Download Krane - Modern Physics 3rd c2012 Solutions ISM PDF

TitleKrane - Modern Physics 3rd c2012 Solutions ISM
File Size1.6 MB
Total Pages326
Table of Contents
                            cover (txtbk)
title & copyright
preface
contents
1. The Failures of Classical Physics
2. The Special Theory of Relativity
3. The Particlelike Properties of Electromagnetic Radiation
4. The Wavelike Properties of Particles
5. The Schrödinger Equation
6. The Rutherford-Bohr Model of the Atom
7. The Hydrogen Atom in Wave Mechanics
8. Many-Electron Atoms
9. Molecular Structure
10. Statistical Physics
11. Solid-State Physics
12. Nuclear Structure and Radioactivity
13. Nuclear Reactions and Applications
14. Elementary Particles
15. Cosmology: The Origin and Fate of the Universe
                        
Document Text Contents
Page 1

Typewritten Text
INSTRUCTOR SOLUTIONS MANUAL

Page 2

Instructor’s Manual


to accompany



Modern Physics, 3rd Edition



Kenneth S. Krane
Department of Physics

Oregon State University













©2012 John Wiley & Sons

Page 163

154

(a) 3,0,0 (b) 3,2,-2 (c) 2,1,+1 (d) 2,2,-1

8. Relative to the z axis, how many possible directions are there in space for the orbital

angular momentum vector that represents an electron in a 4f state (n = 4, l = 3)?
(a) 2 (b) 3 (c) 5 (d) 7 (e) 9

Answers 1. d 2. c 3. d 4. a 5. d 6. d 7. d 8. d


B. Conceptual

1. Excluding cases in which the angular momentum is zero, is the length of the angular

momentum vector that describes an electron in an atom always equal to, either
greater than or equal to, or always greater than the maximum possible z component
of the angular momentum? EXPLAIN YOUR ANSWER.


2. For a certain electronic state in hydrogen, the angular part of the wave function is

( ) sin cosCθ θ θΘ = . Is the electron described by this wave function most likely to be
found close to the z axis, close to the xy plane, or somewhere in between? EXPLAIN
YOUR ANSWER. (θ is the polar angle between r and the z axis.)


3. The angular part of one of the 2p wave functions in atomic hydrogen is A sin θ, where

A is a constant. (θ is the polar angle between the z axis and the line connecting the
electron to the origin.) Does this electron have a greater probability to be found near
the z axis or near the xy plane? EXPLAIN YOUR ANSWER.


Answers 1. always greater than 2. somewhere in between 3. near the xy plane


C. Problems

1. The radial part of the 2p wave function of atomic hydrogen is Cre-r/2a0 where C =

(24a05)-1/2. Consider two very thin spherical shells, each of thickness dr. One shell
has radius 2a0 and the other has radius 4a0. Find the ratio of the probability to find
the electron in the larger shell to the probability to find it in the smaller shell.

(Here “in the shell” means between r and r + dr.)

2. The 2p (l = 1) radial wave function of an electron in atomic hydrogen is

0/ 2

0

( ) r a
r

R r A e
a

−=

where A is a constant.
(a) Find the most probable value of r, that is, the most probable distance between the
electron and the nucleus.

(b) List all possible sets of quantum numbers that can describe an electron in this state

Page 164

155

3. (a) The radial part of the wave function of the n = 2, l = 1 electron in a hydrogen atom is
0/ 2

3/ 2
00

1
( )

3(2 )
r arR r e

aa
−=

Find all maxima and minima of the radial probability density and sketch the radial
probability density as a function of r.
(b) List all possible l values for an electron in hydrogen with n = 2. For each l value,
list all possible values of ml.


4. (a) The radial part of the 2p hydrogen wave function is

0/ 2( ) r aR r Are−=
where A is a constant. Find the most probable distance of the electron from the nucleus.

(b) The complete 2p wave function for a particular value of ml is
0/ 2

1/ 2 3/ 2
0 0

1
( , , ) cos

(4 ) (2 )
r arr e

a a
ψ θ φ θ

π
−=

Describe the angular part of the electron probability density. Consider both the θ and
φ dependences. Include in your discussion a sketch of the angular part of the
probability density.


5. The radial part of the 3d wave function of atomic hydrogen is

0/37 / 2 2
0( )

r aR r Ca r e−−=

where 3(2 / 81) 2 /15 9.016 10C −= = × .
(a) Find the locations of the maxima and minima of the radial probability density.
Sketch the radial probability density as a function of r.
(b) Independent of the angular coordinates, what is the probability to find the electron
in the region between r = a0 and r = 1.01a0?

