##### 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.

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.