Important Formulae: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

 Page 1


Modern Physics – II 
Important Formulae 
1.  Radioactivity 
  (i) ?-emission 
emission A A 4
Z2
Z
XY
?? ?
? ? ? ? ? ?
 
 
  (ii) ? - emission n P e v
?
? ? ? ? ?  
   
emission AA
Z1
Z
XY
??
? ? ? ? ? ? 
(iii) Rutherford and Soddy law 
    
dN dN
N or N
dt dt
? ? ? ? ? 
  (iv) N = N
0
e
-lt
 
  Here, N are the number of remaining nuclei.  
  (v) N
d
 =N
0
(1 – e
- ?t
) 
   Here, N
d
 axe the number of decayed nuclei 
     
  (vi) ? = decay constant 
   
1
?
 = mean life or average life = t
av
 
   
1/ 2
1 0.693
t (ln 2)
??
??
??
??
??
 
(vii) t
av
 > t
1/2
 
 
   
1/ 2 av
ln 2 0.693
t 0.693t ? ? ?
??
 
   
av 1/ 2
1
t 1.44t ??
?
 
  (viii) Probability of survival of a nucleus upto time t 
    
    
t
s
Pe
??
? 
Page 2


Modern Physics – II 
Important Formulae 
1.  Radioactivity 
  (i) ?-emission 
emission A A 4
Z2
Z
XY
?? ?
? ? ? ? ? ?
 
 
  (ii) ? - emission n P e v
?
? ? ? ? ?  
   
emission AA
Z1
Z
XY
??
? ? ? ? ? ? 
(iii) Rutherford and Soddy law 
    
dN dN
N or N
dt dt
? ? ? ? ? 
  (iv) N = N
0
e
-lt
 
  Here, N are the number of remaining nuclei.  
  (v) N
d
 =N
0
(1 – e
- ?t
) 
   Here, N
d
 axe the number of decayed nuclei 
     
  (vi) ? = decay constant 
   
1
?
 = mean life or average life = t
av
 
   
1/ 2
1 0.693
t (ln 2)
??
??
??
??
??
 
(vii) t
av
 > t
1/2
 
 
   
1/ 2 av
ln 2 0.693
t 0.693t ? ? ?
??
 
   
av 1/ 2
1
t 1.44t ??
?
 
  (viii) Probability of survival of a nucleus upto time t 
    
    
t
s
Pe
??
? 
(ix) Probability of decay of a nucleus in time t 
    
   
t
d
P 1 e
??
?? 
  (x) Activity of radioactive substance, 
    
   
tt
00
dN
R N N e R e
dt
? ? ? ?
? ? ? ? ? ? ? 
    Here, R
0 
= ?N
0 
= initial activity 
  (xi) Units of activity  
  (a)1 Curie = 1Ci = 3.7 × 10
10
 dPS  
  (b)1 Rutherford = 1 rd = 10
6
 dPS 
   (c) 1 Becquerel = 1 Bq = 1 dPS 
  (xii) If a nucleus decays in two modes 
   
12
12
12
TT
and T
TT
? ? ? ? ? ?
?
 
   Here, T is representing the half-life.  
(xiii) 
1/ 2 1/ 2
n
tt
00
00
NN 1
N .......... N
2 4 2
??
? ? ? ? ? ?
??
??
 
   
n
1 1 1
1 ..........
2 4 2
??
? ? ? ? ? ?
??
??
 
   
n
1
100% 50% 25%..........100
2
??
? ? ? ? ? ?
??
??
 
  Here, n = number of half-lives 
1/ 2
t
t
?    
  (xiv) Successive radioactivity. 
    
Page 3


Modern Physics – II 
Important Formulae 
1.  Radioactivity 
  (i) ?-emission 
emission A A 4
Z2
Z
XY
?? ?
? ? ? ? ? ?
 
