# Chapter 25 Nuclei¶

## Example 25.1 Page no 1312¶

In [17]:
#Given
R0=1.2*10**-15                       #m
A=208
A1=16

#calculation
R=R0*A**0.33
R1=R0*A1**0.33

#Result
print"Nuclear radius of oxygen is", round(R1*10**15,0),"fm"

Nuclear radius of lead is 7.0 fm
Nuclear radius of oxygen is 3.0 fm


## Example 25.2 page no1312¶

In [27]:
#Given
me=9.1*10**-31
c=3*10**8
e=1.6*10**-19
mp=1.673*10**-27
mn=1.675*10**-27

#Calculation
E=(me*c**2)/e
E1=(mp*c**2)/e
E2=(mn*c**2)/e

#Result
print"(i) Equivalent energy of electron is",round(E*10**-6,2),"Mev"
print"(ii) Equivalent energy of proton is",round(E1*10**-6,1),"Mev"
print"(iii) Equivalent energy of neutron is",round(E2*10**-6,1),"Mev"

(i) Equivalent energy of electron is 0.51 Mev
(ii) Equivalent energy of proton is 941.1 Mev
(iii) Equivalent energy of neutron is 942.2 Mev


## Example 25.3 Page no 1312¶

In [30]:
#Given
m=10**-3                   #m
c=3*10**8                      #m/s
a=3.6*10**6                       #J

#Calculation
E=(m*c**2)/a

#Result
print E*10**-7,"*10**7 KWh"

2.5 *10**7 KWh


## Example 25.4 Page no 1313¶

In [33]:
#Given
Z=17
A=35
Z1=92
A1=235
Z2=4
A2=9

#Calculation
n=A-Z
n1=A1-Z1
n2=A2-Z2

#Calculation
print"Number of neutron in 17Cl35 is",n
print"Number of neutron in 92U235 is",n1
print"Number of neutron in 4Be9 is",n2

Number of neutron in 17Cl35 is 18
Number of neutron in 92U235 is 143
Number of neutron in 4Be9 is 5


## Example 25.5 Page no 1313¶

In [39]:
#Given
A2=235
A1=16.0
R1=3*10**-15                  #m

#Calculation
R=(A2/A1)**0.33
R2=R*R1

#Result

Nuclear radius is 7.281 fermi


## Example 25.6 Page no 1313¶

In [45]:
#Given
me=55.85
u=1.66*10**-27                 #Kg
R=1.2*10**-15

#Calculation
import math
m=me*u
a=(3*u)/(4.0*math.pi*R**3)

#Result
print"Nuclear density is", round(a*10**-17,2)*10**17,"Kg/m**3"

Nuclear density is 2.29e+17 Kg/m**3


## Example 25.7 Page no 1317¶

In [61]:
#Given
M=4.001509                    #a.m.u
N=1.008666
N1=1.007277
a=1.66*10**-27
c=3*10**8
e=1.6*10**-19
n=4.0

#Calculation
A=2*N1+2*N
M1=A-M
Eb=M1*a*c**2/e
B=Eb/n

#Result
print"(i)  Mass defect is",M1,"a.m.u"
print"(ii) Binding energy is",round(Eb*10**-6,1),"Mev"
print" Binding energy per nucleon is",round(B*10**-6,2),"Mev"

(i)  Mass defect is 0.030377 a.m.u
(ii) Binding energy is 28.4 Mev
Binding energy per nucleon is 7.09 Mev


## Example 25.8 Page no 1317¶

In [4]:
#Given
ma=1.00893
m1=1.00813
m2=2.01473
a=931.5
a1=4.00389

#Calculation
m=ma+m1-m2
Eb=m*a
m3=2*ma+2*m1-a1
Eb1=m3*a

#Result
print"(i) Binding energy when one neutron and one proton combined together is",  round(Eb,2),"Mev"
print"(ii) Binding energy when two neutrons and two protons are combined is",round(Eb1,1) ,"Mev"

(i) Binding energy when one neutron and one proton combined together is 2.17 Mev
(ii) Binding eergy when two neutrons and two protons are combined is 28.2 Mev


## Example 25.10 Page no 1318¶

In [14]:
#Given
a=1.66*10**-27             #Kg
c=3*10**8
mp=1.00727
mn=1.00866
mo=15.99053

#Calculation
E=(a*c**2)/1.6*10**-19
m1=8*mp+8*mn-mo
a1=m1*E

#Result
print"Energy equivalent of one atomic mass unit is", round(a1*10**32,1),"Mev/c**2"

Energy equivalent of one atomic mass unit is 127.8 Mev/c**2


## Example 25.11 Page no 1318¶

In [21]:
#Given
mp=1.007825
mn=1.008665
m=39.962589
a2=931.5
Z=40.0

#Calculation
E=20*mp+20*mn
m1=E-m
Eb=m1*a2
B=Eb/Z

#Result
print"Binding energy per nucleon is", round(B,3),"Mev/nucleon"

