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Exam (elaborations)

Quantum Physics 3rd Edition by Stephen Gasiorowicz – Chapters 1–19 Solution Manual | Complete Worked Solutions

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This document provides detailed, step-by-step solutions for Chapters 1 through 19 of Quantum Physics (3rd Edition) by Stephen Gasiorowicz. It covers fundamental quantum mechanics concepts, mathematical derivations, problem-solving techniques, and applied examples used in undergraduate and graduate physics courses. The material is ideal for exam preparation, homework practice, and deepening understanding of quantum physics principles.

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Solution Manual For
Solution manual to Quantum Physics 3rd Edition by Stephen Gasiorowicz
Chapter 1-19business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law
exams include:Essay




CHAPTER 1
1. The energy contained in a volume dV is

U(,T )dV  U (,T )r 2dr sindd

when the geometry is that shown in the figure. The energy from this source that emerges
through a hole of area dA is

dAcos
dE(,T )  U (,T )dV
4r 2

The total energy emitted is

 /2 dA
dE(,T )   dr d  dU (,T ) sin cos
c t 2


0 0 0 4
dA  /2
 2ctU(,T ) d sin cos
. 4 0


1
 ctdAU (,T )
4




By definition of the emissivity, this is equal to EtdA . Hence

c
E(,T )  U (,T )
4

2. We have

c c 8hc 1
w(,T )  U (,T ) | d / d | 
U( ) 2
  5 ehc/kT  1

This density will be maximal when dw(,T ) / d  0. What we need is

, d 1 1  1 1 eA / A 1
 (5  ( )) 0
d 5 eA /  1 6 5 eA/  1 2 eA /  1

Where A  hc / kT . The above implies that with x  A / , we must have

5  x  5exbusiness exams, often requiring students to demonstrate knowledge of legal statutes, case
law, and their ability to apply legal reasoning. Common types of law exams include:Essay

A solution of this is x = 4.965 so that




3. The relationship is

h  K  W

where K is the electron kinetic energy and W is the work function. Here
hc (6.626  1034 J .s)(3 108 m / s)  19
h  5.68  10 J  3.55eV
  350  109 m

With K = 1.60 eV, we get W = 1.95 eV

4. We use
hc hc
 K K
  1 2
1 2


since W cancels. From ;this we get

1 12
h (K  K ) 
2
c 2  1 1

9 9
 (200  10 m)(258  10 m)  (2.3 0.9)eV  (1.60  10 19 )J / eV
(3 108 m / s)(58  109 m)

,  6.64  1034 J .s
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay

, 5. The maximum energy loss for the photon occurs in a head-on collision, with the
photon scattered backwards. Let the incident photon energy be h , and the backward-
scattered photon energy be h' . Let the energy of the recoiling proton be E. Then its
recoil momentum is obtained from E  p2c 2  m 2c 4 . The energy conservation
equation readsbusiness exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning.
Common types of law exams include:Essay



h  mc2  h' E

and the momentum conservation equation reads

h h'
 p
c c




6. Let h be the incident photon energy, h' the final photon energy and p the outgoing
electron momentum. Energy conservation reads

h  mc2  h' p2c2  m2c4

We write the equation for momentum conservation, assuming that the initial photon
moves in the x –direction and the final photon in the y-direction. When multiplied by c it
read

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