2019-03-01_Physics_Times

logR  t logR 0 

From the graph

1 0.693 3.0min

2

t



 

The power generated in the rod is

P 5 10 /^8 W m^3 r L^2 dQ dmc T

dt dt

     

Where c is the heat capacity of the coolent and

T is the change in temperature of the coolent.

Given that r 4 10 , 0.2^3 m L m

c 4 10 /. ,^3 J kg kdm 0.2 /kg s

dt

  

   T 6 C
The given nuclear reaction is
12 11
6 C 5 B e ve 0.96MeV
   
Energy released in every reaction is
0.96MeV 1.02MeV (Due to annihilation)
1.98MeV
The initial number of^12 C are
19
16

(^027)
10
4.9 10
12 1.67 10
kg
N

     
The number of^12 C left after 2 t 0 time are equal to
(^0) 1.22 10 16
4
N
N  
The number of nuclei disintegrated are equal to
3.67 10^16
Total energy liberated 3.67 10 1.98^16 
(^) 7.27 10 MeV^16
The K.E of particle is
(^12)
2
KE mv
0.05 1.6 10^191 1.6 10^262
2
       v
0.05 2 10 ^7   v^2 v 1000 / secm
Time taken to travel a distance of 1 m is
1
0.001sec
1000

t1/26.9sec.
1/2
0.693 0.693
0.1
6.9
t     

Fraction of particle decay in 0.001 sec
0.1^1
  1 et  1 e ^1000 0.0001
As (B.E/A) of B^10 is more than B^11
so formation of 5 B^10 will release energy
E 1  Binding energy of 5 B^11 7.5 11 MeV
(^) 82.5MeV
E 2  Binding energy of
10
5 B 8.0 10 MeV
(^)  80 MeV
Energy given   E E 2 82.5 80
= 2.5 MeV
If the nuclear reaction has to take place then the
mass of products must be less than the mass of
reactants.
If Y concentration remains constant then
rate of decay of X = rate of decay of Y
,
x Y
x x Y Y
x Y
N N
N N
T T
    
The half life of B is
3
2
3
2
A
A
B
A
A

 
 

 
3
7
B
A



Given T=30 minutes.
2.Sol:
3.Sol:
4.Sol:
5.Sol:
6.Sol:
7.Sol:
8.Sol:
9.Sol:
120
dN counts
K
dt sec

dN
dt
at 5 P.M. will be equal to activity remaining
after four half lives.
5
1
at P.M.^16
dN
th
dt
   
   
   of the initial activity
5 3
1
P.M.^16 P.M
dN dN
dt dt
    
    
    