1. S
SO
OL
LV
VE
ED
D O
OB
BJ
JE
EC
CT
TI
IV
VE
E P
PR
RO
OB
BL
LE
EM
MS
S
Problem 1. The ratio of minimum to maximum wavelengths of radiation that an excited electron in a
hydrogen atom can emit while going to the ground state is
(a) 1/2 (b) Zero
(c) 3/4 (d) 27/32.
Ans. (c)
Solution: Energy of radiation that corresponds to the energy difference between two energy levels 1
n and
2
n is given as
2 2
1 2
1 1
13.6
E eV
n n
 
 
 
 
E is minimum when 1 1
n  & 2 2
n 
 min
1 1 3
13.6 13.6
1 4 4
E eV eV
 
   
 
 
E is maximum when 1 1
n  & 2
n   (the atom is ionized, that is known as ionization energy)
 max
1
13.6 1 13.6
E eV
 
  
 

 
.
 min
max
3
4
E
E
  max
min
/ 3
/ 4
hc
hc



 min
max
3
4



Problem 2. The wavelength of α
K X-ray produced by an X-ray tube is 0.76 Å. The atomic number of
anticathode material is
(a) 82 (b) 41
(c) 20 (d) 10.
Ans. (b)
Solution: For K X-ray line,
2 2
2 2
1 1 1 1
( 1) ( 1) 1
4
1 2
R Z R Z

   
     
   
    
 2
1 3
( 1)
4
R Z

 

…(i)
With reference to given data,
0.76

  Å = 0.76 10
10
 m
7
1.097 10
R   m
Putting these values in equation (i)
2
10 7
4 1
( 1)
3 0.76 10 1.097 10
Z 
 
  
1600

 1 40
Z  
 41
Z  .
Problem 3. If the stationary proton and α - particle are accelerated through same potential difference, the
ratio of de Broglie’s wavelength will be
(a) 2 (b) 1
(c) 2 2 (d) none of these.

2. Ans. (c)
Solution: The gain in K.E. of a charged particle after moving through a potential difference of V is given by
eV, that is also equal to 2
1
2
mv where v is the velocity of the charged particle. Disregarding the
relativistic effect,
2
1
2
mv qV
 
2qV
v
m

 2
mv mqV

 de Broglie wavelength
2
h h
mv mqV
   
 p
p p p
m q V
m q V
  




Putting
(4)(2)
, 2 2
(1)(1)
p
p
V V



  

.
Problem 4. There are two radioactive substances A and B . Decay constant of B is two times that of A .
Initially, both have equal number of nuclei. After n half lives of A , rate of disintegration of
both are equal. The value of n is
(a) 4 (b) 2
(c) 1 (d) 5.
Ans. (c)
Solution: Let A
    2
B
  
If 0
N is total no. of atoms in A and B at 0
t  , then initial rate of disintegration of 0
A N
  , and
initial rate of disintegration of 0
2
B N
 
As 2
B a
  

1
2
B A
T T

i.e. half life of B is half the half life of A .
After one half life of A
0
2
A
N
dN
dt

 
 
 
 
Equivalently, after two half lives of B
0 0
2
4 2
B
N N
dN
dt
 
 
  
 
 
Clearly,
A B
dN dN
dt dt
   
  
   
   
,
after 1
n  , i.e., one half life of A .
Problem 5. A hydrogen atom is in an excited state of principal quantum number ( n) it emits a photon of
wavelength ( λ ), when it returns to the ground state. The value of n is
(a)
λR
λR - 1
(b)
(λR - 1)
λR
(c) λ(R - 1) (d)
λR
λR - 1
.
Ans. (d)

3. Solution: As 2 2
1 2
1 1 1
R
n n
 
 
 
  
 2 2
1 1 1
1
R
n
 
 
 
  
Multiply both sides by 
2
1
1 1
R
n
 
  
 
 
2
1 1
1
R n
 

or 2
1 1 1
1
R
R R
n
 
  
 
1
R
n
R


 
.
Problem 6. The weight based ratio of 238
U and 226
Pb in a sample of rock is 4 : 3 . If the half life of 238
U is
9
4.5× 10 years, then the age of rock is
(a) 9
9.0×10 years (b) 9
6.3×10 years
(c) 9
4.5× 10 years (d) 9
3.78× 10 years.
Ans. (d)
Solution: Let initial no. of U-atoms = 0
N
After time t , (age of rock), let no. of atoms remaining undecayed N
 .

