4
The Particle Nature of Matter
4-1
F corresponds to the charge passed to deposit one mole of monovalent element at a cathode.
96 500 C
19
C.
As one mole contains Avogadro’s number of atoms, e 
23  1.60  10
6.02  10
4-2
(a)
Total charge passed  i *t  1.00 A3 600 s   3 600 C . This is
3 600 C
1.60  10
19
C
 2.25  1022 electrons.
As the valence of the copper ion is two, two electrons are required to deposit each ion
as a neutral atom on the cathode.
The number of Cu atoms 
number of electrons
2
(b)
So the weight (mass) of a Cu atom is:
(c)
m q
molar weight
96 500
 1.125 10
1.185 g
1.125  10
22
atoms
22
Cu atoms.
 1.05  10 22 g .
2  or
2
2
 63.53 g .
molar weight  m 96 500  1.185 g 96 500 C 
q
3 600 C
4-3
Thomson’s device will work for positive and negative particles, so we may apply
(a)
q V
0.20 radians

 2 000 V
m B2 ld
4.57  10 2 T


2
0.10 m 0.02 m   9.58  10 7
V
q
 2 .
m B ld
C kg
(b)
As the particle is attracted by the negative plate, it carries a positive charge and is a
19
 q

C
1.60  10
7
proton. 

 9.58  10 C kg 
27
kg
m p 1.67  10

(c)
vx 


2 000 V
E V


4.57  10 2 T  2.19  106 m s
B dB 0.02 m
(d)
4-6
As v x ~ 0.01c there is no need for relativistic mechanics.
The velocity of fall v 
(a)
y 0.004 m
4

 2.52  10 m s .
t
15.9 s
The radius, a, is given by
12
9 v 
a  

2  g




12
9 1.81  10 5 kg m s 2.52  10 4 m s 

 
3
2


2 800 kg m 9.81 m s




 1.62  10 6 m .
The mass, m, is given by
4  3
3
6
m  V     a  1.33 800 kg m   1.62  10 m
3 

(b)
q1 
mg
E
v  v 1
v
 

3
 1.42  10
14
kg .
where v is the velocity of fall and v  is the velocity of rise. The
electric field is given by E 
4 000 V
V

 200 000 V m , v  2.52  10 4 m s ,
d 0.020 0 m
0.004 m
4
4
 1.11  10 4 m s , v 2  2.31  10 m s , v 3  1.67  10 m s ,
36.0 s
4
4
v 4  3.51 10 m s , v 5  5.31 10 m s
v1 


19
C 2.52  1.11
mg v  v1 6.97  10
 10.0  10 19 C
q1   
 
2.52
 E  v 
mg v  v 2
q2   
 13.4  10 19 C
 E  v 

(c)
q3  11.6  10
19
q4  16.7  10
19
q5  21.6  10
19
C
C
C
By inspection we choose integers that will yield an elementary charge between 1.5
and 2.0  10 19 C
q1
6
q2
8
q3
7
q4
10
q5
13
 1.67  10
19
C
 1.68  10 19 C
 1.66  10 19 C
 1.67  10
19
C
 1.66  10 19 C
The amount of charge gained or lost:
q2  q1  3.4  10 19 C
q3  q1  1.6  10 19 C
q2  q3  1.8  10 19 C
q4  q1  6.7  10 19 C
q4  q3  5.1  10
19
q4  q2  3.4  10
19
q5  q1  11.6  10
q5  q4  4.9  10
C
19
19
q5  q3  10.0  10
C
C
C
19
C
By inspection we again find integers that yield a value of e between 1.5 and
19
2.0  10
C
q2  q1
2
q3  q1
1
q2  q3
1
q4  q1
4
q4  q3
3
q4  q2
2
q5  q1
7
q5  q4
3
q5  q3
6
 1.70  10
19
C
 1.60  10 19 C
 1.80  10 19 C
 1.68  10
19
C
 1.70  10 19 C
 1.65 10 19 C
 1.66  10 19 C
 1.63  10
19
C
 1.67  10 19 C
Average of all values  1.67  10 19 C .
4
4-8
(a)
sin  
From Equation 4.16 we have  n

 2 
or  n2   n1
  . Thus the number
 
sin  1 4
2
sin  4
2
2
of ’s scattered at 40 degrees is given by
4
4
sin 20

sin 10 
2 
 n2  100 cpm 


100
cpm

 6.64 cpm .


