Modern Physics with Modern Computational Methods for Scientists and Engineers – 3rd Edition by Morrison | Solution Manual Chapters 1–15

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

,Table of contents go go




1. The Wave-Particle Duality
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2. The Schrödinger Wave Equation
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3. Operators and Waves
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4. The Hydrogen Atom
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5. Many-Electron Atoms
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6. The Emergence of Masers and Lasers
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7. Diatomic Molecules
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8. Statistical Physics
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9. Electronic Structure of Solids
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10. Charge Carriers in Semiconductors
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11. Semiconductor Lasers
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12. The Special Theory of Relativity
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13. The Relativistic Wave Equations and General Relativity
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14. Particle Physics
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15. Nuclear Physics
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,1

The Wave-Particle Duality - Solutions
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1. The energy of photons in terms of the wavelength of light is gi
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ven by Eq. (1.5). Following Example 1.1 and substituting λ = 2
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00 eV gives:
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hc 1240 eV · nm
= = 6.2 eV
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Ephoton = λ
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200 nm go go




2. The energy of the beam each second is:
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power 100 W
= = 100 J
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Etotal = time
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1s go go




The number of photons comes from the total energy divided by t
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he energy of each photon (see Problem 1). The photon’s energy
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must be converted to Joules using the constant 1.602 × 10−19 J/e
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V , see Example 1.5. The result is:
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N =Etotal = 100 J = 1.01 × 1020 go g o go




photons E
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pho
ton 9.93 × 10−19 go go




for the number of photons striking the surface each second.
g o g o g o g o g o g o g o g o g o




3.We are given the power of the laser in milliwatts, where 1 mW =
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10−3 W . The power may be expressed as: 1 W = 1 J/s. Followin
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g Example 1.1, the energy of a single photon is:
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1240 eV · nm
hc = 1.960 eV
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Ephoton = 632.8 nm
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g go
g o



=
λ
o


g o




We now convert to SI units (see Example 1.5):
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1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
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Following the same procedure as Problem 2: go go go go go go




1 × 10−3 J/s 15 photons go go go
g o



Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
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g o
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, 2

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




Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 eV
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and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .
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5.This problem again concerns the photoelectric effect. As in Prob
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lem 4, we use Eq. (1.6):
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hc − go



(KE)max = go




Wλ
go




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where W is the work function of the material and the term hc/
g o g o g o g o g o g o g o g o g o g o g o g o




λ describes the energy of the incoming photons. Solving for the latte
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r:
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
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g o




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




1240 eV · nm
λ=
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= 387.5 nm go



3.2 e
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g o



V
6. A potential energy of 0.72 eV is needed to stop the flow of electrons.
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Hence, (KE)max of the photoelectrons can be no more than 0.72 eV.
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Solving Eq. (1.6) for the work function: go go go go go go




hc 1240 eV · n — 0.72 eV = 1.98 eV
W= —
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λ m
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go g o




(KE)max g



= o




460 nm go




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




hc 1240 eV · n
(KE)max = − W
g o




— 1.98 eV = 3.19 eV
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m
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=
λ
240 nm go




Hence, a stopping potential of 3.19 eV prohibits the electrons from
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reaching the anode. go go




8. Just at threshold, the kinetic energy of the electron is
g o g o g o g o g o g o g o g o g o g o




zero. Setting (KE)max = 0 in Eq. (1.6),
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hc 1240 eV · n
W= = = 3.44 eV
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λ0
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m
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360 nm go




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