SOLUTION MANUAL Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by Morrison 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 AQ AQ




1. The Wave-Particle Duality
AQ AQ AQ




2. The Schrödinger Wave Equation
AQ AQ AQ AQ




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
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ



is given by Eq. (1.5). Following Example 1.1 and substitut
AQ AQ AQ AQ AQ AQ AQ A Q AQ AQ



ing λ = 200 eV gives:
AQ AQ AQ AQ AQ




hc 1240 eV · nm
= = 6.2 eV
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Ephoton = λ
AQ AQ

200 nm AQ AQ




2. The energy of the beam each second is:
AQ A Q A Q A Q AQ A Q A Q




power 100 W
=
AQ

= 100 J
Etotal = time
AQ AQ

1 s AQ AQ




The number of photons comes from the total energy divided
AQ AQ AQ AQ AQ AQ AQ AQ AQ



by the energy of each photon (see Problem 1). The photon’s
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ A



Qenergy must be converted to Joules using the constant 1.60
AQ AQ AQ AQ AQ AQ AQ AQ AQ



2 × 10−19 J/eV , see Example 1.5. The result is:
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ




N =Etotal = 100 J = 1.01 × 1020 AQ A Q AQ


photons E
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pho
ton 9.93 × 10−19 AQ AQ




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
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ



mW = 10−3 W . The power may be expressed as: 1 W = 1
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ



J/s. Following Example 1.1, the energy of a single photon
AQ AQ AQ AQ AQ AQ AQ AQ AQ A



is:
Q




1240 eV · nm
hc = 1.960 eV
AQ A Q AQ

AQ AQ AQ

Ephoton = 632.8 nm A Q AQ


=
λ
A Q

A Q



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




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:
AQ AQ AQ AQ AQ AQ




1 × 10−3 J/s 15 photons AQ AQ AQ
A Q


Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
AQ A Q AQ AQ AQ AQ AQ
A Q
AQ AQ AQ

, 2

4.The maximum kinetic energy of photoelectrons is found
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using Eq. (1.6) and the work functions, W, of the metals ar
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ



e given in Table 1.1. Following Problem 1, Ephoton = hc/λ =
AQ AQ AQ AQ AQ AQ A Q A Q AQ AQ AQ A



6.20 eV . For part (a), Na has W = 2.28 eV :
Q A Q AQ A Q A Q A Q A Q A Q A Q A Q AQ A Q AQ




(KE)max = 6.20 eV − 2.28 eV = 3.92 eVAQ AQ AQ AQ AQ AQ AQ AQ AQ




Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ A Q AQ AQ AQ



12 eV AQ



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




5.This problem again concerns the photoelectric effect. As in
AQ AQ AQ AQ AQ AQ AQ AQ AQ



Problem 4, we use Eq. (1.6): AQ AQ AQ AQ AQ




hc − AQ

(KE)max = AQ




Wλ
AQ



AQ




where W is the work function of the material and the term
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hc/λ describes the energy of the incoming photons. Solving for
A Q A Q AQ AQ AQ AQ AQ AQ AQ AQ A



Qthe latter: AQ




hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
AQ AQ AQ AQ AQ AQ A Q AQ AQ A Q AQ AQ

A Q




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




1240 eV · nm
λ=
AQ A Q AQ


= 387.5 nm AQ


3.2 e
AQ AQ

AQ


V
6. A potential energy of 0.72 eV is needed to stop the flow of elec
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ



trons. Hence, (KE)max of the photoelectrons can be no more tha
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ



n 0.72 eV. Solving Eq. (1.6) for the work function:
AQ AQ AQ AQ AQ AQ AQ AQ AQ




hc 1240 eV · — 0.72 eV = 1.98 eV
W = —
AQ A Q AQ



λ
AQ AQ A Q AQ AQ

nm
AQ A Q



(KE)max A


= Q


460 nm AQ




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




hc
− W 1240 eV · — 1.98 eV = 3.19 eV
AQ


(KE)max =
AQ A Q AQ
AQ A



nm
AQ AQ AQ A Q AQ AQ


= Q


λ
240 nm AQ




Hence, a stopping potential of 3.19 eV prohibits the electrons f
AQ AQ AQ AQ AQ AQ AQ AQ AQ AQ



rom reaching the anode.
AQ AQ AQ




8. Just at threshold, the kinetic energy of the electron
A Q A Q A Q A Q A Q A Q A Q A Q A



Q is zero. Setting (KE)max = 0 in Eq. (1.6),
A Q A Q AQ AQ AQ A Q A Q A Q




hc
W = = 1240 eV · = 3.44 eV
AQ
AQ A Q AQ



λ0 nm
AQ AQ




360 nm AQ