 
  (ii) ? - emission n P e v
?
? ? ? ? ?  
   
emission AA
Z1
Z
XY
??
? ? ? ? ? ? 
(iii) Rutherford and Soddy law 
    
dN dN
N or N
dt dt
? ? ? ? ? 
  (iv) N = N
0
e
-lt
 
  Here, N are the number of remaining nuclei.  
  (v) N
d
 =N
0
(1 – e
- ?t
) 
   Here, N
d
 axe the number of decayed nuclei 
     
  (vi) ? = decay constant 
   
1
?
 = mean life or average life = t
av
 
   
1/ 2
1 0.693
t (ln 2)
??
??
??
??
??
 
(vii) t
av
 > t
1/2
 
 
   
1/ 2 av
ln 2 0.693
t 0.693t ? ? ?
??
 
   
av 1/ 2
1
t 1.44t ??
?
 
  (viii) Probability of survival of a nucleus upto time t 
    
    
t
s
Pe
??
? 
(ix) Probability of decay of a nucleus in time t 
    
   
t
d
P 1 e
??
?? 
  (x) Activity of radioactive substance, 
    
   
tt
00
dN
R N N e R e
dt
? ? ? ?
? ? ? ? ? ? ? 
    Here, R
0 
= ?N
0 
= initial activity 
  (xi) Units of activity  
  (a)1 Curie = 1Ci = 3.7 × 10
10
 dPS  
  (b)1 Rutherford = 1 rd = 10
6
 dPS 
   (c) 1 Becquerel = 1 Bq = 1 dPS 
  (xii) If a nucleus decays in two modes 
   
12
12
12
TT
and T
TT
? ? ? ? ? ?
?
 
   Here, T is representing the half-life.  
(xiii) 
1/ 2 1/ 2
n
tt
00
00
NN 1
N .......... N
2 4 2
??
? ? ? ? ? ?
??
??
 
   
n
1 1 1
1 ..........
2 4 2
??
? ? ? ? ? ?
??
??
 
   
n
1
100% 50% 25%..........100
2
??
? ? ? ? ? ?
??
??
 
  Here, n = number of half-lives 
1/ 2
t
t
?    
  (xiv) Successive radioactivity. 
    
   
Suppose A and B are radioactive and C is stable. Let us further assume that initially there 
are only nuclei of A Then, number of nuclei of A, B and C vary with time as shown 
below 
    
  N
B
 are maximum when ?
A
N
A
 = ?
B
N
B
 
 
2.  Fusion and Fission 
  (i) E = mc
2
  
  (ii) 1 amu = 931.48 
2
MeV
c
 
  (iii) Mass defect = [Zm
p
 + (A - Z)m
N
 - m
X
] 
  Here m
X
 is mass of nucleus. 
  (iv) Binding energy E
1
 =( ?m) c
2
  
If ?m is represented in amu, then E
1
 can be obtained in MeV by multiplying it 
with 931.48.  
  (v) Binding energy per nucleon 
   
1
2
E
E
Total number of nucleons
? 
(vi) For stability of a nucleus binding energy per nucleon is more important rather than 
the total binding energy. 
(vii) Binding energy per nucleon is of the order of few MeV (2 MeV-10 MeV). 
(viii) During formation of a nucleus some mass is lost. This is called mass defect. 
Equivalent to that mass some energy is liberated. This is called binding energy of the 
nucleus. 
(ix) In any nuclear process energy is released if total binding energy of the daughter 
nuclei is more than the total binding energy of the parent nuclei.  
(x) Binding energy per nucleon (E
2
) versus number of nucleon (A) graph 
 
  From the graph it is clear that binding energy per nucleon is maximum near iron nucleus. 
Page 4


Modern Physics – II 
Important Formulae 
1.  Radioactivity 
  (i) ?-emission 
emission A A 4
Z2
Z
XY
?? ?
? ? ? ? ? ?
 
 
  (ii) ? - emission n P e v
?
? ? ? ? ?  
   
emission AA
Z1
Z
XY
??
? ? ? ? ? ? 
(iii) Rutherford and Soddy law 
    
dN dN
N or N
dt dt
? ? ? ? ? 
  (iv) N = N
0
e
-lt
 
  Here, N are the number of remaining nuclei.  
  (v) N
d
 =N
0
(1 – e
- ?t
) 
   Here, N
d
 axe the number of decayed nuclei 
     
  (vi) ? = decay constant 
   
1
?
 = mean life or average life = t
av
 
   
1/ 2
1 0.693
t (ln 2)
??
??
??
??
??
 