0
238 4
26( ) 3
N
N N


 0
1.79
N
N

0
log /
log 2
T N N
t 
9
4.5 10 log1.79
0.301
 

= 9
3.78 10
 years.
Problem 7. Binding energy per nucleon vs mass number curve for
nuclei is shown in the figure. W, X, Y and Z are four
nuclei indicated on the curve. The process that would
release energy is
(a) Y  2Z (b) W  X + Z
(c) W  2Y (d) X  Y + Z.
Binding
Energy/Nucleon
(in
eV)
Mass number of nuclei
5.0
7.5
8.0
8.5
Z
Y
X
W
0 30 60 90 120
Ans. (c)
Solution: If it mass A
2
Y Z

Reactant: 60 8.5 510
R    MeV
Product: 2 30 5 300
P     MeV
210
E
   MeV
ENDOTHERMIC
If it was B

4. W X Z
 
120 7.5 900
R    MeV
90 8 30 5 870
P      MeV
30
E
   MeV
ENDOTHERMIC
If it was C
2
W Y

120 7.5 900
R    MeV
2 60 8.5 1020
P     MeV
120
E
  Mev
EXOTHERMIC
If it was D
X Y Z
 
90 8.0 720
R    MeV
60 8.5 30 5.0 660
P      MeV
60
E
   MeV
ENDOTHERMIC
Problem 8. In the sun about 4 billion kg of matter is converted to energy each second. The power output of
the sun in watt is
(a) 26
3.6 ×10 (b) 26
0.36 ×10
(c) 26
36 ×10 (d) 26
0.036 × 10 .
Ans. (b)
Solution: 8
4 10
m
t
  kgs-1
2
E mc

 2
E m
c
t t
 
  
 
 8 16
4 10 9 10
E
t
   
 25
3.6 10
E
t
  Js-1
 25
3.6 10
E
W
t
  .
Problem 9. A star initially has 40
10 deutrons. It produces energy via the processes 
2 2 3
1 1 1
H + H H + p
and 
2 3 4
1 1 2
H + H He + n . If the average power radiated by the star is 16
10 W , the deuteron
supply of the star is exhausted in a time of the order of
(a) 6
10 s (b) 8
10 s
(c) 12
10 s (d) 16
10 s .
(The masses of nuclei are: 2
m(H ) = 2.014amu, m(p) = 1.007amu ,
4
m(n) = 1.0084amu, m(He ) = 4.001amu )
Ans. (c)
Solution: 2 2 3
1 1 1
H H H p
  
2 3 4
1 1 2
H H He n
  
 2 4
1 2
3 H He p n
  
4 2
2 1
( ) ( ) ( ) 3 ( )
m m He m p m n m H
    
 [4.001 1.007 1.008 3(2.014)]
m amu
    

5.  0.026
m amu
  
 2
| | | |
E c m
  
 16 27
(9 10 )(0.026 1.67 10 )
E 
    
 (931.5)(0.026)
E
  MeV
 12
3.87 10 J
E 
  
As each reaction involves 3 deuterons, so total number of reactions involved in the process
40
10
3
 . If each reaction produces an energy E
 , then
40
28
10
1.29 10 J
3
total
E E
   
total
E Pt

Time of exhaustion of the star
28
16
1.29 10
10
t


 12
1.29 10
t s
  .
Problem 10. Consider the following reaction 
2 2 4
1 1 2
H + H He + Q . If 2
1
m( H ) = 2.0141amu ;
4
2
m( He ) = 4.0024amu . The energy Q released (in MeV) in this fusion reaction is
(a) 12 (b) 6
(c) 24 (d) 48.
Ans. (c)
Solution: 2 2 4
1 1 2
H H He Q
  
 4 2
2 1
( ) 2 ( )
m m He m H
  
 4.0024 2(2.0141)
m
  
 0.0258
m amu
  
Since, 2
Q c m
 
 (0.0258)(931.5)
Q  MeV
 24
Q MeV.