P41.5
(a)
We can draw a diagram that parallels our treatment of
standing mechanical waves. In each state, we measure the
distance d from one node to another (N to N), and base our
solution upon that:
h

Since
dN to N  and  
p
2
h h
p 
 2d
In state 1,
2
34
p2
h2
1   6.626  10 J s 


K


2me 8med2 d2  8  9.11 1031 kg  


38
2
19
6.02  10 J m
3.77  10 eV  m 2
K
K
2
d
d2
K1  37.7 eV
d  1.00 1010 m
In state 2,
d  5.00 1011 m
Next,
Evaluating,
In state 3,
In state 4,
(b)
11
d  3.33 10
11
d  2.50 10
K2  151 eV
m
K3  339 eV
m
K4  603 eV
When the electron falls from state 2 to state 1, it puts out energy
hc
E  151 eV  37.7 eV  113 eV  hf 

into emitting a photon of wavelength
34
8
hc  6.626  10 J s 3.00  10 m s
 
 11.0 nm
E
113 eV  1.60  1019 J eV 
The wavelengths of the other spectral lines we find similarly:
Transition
E eV 
  nm 
P41.31
T  e2CL where C 
2CL 
(a)
(b)
43
264
4 2
452
41
565
32
188
31
302
21
113
4.71
2.75
2.20
6.60
4.12
11.0
2m U  E
2 2  9.11 1031  8.00  1019 
1.055  1034
 2.00  10   4.58
10
T  e4.58  0.010 3 , a 1% chance of transmission.
R  1  T  0.990 , a 99% chance of reflection.
P42.13
The reduced mass of positronium is less than hydrogen, so the photon energy will be
less for positronium than for hydrogen. This means that the wavelength of the emitted
photon will be longer than 656.3 nm. On the other hand, helium has about the same
reduced mass but more charge than hydrogen, so its transition energy will be larger,
corresponding to a wavelength shorter than 656.3 nm.
All the factors in the given equation are constant for this problem except for the
reduced mass and the nuclear charge. Therefore, the wavelength corresponding to the
energy difference for the transition can be found simply from the ratio of mass and
charge variables.
mpme
For hydrogen,

Its wavelength is
  656.3 nm
(a)
mp  me
For positronium,

 me
The photon energy is
E  E3  E2
where

c hc

f E
meme
m
 e
me  me
2
so the energy of each level is one half as large as in hydrogen, which we could
call “protonium.” The photon energy is inversely proportional to its
wavelength , so for positronium,
32  2 656.3 nm   1.31 m (in the infrared region)
(b)
  me , q1  e , and q2  2e
For He ,
so the transition energy is 22  4 times larger than hydrogen
32  
656 
 nm  164 nm (in the ultraviolet region)
 4 
Then,
P42.41
P42.47
3
 42  12 13.6 eV   1.71 104 eV  2.74 1015 J
4
f  4.14  1018 Hz
Following Example 42.5
E 
and
  0.072 5 nm
E  P t   1.00  106 W  1.00  108 s  0.010 0 J
E  hf 
hc
 6.626  10  3.00  10 

34
8
J  2.86  1019 J

694.3  10
0.010 0
E
N

 3.49  1016 photons
E 2.86  1019
9
P44.21
(a)
R  R0 e  t ,
From
 R0   1   10.0 
2
1
5 1
 ln 
  5.58  10 h  1.55  10 s

 R   4.00 h   8.00 
ln 2

 12.4 h
1
t
  ln 
T1 2
P44.28
R0
10.0  103 Ci  3.70  1010 s 
13

  2.39  10 atoms
1.55  105 s 
1 Ci

(b)
N0 
(c)
R  R0e t  10.0 mCi  exp  5.58 102  30.0  1.88 mCi
(a)
A gamma ray has zero charge and it contains no protons or neutrons. So for a
gamma ray Z  0 and A  0 . Keeping the total values of Z and A for the
system conserved then requires Z  28 and A  65 for X. With this atomic
number it must be nickel, and the nucleus must be in an exited state, so it

is 65
28 N i .
(b)
  42 H e has
so for X
for Pb
(c)



Z  2 and
A4
Z  84  2  82
A  215  4  211 , X 
we require
and
211
82
Pb
A positron e  01 e has charge the same as a nucleus with Z  1 . A neutrino 00
has no charge. Neither contains any protons or neutrons. So X must have by
55
Co .
conservation Z  26  1  27 . It is Co. And A  55  0  55 . It is 27
Similar reasoning about balancing the sums of Z and A across the reaction
reveals:
(d)
(e)
P44.29
0
1
e
H (or p). Note that this process is a nuclear reaction, rather than radioactive
decay. We can solve it from the same principles, which are fundamentally
conservation of charge and conservation of baryon number.
1
1
Q   M U-238  M Th-234  M He-4   931.5 MeV u 
Q   238.050 783  234.043 596  4.002 603 u  931.5 MeV u   4.27 MeV
P44.36
(a)
Let N be the number of
238
U nuclei and N  be
206
Pb nuclei.
Then N  N 0 e  t and N 0  N  N  so N   N  N  e t or e t  1 
N 
 where
N 
Taking logarithms,
 t  ln  1 
Thus,
 T1 2  
N
t 
 ln  1 

ln
2
N 

 
If
N
 1.164 for the
N

238

N
.
N
ln 2
T1 2
U  206 Pb chain with T1 2  4.47  109 yr , the age is:
 4.47  109 yr  
1 
9
t 
 ln  1 
  4.00  10 yr
ln
2
1.164




(b)
From above, e t  1 
N
N
N
e  t
. Solving for
gives
.

N
N
N  1  e  t
With t  4.00  109 yr and T1 2  7.04  108 yr for the
235
U  207 Pb chain,
 ln 2 
 ln 2  4.00  109 yr 
 3.938 and
t 
 T1 2 
7.04  108 yr


t 
N
 0.019 9
N
With t  4.00  109 yr and T1 2  1.41 1010 yr for the
t 
P44.39
(a)
(b)
For X,
A  24  1  4  21
and
Z  12  0  2  10 ,
1.41 1010 yr
Th  208 Pb chain,
 0.196 6 and
so X is
21
10
so X is
144
54
N
 4.60
N
Ne
A  235  1  90  2  144
and
(c)
 ln 2  4.00  109 yr 
232
Z  92  0  38  0  54 ,
Xe
A  2 2  0
and
Z  2  1  1 ,
As it is ejected, so is a neutrino:
so X must be a positron.
X  01e and X  00