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

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

,Table of contents TV TV




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
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given by Eq. (1.5). Following Example 1.1 and substituting λ
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= 200 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 TV TV




2. The energy of the beam each second is:
T V T V T V T V T V T V T V




power 100 W
= = 100 J
T V




Etotal = time
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1s TV TV




The number of photons comes from the total energy divided b
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y the energy of each photon (see Problem 1). The photon’s ener
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gy must be converted to Joules using the constant 1.602 × 10−1
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9 J/eV , see Example 1.5. The result is:
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N =Etotal = 100 J = 1.01 × 1020 TV T V TV




photons E
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pho
ton 9.93 × 10−19 TV TV




for the number of photons striking the surface each second.
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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. Follo
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wing Example 1.1, the energy of a single photon is:
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1240 eV · nm
hc = 1.960 eV
T V T V TV




Ephoton = 632.8 nm
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T V T V




=
λ
T V


T V




We now convert to SI units (see Example 1.5):
T V T V T V T V T V T V T V T V




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: TV TV TV TV TV TV




1 × 10−3 J/s 15 photons TV TV TV
T V



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

4. The maximum kinetic energy of photoelectrons is found usi
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ng Eq. (1.6) and the work functions, W, of the metals are give
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n in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV .
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For part (a), Na has W = 2.28 eV :
T V T V T V T V T V T V T V TV T V TV




(KE)max = 6.20 eV − 2.28 eV = 3.92 eV TV TV TV TV TV TV T V TV TV




Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 e
T V TV TV TV TV TV T V T V TV TV T V TV TV TV TV




V
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 Pro
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blem 4, we use Eq. (1.6): TV TV TV TV TV




hc − TV



(KE)max = TV




λ
T V




W TV




where W is the work function of the material and the term hc
T V T V T V T V T V T V T V T V T V T V T V T V




/λ describes the energy of the incoming photons. Solving for the la
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tter:
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
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T V




Solving Eq. (1.5) for the wavelength: TV T V T V TV T V




1240 eV · nm
λ=
T V T V TV




= 387.5 nm TV



3.2 e
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T V



V
6. A potential energy of 0.72 eV is needed to stop the flow of electron
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s. Hence, (KE)max of the photoelectrons can be no more than 0.72 e
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V. Solving Eq. (1.6) for the work function:
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hc 1240 eV · n — 0.72 eV = 1.98 eV
W= —
T V T V TV




λ m
TV T V T V TV T V
TV T V




(KE) max T



= V




460 nm TV




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




hc 1240 eV · n
(KE)max = − W
T V




— 1.98 eV = 3.19 eV
T V T V TV
TV T V




m
T V TV T V T V TV T V




=
λ
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Hence, a stopping potential of 3.19 eV prohibits the electrons fro
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m reaching the anode.
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8. Just at threshold, the kinetic energy of the electron is
T V T V T V T V T V T V T V T V T V




T V zero. Setting (KE)max = 0 in Eq. (1.6),
T V TV TV TV T V T V T V




hc 1240 eV · n
W= = = 3.44 eV
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λ0
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m
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360 nm TV