SERWAY HOMEWORK SOLUTIONS

ASSIGNMENT 6

4-3 Thomson’s device will work for positive and

negative particles, so we may apply θ

≈2

q V

m B ld.

(a) �� ≈ ���� = � �� .�����

�.���� ��� .� ��� . ��� = 9.58�10� /"#

(b) As the particle is attracted by the negative

plate, it carries a positive charge and is a

proton. −

−

×= = × ×

197

27

1.60 10 C9.58 10 C kg

1.67 10 kgp

q

m

(c) $� = %� = �

�� = � � . ����.���� ��� = 2.19�10'(/)

(d) As ~ 0.01xv c there is no need for relativistic

mechanics.

4-12 �* = + , �

-.� −�-0�1 where = K2, 3, 4, n and

−= × 7 11.097 373 2 10 mR ;

Therefore

2 = +3� 41 − 156�73�

( ) ( )

( )

( ) ( )

λ

λ

λ

−− −

−− −

−− −

= = − = × =

= = − = × =

= = − = × =

11 7

2

11 7

2

11 7

2

1For 2 : 1 1.215 02 10 m 121.502 nm UV

2

1For 3 : 1 1.025 17 10 m 102.517 nm UV

3

1For 4 : 1 1.972 018 10 m 97.201 8 nm UV

4

n R

n R

n R

4-18 (a) ( ) ∆ = − = 2 2

1 113.6 eV 12.1 eV

1 3E

(b) Either ∆ = 12.1 eVE or

( )( )∆ = − =2

1 113.6 eV 10.2 eV

1 2E

and ( ) ∆ = − = 2 2

1 113.6 eV 1.89 eV

2 3E .

4-21 (a) For the Paschen series; λ

= −

2 2

i

1 1 1

3R

n ; the

maximum wavelength corresponds to =i 4n ,

λ

= − 2 2

max

1 1 1

3 4R ; λ =max 1 874.606 nm . For

minimum wavelength, → ∞in , λ

= − ∞2

min

1 1 1

3R ;

λ = =min

9820.140 nm

R.

(b) ( )

λ −= =

×

1 874.606 nm

19min

0.662 7 nm1.6 10 J eV

hchc

( )

λ −= =

×

820.140 nm19

min

1.515 nm1.6 10 J eV

hchc

Or

(b) 8 = �.�9'�� �:;�9�� <��=���� �> = 0.662@A

8 = �.�9'�� �:;�9�� <�=� .���� �> = 1.513@A

4-23 (a) ( )= =21 0.052 9 nm 0.052 9 nmr n (when = 1n )

(b) ( )( )

( )−

−

−

−

=

× ×= × ×

×

= ×

1 22

1 231 9 2 219

11

24

9.1 10 kg 9 10 Nm C1.6 10 C

5.29 10 m

1.99 10 kg m s

e ee

e

e

kem v m

m r

m

M v

For the first equation in (b) use

(C$ = (C 4"@�

(CD7� �E = 4(C�"@�(CD 7

� �E = ,(C"D 1� �E @

Second equation in (b) v is missing on left.

(c) ( )( )− −= = × ×24 111.99 10 kg m s 5.29 10 meL m vr ,

( )−= × = h34 21.05 10 kg m sL

(d) = = 13.6 eVK E

Or from Eq. 4.24

$ = 5ℏ(CD= �1��1.055�1039�G)�

�9.1�1039�"#��0.0529�103H(� = 2.192�10'(/)

I =12($�

I =12 �9.1�1039�"#��2.192�10'(/)��

I = 2.185�103�=G = 13.6@A

(e) = − = −2 27.2 eVU K

Or from Physics 208

J = 14LM

N�N�D

J = 9�10H �1.6�103�H��−1.6�103�H�0.0529�103H

J = �4.355�103�=G� = 27.2@A

(f) = + = −13.6 eVE K U

4-25 (a) ( )

∆ = = −

2 2f i

1 113.6 eVE hf

n n or

( )−

− = = ×

×

1 1149 16

1513.6 eV 1.60 10 Hz

4.14 10 eV sf

(b) π

=2 nrTv so π

= =rev

1

2 n

vf

T r . Using

=

1 22

e n

kev

m r ,

=

1 22

revn

kef

mr. For = 3n , ( )= 2

3 03r a and

( )( )( )( )( )

( )( )( )( )

( )

( )

− −

−

−

× ×

×

× ×=

= × =

= × =

1 231 11

11

29 2 2 19

rev9.11 10 kg 9 5.29 10 m

2 3.14 9 5.29 10 m

14rev

14rev

8.99 10 Nm C 1.60 10 C

2.44 10 Hz 3

1.03 10 Hz 4

f

f n

f n

Thus the photon frequency is about halfway

between the two frequencies of the

revolution.

4-28 Call the energy available from the = 2n to = 1n transition ∆E :

( ) ( )( ) ( ) ∆ − = − = =

2 22 2f i

1 1 1 113.6 eV 13.6 24 5 875 eV 5.875 keV

1 4E Z

n n.

(put in = after Δ8)

The kinetic energy K of the Auger electron is

equal to 5.875 keV minus the energy required to

ionize an electron in the = 4n state, ionizationE .

Thus,

( )= − = − =2

ionization5.875 keV 5.875 keV 13.6 eV 5.385 keV4

ZK E .

(Z=24 and the 4 needs to be squared.)

4-42 λ

= = = 2

1 22 2

2

hcE hf

mv . So each atom gives up

its kinetic energy in emitting a photon.

( )− −

−

×= × = ×

×

= ×

834 18

2 7

6

3 10 m s16.626 10 Js 1.63 10 J

2 1.216 10 m

1.89 10 m s

mv

v