(vii) t
av
 > t
1/2
 
 
   
1/ 2 av
ln 2 0.693
t 0.693t ? ? ?
??
 
   
av 1/ 2
1
t 1.44t ??
?
 
  (viii) Probability of survival of a nucleus upto time t 
    
    
t
s
Pe
??
? 
(ix) Probability of decay of a nucleus in time t 
    
   
t
d
P 1 e
??
?? 
  (x) Activity of radioactive substance, 
    
   
tt
00
dN
R N N e R e
dt
? ? ? ?
? ? ? ? ? ? ? 
    Here, R
0 
= ?N
0 
= initial activity 
  (xi) Units of activity  
  (a)1 Curie = 1Ci = 3.7 × 10
10
 dPS  
  (b)1 Rutherford = 1 rd = 10
6
 dPS 
   (c) 1 Becquerel = 1 Bq = 1 dPS 
  (xii) If a nucleus decays in two modes 
   
12
12
12
TT
and T
TT
? ? ? ? ? ?
?
 
   Here, T is representing the half-life.  
(xiii) 
1/ 2 1/ 2
n
tt
00
00
NN 1
N .......... N
2 4 2
??
? ? ? ? ? ?
??
??
 
   
n
1 1 1
1 ..........
2 4 2
??
? ? ? ? ? ?
??
??
 
   
n
1
100% 50% 25%..........100
2
??
? ? ? ? ? ?
??
??
 
  Here, n = number of half-lives 
1/ 2
t
t
?    
  (xiv) Successive radioactivity. 
    
   
Suppose A and B are radioactive and C is stable. Let us further assume that initially there 
are only nuclei of A Then, number of nuclei of A, B and C vary with time as shown 
below 
    
  N
B
 are maximum when ?
A
N
A
 = ?
B
N
B
 
 
2.  Fusion and Fission 
  (i) E = mc
2
  
  (ii) 1 amu = 931.48 
2
MeV
c
 
  (iii) Mass defect = [Zm
p
 + (A - Z)m
N
 - m
X
] 
  Here m
X
 is mass of nucleus. 
  (iv) Binding energy E
1
 =( ?m) c
2
  
If ?m is represented in amu, then E
1
 can be obtained in MeV by multiplying it 
with 931.48.  
  (v) Binding energy per nucleon 
   
1
2
E
E
Total number of nucleons
? 
(vi) For stability of a nucleus binding energy per nucleon is more important rather than 
the total binding energy. 
(vii) Binding energy per nucleon is of the order of few MeV (2 MeV-10 MeV). 
(viii) During formation of a nucleus some mass is lost. This is called mass defect. 
Equivalent to that mass some energy is liberated. This is called binding energy of the 
nucleus. 
(ix) In any nuclear process energy is released if total binding energy of the daughter 
nuclei is more than the total binding energy of the parent nuclei.  
(x) Binding energy per nucleon (E
2
) versus number of nucleon (A) graph 
 
  From the graph it is clear that binding energy per nucleon is maximum near iron nucleus. 
In any nuclear process, if the products are towards the peak of this graph then binding 
energy per nucleon, hence total binding energy will increase. Therefore, energy will be 
released. 
In fusion reaction two or more lighter nuclei combine to make a relatively heavier 
nucleus. This heavy nucleus is towards peak of this graph. Therefore, energy will be 
released. 
In fission reaction a heavy nucleus breaks into two or more lighter nuclei. These lighter 
nuclei again lie towards peak of this graph. Therefore, energy will be released. 
 
(xi) Normally, fusion reaction is more difficult compared to fission reaction. Because to 
combine two or more positively charged nuclei is difficult compared to break it. 
(xii) Radius of a nucleus R = R
0
A
1/3
 
   Here, R
0
 =1.3 fm = 1.3 × 10
-15
 m and A = mass number 
   Thus, R ? A
1/3
 
  (xiii) The density of any nucleus is independent of A and is of the order of 10
17
kgm
-3
.