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 withModern
n n n n n




n Computational Methods: for Scientists and Engineers
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n 3rd Edition by Morrison Chapters 1- 15
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,Table of contents n n




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
n n n n




1. Theenergyof photonsinterms of the wavelength of lightis
n n n n n n n n n n n




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




200 eV gives:
n n n




hc 1240 eV · nm
= =6.2eV
n n n




Ephoton = λ 200 nm
n n




n
n




2. The energy of the beam each second is:
n n n n n n n




power 100 W
= =100J
n




Etotal = time 1s
n n




n
n




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




energy of each photon (see Problem 1). The photon’s energy must
n n n n n n n n n n n




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




Example 1.5. The result is:
n n n n n




N =Etotal = 100J =1.01×1020 n n n




photons E
n n n




pho
ton 9.93×10−19 n n




for the number of photons striking the surface each second.
n n n n n n n n n




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




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




Example 1.1, the energy of a single photon is:
n n n n n n n n n




1240 eV · nm
hc =1.960 eV
n n n




Ephoton = 632.8 nm
n n




n
n n




=
λ
n




n




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




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




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




1×10−3J/s 15 photons n n n




Rate of emission = = 3.19× 10
n




3.14×10−19 J/photon s
n n n n n n n


n


n n n

, 2

4. The maximum kinetic energy of photoelectrons is found using
n n n n n n n n




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




n Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV . For part
n n n n n n n n n n n n n




n (a), Na has W = 2.28 eV :
n n n n n n n




(KE)max=6.20eV −2.28 eV =3.92 eV n n n n n n n n n




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




andfor Ag metal in part (c), W =4.73 eV, giving (KE)max=1.47 eV.
n n n n n n n n n n n n n n n n n




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




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




hc−
(KE)max =
n




Wλ
n




n n




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




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




hc
= (KE)max+W = 2.3 eV +0.9 eV = 3.2 eV
λ
n n n n n n n n n n n n




n




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




1240 eV · nm
λ=
n n n




=387.5 nm
3.2
n

n n




eV n




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




Hence,
n




(KE)maxofthephotoelectronscanbenomorethan0.72eV.SolvingE
n n n n n n n n n n n n n




q.(1.6) for the work function:
n n n n n




hc 1240 eV · —0.72 eV = 1.98 eV
W= (KE)max
n n




λ nm
n n n n n
n n




=
n




n




460 nm n




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




hc 1240 eV ·
(KE)max = − W
n




—1.98 eV = 3.19 eV
n n
n




nm
n n n n n n




= n
n




λ
240 nm n




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




reaching the anode.
n n n




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




n zero. Setting (KE)max = 0 in Eq. (1.6),
n n n n n n n




hc
W= = 1240 eV · =3.44 eV n n




λ0
n




nm
n n




n




360 nm n




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