# Chapter 32 Nuclear reactions¶

## Example 32.3 Page no 874¶

In [3]:
#Given
r1=6.015126
r2=1.008665
p1=4.002603
p2=3.016049
a=931

#Calculation
R=r1+r2
P=p1+p2
M=R-P
E=M*a

#Result
print"Energy released is",round(E,2),"Mev"

Energy released is 4.78 Mev


## Example 32.4 Page no 874¶

In [7]:
#Given
P=400*10**6                       #J/s
c=3.0*10**8

#Calculation
E=P*24*60*60
m=E/c**2

#Result
print"U235 is",m*10**3,"g"

U235 is 0.384 g


## Example 32.5 Page no 874¶

In [17]:
#Given
A=6.023*10**23
w=235.0
E=200
e=1.6*10**-13

#Calculation
E1=((A*E)/w)*e
E2=E1/(1000.0*3600.0)

#Result
print"Energy released is", round(E2*10**-4,3),"*10**4 KWh"

Energy released is 2.278 *10**4 KWh


## Example 32.6 Page no 874¶

In [22]:
#Given
E=7.6*10**13                   #J
E1=200.0*1.6*10**-13
h=6.023*10**23
a=235

#Calculation
n=E/E1
M=(a*n)/h

#Result
print"(i) Number of uranium atom is",n
print"(ii) Mass of uranium is",round(M,2),"g"

(i) Number of uranium atom is 2.375e+24
(ii) Mass of uranium is 926.66 g


## Example 32.7 Page no 874¶

In [26]:
#Given
m1=235.043933
m2=140.917700
m3=91.895400
n1=1.0086651
a=931.5

#Calculation
Q=(m1-m2-m3-2*(n1))*a

#Result
print"Energy released is",round(Q,2),"Mev"

Energy released is 198.88 Mev


## Example 32.8 Page no 874¶

In [30]:
#Given
M1=1.007825
M2=4.002603
M3=0.000549
A=931.5
S=4
W=2

#Calculation
Q=((S*M1-M2-W*M3)*A)

#Result
print"The energy is",round(Q,2),"MeV"

The energy is 25.71 MeV


## Example 32.9 Page no 875¶

In [33]:
#Given
M1=1.008                   #a.m.u.
M2=4.004                   #a.m.u.
M3=7.016                   #a.m.u.
A=2
W=931

#Calculation
Q=(M3+M1-A*M2)*W

#Result
print"The initial energy of each alpha-particle is",Q,"MeV"

The initial energy of each alpha-particle is 14.896 MeV


## Example 32.10 Page no 875¶

In [38]:
#Given
M1=235.0439
M2=93.9065
M3=139.9055
M4=1.00866
S=931.5

#Calculation
Q=(M1-M2-M3-M4)*S

#Result
print"Total energy is",round(Q,2),"MeV"
print"All the available energy does not appear as the kinetic energy of fission products"

Total energy is 207.95 MeV
All the available energy does not appear as the kinetic energy of fission products


## Example 32.11 Page no 875¶

In [55]:
#Given
M1=2.014102
M2=3.016049
M3=4.002603
M4=1.008665
S=931.5
q1=1.6*10**-19                        #C
q2=9*10**9
r=2*1.5*10**-15
r1=1.38*10**-23

#Calculation
Q=(M1+M2-M3-M4)*S
Q2=q2*(q1**2/r)
Q3=(2*Q2)/(3*r1)

#Result
print"(a) The amount of energy is",round(Q,2),"MeV"
print"(b) The kinetic energy is",Q2,"J"
print"The temperature is",round(Q3*10**-9,2),"*10**9 K"

(a) The amount of energy is 17.59 MeV
(b) The kinetic energy is 7.68e-14 J
The temperature is 3.71 *10**9 K