4
sin 20 
sin 402 
Similarly
 n at 60 degrees  1.45 cpm
 n at 80 degrees  0.533 cpm
 n at 100 degrees  0.264 cpm
(b)
(c)
1 
2
From 4.16 doubling  m v reduces  n by a factor of 4. Thus  n at
2 
1 
20 degrees   100 cpm   25 cpm .
4 
From 4.16 we find
 nCu
Z 2 N Cu
 Cu
, Z Cu  29 , Z Au  79 .
 nAu
Z 2Au N Au
N Cu  number of Cu nuclei per unit area
 number of Cu nuclei per unit volume
N Au
23


nuclei 
3 6.02  10
22
  8.9 g cm 
t  8.43  10 t
63.54 g






23

6.02  10 nuclei 
22
  19.3 g cm 3 
t  5.90  10 t
197.0
g






So  nCu   n Au 29




2
4-9
*foil thickness
8.43  10
79 2
22

5.90  10
2

2
29  8.43 
 100  
  19.3 cpm .
79   5.90 
The initial energy of the system of  plus copper nucleus is 13.9 MeV and is just the kinetic
energy of the  when the  is far from the nucleus. The final energy of the system may be
evaluated at the point of closest approach when the kinetic energy is zero and the potential
Ze
energy is k2e
where r is approximately equal to the nuclear radius of copper. Invoking
r
2Ze 2
conservation of energy Ei  E f , K  k 
or
r




229 1.60  10 19 8.99  109
2Ze 2

 6.00  10 15 m .
r  k 
6
19
K
13.9  10 eV 1.60  10
J eV

4-11
2

 1
1 
 R 2  2  . For the Balmer series, nf  2 ; n i  3, 4, 5,  . The first three lines in the

n f n i 
1
series have wavelengths given by
1

 1
1 
7
1
 R 2  2  where R  1.097 37  10 m .
2

n
1st line:
1 1   5 
36
 R     R;  
 656.112 nm

4 9  36 
5R
1
1 1   3 
16
 R     R;  
 486.009 nm
4 16  16 
3R
100
1
1 1   21 
 R    
 433.937 nm
3rd line:
R;  
21 R
4 25  100 

2nd line:
4-12
1

1  1
7
1
 R 2  where n  2, 3 , 4,  and R  1.097 373 2  10 m ;
 n 

1
1 
1
1 
1
1 
1
1 
For n  2 :   R 1  2 
 2 
1 
For n  3 :   R 1  2 
 3 
1 
For n  4 :   R 1  2 
 4 
4-13
(a)
(b)
4-20
4-27
  102.6 nm ;
 1.215 02  10 7 m  121.502 nm UV 
 1.025 17  10 7 m  102.517 nm UV 
 1.972 018  10 7 m  97.201 8 nm UV 

1 
R
 R1  2   n 



n
R  1
1


1 2

R 
R
1
102.610 9 m

12
 2.99  3
This wavelength cannot belong to either series. Both the Paschen and Brackett series
lie in the IR region, whereas the wavelength of 102.6 nm lies in the UV region.
 1
1 
E  13.6 eV  2  2 
n f n i 
(a)
1 
 1
E  13.6 eV     0.306 eV
25 16 
(b)
 1
1 
E  13.6 eV     0.166 eV
36 25 
(a)

C2 n
2
n 2
2
2
where R 
(b)
R
so
 1  n 2  2 2  1 1  22 
1 
1 22  1
or
  



 2  2 

2 
 C2  n 2
 
C 2 

n 
 n 
C 2 2
1
22
.
C2
22
36 545.6  10
8
cm
 109 720 cm 1
1 
 1
R 2  2 
2
n 