FGH FGH




11. Semiconductor Lasers
FGH FGH




12. The Special Theory of Relativity
FGH FGH FGH FGH FGH




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




14. Particle Physics
FGH FGH




15. Nuclear Physics
FGH FGH

,1

The Wave-Particle Duality - Solutions
F G H F G H F G H F G H




1. The energy of photons in terms of the wavelength of light is g
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




iven by Eq. (1.5). Following Example 1.1 and substituting λ =
FGH FGH FGH FGH FGH F G H FGH FGH FGH FGH F




200 eV gives:
GH FGH FGH




hc 1240 eV · nm
= = 6.2 eV
FG H F G H FGH




Ephoton = λ
FGH FGH



200 nm FGH FGH




2. The energy of the beam each second is:
F G H F G H F G H F G H F G H F G H F G H




power 100 W
= = 100 J
FGH




Etotal = time
FGH FGH



1s FGH FGH




The number of photons comes from the total energy divided by
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




the energy of each photon (see Problem 1). The photon’s energy
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




must be converted to Joules using the constant 1.602 × 10−19 J/
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




eV , see Example 1.5. The result is:
FGH FGH FGH FGH FGH FGH FGH




N =Etotal = 100 J = 1.01 × 1020 FGH F G H FGH




photons E
FGH FGH FGH




pho
ton 9.93 × 10−19 FGH FGH




for the number of photons striking the surface each second.
F G H F G H F G H F G H F G H F G H F G H F G H F G H




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




10−3 W . The power may be expressed as: 1 W = 1 J/s. Followin
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




g Example 1.1, the energy of a single photon is:
FGH FGH FGH FGH FGH FGH FGH FGH FGH




1240 eV · nm
hc = 1.960 eV
FG H F G H FGH




Ephoton = 632.8 nm
FGH FGH FGH



F G H FGH




=
λ
F G H


F G H




We now convert to SI units (see Example 1.5):
FGH FGH FGH FGH FGH FGH FGH F G H




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




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




1 × 10−3 J/s 15 photons FGH FGH FGH
F G H



Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
FGH F G H FGH F G H FGH FGH FGH
F G H
FGH FGH F G H

, 2

4. The maximum kinetic energy of photoelectrons is found usi
FGH FGH FGH FGH FGH FGH FGH FGH




ng Eq. (1.6) and the work functions, W, of the metals are given
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV .
FGH FGH FGH FGH F G H F G H FGH FGH FGH FGH F G H FGH F G H




For part (a), Na has W = 2.28 eV :
F G H F G H F G H F G H F G H F G H FGH F G H FGH




(KE)max = 6.20 eV − 2.28 eV = 3.92 eV FGH FGH FGH F GH FGH FGH F GH FGH FGH




Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 e FGH FGH FGH FGH FGH FGH FGH F G H FGH FGH F G H FGH FGH FGH FGH




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




5.This problem again concerns the photoelectric effect. As in Pro
FGH FGH FGH FGH FGH FGH FGH FGH FGH




blem 4, we use Eq. (1.6): FGH FGH FGH FGH FGH




hc − FGH


(KE)max = FGH




Wλ
FGH




FGH




where W is the work function of the material and the term
F G H F G H F G H F G H F G H F G H F G H F G H F G H F G H F G H F G




Hhc/λ describes the energy of the incoming photons. Solving for the
F G H FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




latter:
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
FGH FGH FGH F G H FGH FGH F G H FGH F GH F G H FGH FGH


F G H




Solving Eq. (1.5) for the wavelength: FGH F GH FGH FGH FGH




1240 eV · nm
λ=
FG H F G H FGH



= 387.5 nm FGH



3.2 e
FGH FGH


F G H



V
6. A potential energy of 0.72 eV is needed to stop the flow of electron
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




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




. Solving Eq. (1.6) for the work function:
FGH FGH FGH FGH FGH FGH FGH




hc 1240 eV · — 0.72 eV = 1.98 eV
W= —
FG H F G H FGH




λ nm
FGH FG H F G H FGH FGH
FGH F G H




(KE)max F



= G H




460 nm FGH




7. Reversing the procedure from Problem 6, we start with Eq. (1.6): FGH FGH FGH FG H FGH FGH FGH F G H FGH FGH




hc
− W 1240 eV · — 1.98 eV = 3.19 eV
F G H




(KE)max =
FG H F G H FGH
FGH F G H




nm
FGH FGH FG H F G H FGH FGH




=
λ
240 nm FGH




Hence, a stopping potential of 3.19 eV prohibits the electrons fro
FGH FGH FGH FGH FGH FGH FGH FGH FGH FGH




m reaching the anode.
FGH FGH FGH




8. Just at threshold, the kinetic energy of the electron
F G H F G H F G H F G H F G H F G H F G H F G H F




G H is zero. Setting (KE)max = 0 in Eq. (1.6),
F G H F G H FGH FGH FGH F G H F G H F G H




hc
W= = 1240 eV · = 3.44 eV FG H F G H FGH




λ0
FGH



nm
FGH FGH




360 nm FGH