6. (a) The radial part of the 2p (l = 1) wave function of hydrogen is

0/ 25/ 2
0( )

r aR r Aa re−−=

where A = 1/ 24 = 0.2041. What is the probability to find the electron in the
entire region of the thin spherical shell between 0.99a0 and 1.01a0?

(b) The angular part of this wave function is ( ) sinBθ θΘ = , where B = 12 3 .
Where would you expect the probability to find the electron to be larger: closer to
the z axis or closer to the xy plane? Explain your answer. (θ is the polar angle
between the z axis and a line connecting the electron’s volume element to the
origin.)


Answers 1. 2.16 2. (a) 4a0 (b) (2,1,±1,±1/2), (2,1,0,±1/2)
3. (a) min: 0,∞; max: 4a0 (b) l = 0 (ml = 0), l = 1 (ml = 0,±1)
4. (a) 4a0 (b) Independent of φ; max for θ = 0 and π, zero for θ = π/2
5. (a) minima at 0,∞; maximum at 9a0 (b) 4.17 × 10-7
6. (a) 3.07 × 10-4 (b) near xy plane

Page 325

316

(e) If the radiation were distributed uniformly over a spherical surface at the Earth’s
distance d = 104 light years = 9.46 × 1019 m, the fraction of the power received by the
antenna of area A would be



2

25 14
received transmitted 2 19 2

10 m
(6.98 10 W) 6.21 10 W

4 4 (9.46 10 m)
A

P P
dπ π

−= = × = ×
×




26. (a) With i j kt h c G∝ , we have [ ] [ ] [ ] [ ]i j kt h c G= where [ ] indicates dimensions.

Inserting the appropriate units, we have




1 2 2 2 1 1 3 2 1

2 3 2

s (J s) (m s ) (N m kg ) (kg m s ) (m s ) (m s kg )

(kg) (m) (s)

i j k i j k

i k i j k i j k

− − − − − −

− + + − − −

= ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅

=



Equating powers of the units on both sides of the equation, we obtain

0 2 3 0 2 1i k i j k i j k− = + + = − − − =

which can be solved to give 1/ 2, 5 / 2, 1/ 2.i j k= = − =


(b)
34 11 2 2

43
5 8 5

(6.626 10 J s)(6.67 10 N m /kg )
1.3 10 s

(3.00 10 m/s)
hG

t
c

− −
−× ⋅ × ⋅= = = ×

×



(c) 8 43 35(3.00 10 m/s)(1.3 10 s) 3.9 10 mR ct − −= = × × = ×


27. By taking differentials of Equations 2.23a and 2.23d, we obtain



2

2 2 2 2

( / )
and

1 / 1 /
dx u dt dt u c dx

dx dt
u c u c
− −

′ ′= =
− −




Making the substitutions in the expression for 2( )ds′ ,




2 2

2 2 2

2 2 2 2

2 2 2 2 2 2 2 2

2 2

2 2 2 2 2 2 2
2 2 2 2

2 2

( / )
( ) ( ) ( )

1 / 1 /

( ) 2 ( / )( ) ( ) 2 ( )
1 /

(1 / )( ) (1 / )( )
( ) ( ) ( )

1 /

c dt u c dx dx u dt
ds c dt dx

u c u c

c dt u dx dt u c dx dx u dx dt u dt
u c

c u c dt u c dx
c dt dx ds

u c

⎡ ⎤ ⎡ ⎤− −
′ ′ ′= − = −⎢ ⎥ ⎢ ⎥

− −⎣ ⎦ ⎣ ⎦

− + − + −
=



− − −
= = − =



Page 326

317

28. (a) For small angles, the distance between I and the horizontal axis is Sdθ , the distance
between S and the axis is Sdβ , and the distance between I and S is ( )S Ld dα − . Thus
( )S S S Ld d d dθ β α= + −

From Equation 15.22, after inserting the extra factor of 2 to account for the difference
between special and general relativity for the deflection angle, we have (again for small
angles)

2 2
4 4

L

GM GM
bc d c

α
θ

= =

so

2
4

( )S S S L
L

GM
d d d d

d c
θ β

θ
= + −

(b) We can rewrite this equation as


2

2

4
( ) ES L

L S

GM
d d

d d c
θ

θ β β
θ θ

= + − = +

or
2 2 0Eθ βθ θ− − =

with 24 ( ) /E S L S LGM d d c d dθ = − . Using the quadratic formula then gives the solutions


( )2 212 4 Eθ β β θ± = ± +

and the difference between the angular positions of the two images is

2 24 Eθ θ θ β θ+ −Δ = − = +

(c) For β = 0, we have Eθ θ= and the problem has rotational symmetry about the
horizontal axis. The image of the star is thus a circle.

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