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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$16.49
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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 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 for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3...
SOLUTION MANUAL Modern Physics with Modern
Computational Methods: for Scientists and
Engineers 3rd Edition by Morrison Chapters 1- 15
,Table of contents cd cd
1. The Wave-Particle Duality
cd cd cd
2. The Schrödinger Wave Equation
cd cd cd cd
3. Operators and Waves
cd cd cd
4. The Hydrogen Atom
cd cd cd
5. Many-Electron Atoms
cd cd
6. The Emergence of Masers and Lasers
cd cd cd cd cd cd
7. Diatomic Molecules
cd cd
8. Statistical Physics
cd cd
9. Electronic Structure of Solids
cd cd cd cd
10. Charge Carriers in Semiconductors
cd cd cd cd
11. Semiconductor Lasers
cd cd
12. The Special Theory of Relativity
cd cd cd cd cd
13. The Relativistic Wave Equations and General Relativity
cd cd cd cd cd cd cd
14. Particle Physics
cd cd
15. Nuclear Physics
cd cd
,1
The Wave-Particle Duality - Solutions
c d c d c d c d
1. The energy of photons in terms of the wavelength of light is
cd cd cd cd cd cd cd cd cd cd cd cd
given by Eq. (1.5). Following Example 1.1 and substituting λ
cd cd cd cd cd c d cd cd cd cd
= 200 eV gives:
cd cd cd
hc 1240 eV · nm
= = 6.2 eV
cd c d cd
Ephoton = λ
cd cd
200 nm cd cd
2. The energy of the beam each second is:
cd cd cd cd cd cd cd
power 100 W
= = 100 J
cd
Etotal = time
cd cd
1s cd cd
The number of photons comes from the total energy divided b
cd cd cd cd cd cd cd cd cd cd
y the energy of each photon (see Problem 1). The photon’s ener
cd cd cd cd cd cd cd cd cd cd cd
gy must be converted to Joules using the constant 1.602 × 10−1
cd cd cd cd cd cd cd cd cd cd cd
9 J/eV , see Example 1.5. The result is:
cd cd cd cd cd cd cd cd
N =Etotal = 100 J = 1.01 × 1020 cd c d cd
photons E
cd cd cd
pho
ton 9.93 × 10−19 cd cd
for the number of photons striking the surface each second.
c d c d c d c d c d c d c d c d c d
3. We are given the power of the laser in milliwatts, where 1 mW
cd cd cd cd cd cd cd cd cd cd cd cd cd
= 10−3 W . The power may be expressed as: 1 W = 1 J/s. Follo
cd cd cd cd cd cd cd cd cd cd cd cd cd cd cd
wing Example 1.1, the energy of a single photon is:
cd cd cd cd cd cd cd cd cd
1240 eV · nm
hc = 1.960 eV
cd c d cd
cd cd cd
Ephoton = 632.8 nm cd cd
=
λ
c d
c d
We now convert to SI units (see Example 1.5):
cd cd cd cd cd cd cd cd
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
cd cd cd cd cd cd cd cd cd cd cd
Following the same procedure as Problem 2: cd cd cd cd cd cd
1 × 10−3 J/s 15 photons cd cd cd
c d
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
cd c d cd cd cd cd cd
c d
cd cd cd
, 2
4. The maximum kinetic energy of photoelectrons is found usi
cd cd cd cd cd cd cd cd
ng Eq. (1.6) and the work functions, W, of the metals are give
cd cd cd cd cd cd cd cd cd cd cd cd
n in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV .
cd cd cd cd cd c d c d cd cd cd cd c d cd
For part (a), Na has W = 2.28 eV :
c d c d c d c d c d c d c d cd c d cd
(KE)max = 6.20 eV − 2.28 eV = 3.92 eV cd cd cd cd cd cd cd cd cd
Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 e
cd cd cd cd cd cd cd cd cd cd c d cd cd cd cd
V
and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .
cd cd cd cd cd cd cd cd cd cd cd cd cd cd cd cd cd
5. This problem again concerns the photoelectric effect. As in Pro
cd cd cd cd cd cd cd cd cd
blem 4, we use Eq. (1.6): cd cd cd cd cd
hc − cd
(KE)max = cd
λ
cd
W cd
where W is the work function of the material and the term hc
c d c d c d c d c d c d c d c d c d c d c d c d
/λ describes the energy of the incoming photons. Solving for the la
c d cd cd cd cd cd cd cd cd cd cd
tter:
hc
= (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV
λ
cd cd cd cd cd cd cd cd cd c d cd cd
c d
Solving Eq. (1.5) for the wavelength: cd cd cd cd cd
1240 eV · nm
λ=
cd c d cd
= 387.5 nm cd
3.2 e
cd cd
cd
V
6. A potential energy of 0.72 eV is needed to stop the flow of electron
cd cd cd cd cd cd cd cd cd cd cd cd cd
s. Hence, (KE)max of the photoelectrons can be no more than 0.72 e
cd cd cd cd cd cd cd cd cd cd cd cd
V. Solving Eq. (1.6) for the work function:
cd cd cd cd cd cd cd
hc 1240 eV · n — 0.72 eV = 1.98 eV
W= —
cd c d cd
λ
cd cd c d cd cd
m
cd c d
(KE) max c
= d
460 nm cd
7. Reversing the procedure from Problem 6, we start with Eq. (1.6): cd cd cd cd cd cd cd cd cd cd
hc 1240 eV · n
(KE)max = − W
cd
— 1.98 eV = 3.19 eV
cd c d cd
cd cd
m
cd cd cd c d cd cd
=
λ
240 nm cd
Hence, a stopping potential of 3.19 eV prohibits the electrons fro
cd cd cd cd cd cd cd cd cd cd
m reaching the anode.
cd cd cd
8. Just at threshold, the kinetic energy of the electron is
c d c d c d c d c d c d c d c d c d
c d zero. Setting (KE)max = 0 in Eq. (1.6),
c d cd cd cd c d c d c d
hc 1240 eV · n
W= = = 3.44 eV
cd c d cd
λ0
cd
m
cd cd
360 nm cd
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