Binding energy per nucleon is 8.551 Mev/nucleon


## Example 25.12 Page no 1330¶

In [42]:
#Given
t=5000                   #Days
t1=2000.0
a=0.693

#Calculation
import math
dt=(a*t)/t1
N=math.log10(dt)
l=a*N/(t1)

#Result
print"(i) The fraction remaining after 5000 days is", round(N,3)
print"(ii) The activity of sample after 5000 days is",round(l*10**5,1),"*10**8 Bq"

(i) The fraction remaining after 5000 days is 0.239
(ii) The activity of sample after 5000 days is 8.3 *10**8 Bq


## Example 25.13 Page no 1330¶

In [48]:
#Given
N=3.67*10**10              #dis/second
r=226.0
A=6.023*10**23

#Calculation
n=A/r
l=N/n
D=0.693/l
a=D/(3600.0*24.0*365.0)

#Result
print" Half life of radium is",round(a,0),"years"

 Half life of radium is 1596.0 years


## Example 25.14 page no 1330¶

In [55]:
#Given
N0=475
N=270.0
t=5.0

#Calculation
import math
a=N0/N
l=math.log(a)/t
T=1/l
T1=0.693/l

#Result
print"(i) The decay constant is",round(l,3),"/minute"
print"(ii) Mean life is",round(T,2),"minute"
print"(iii) Half life is",round(T1,2),"minute"

(i) The decay constant is 0.113 /minute
(ii) Mean life is 8.85 minute
(iii) Half life is 6.13 minute


## Example 25.15 page no 1331¶

In [61]:
#Given
t=1500
N=0.01
N0=0.999

#Calculation
import math
T=t*math.log(N)/math.log(0.5)
T1=t*math.log(N0)/math.log(0.5)

#Result
print"(i) Years will reduce to 1 centigram is",round(T,1),"years"
print"(ii) Years will lose 1 mg is",round(T1,2),"years"

(i) Years will reduce to 1 centigram is 9965.8 years
(ii) Years will lose 1 mg is 2.17 years


## Example 25.16 page no 1331¶

In [66]:
#Given
a=2*10**12
b=9.0*10**12
T=80

#Calculation
import math
c=math.log(a/b)
t=-(c*T)/0.693

#Result
print"Time required is",round(t,0),"second"

Time required is 174.0 second


## Example 25.17 page no 1332¶

In [80]:
#Given
T=6.0
A=6.023*10**23
W=99.0

#Calculation
import math
l=0.693/T
N0=A*10**-12/W
A0=l*N0
N=N0*(1/math.log10(l))
A1=-(l*N)

#Result
print" Activity in the beginning and after one hour",round(A1*10**-8,3),"/h"

 Activity in the beginning and after one hour 7.496 /h


## Example 25.18 page no 1332¶

In [93]:
#Given
T=30.0

#Calculation
import math
l=0.693/T
T1=1/l
t=math.log(4)/l
t1=math.log(8)/l

#Result
print"(i) average life is",round(l,4),"/day"
print"(ii) The time taken for 3/4 of the original no. to disintegrate is",round(T1,2),"days"
print"(iii) Time taken is",round(t,0),"days"
print"(iv) Time taken is",round(t1,0),"days"

(i) average life is 0.0231 /day
(ii) The time taken for 3/4 of the original no. to disintegrate is 43.29 days
(iii) Time taken is 60.0 days
(iv) Time taken is 90.0 days


## Example 25.19 page no 1332¶

In [99]:
#Given
l=1620.0
l1=405.0

#Calculation
import math
T=(1/l)+(1/l1)
t=math.log(4)/T

#Result
print"The time during which three-fourths of a sample will decay is",round(t,0),"years"

The time during which three-fourths of a sample will decay is 449.0 years


## Example 25.20 page no 1333¶

In [105]:
#Given
C=3.7*10**10                   #disintegrations/s
A=6.02*10**23
B=234

#Calculation
D=(C*B)/A

#Result
print"Mass ofuranium atoms disintegrated per second is",round(D*10**11,3)*10**-11,"g"

Mass ofuranium atoms disintegrated per second is 1.438e-11 g


## Example 25.21 page no 1333¶

In [122]:
#Given
M=0.075              #kg /mol
m=1.2*10**-6         #kg
A=6.0*10**23         #/mol
t=9.6*10**18
N=170

#Calculation
n=(A*m)/M
l=N/t
T=0.693/l

#Result
print"Number of K-40 atoms in the sample is", n
print"Half life of K-40 is", round(T/(24.0*3600.0*365)*10**-9,3),"*10**9 years"

Number of K-40 atoms in the sample is 9.6e+18
Half life of K-40 is 1.241 *10**9 years


## Example 25.22 Page no 1337¶

In [139]:
#Given
mp=232.03714
mn=228.02873
m0=4.002603
a=931.5
A=232.0
e=1.6*10**-19
m=1.66*10**-27

#Calculation
M=mp-mn-m0
Q=M*a
K=(A-4)*Q/A
S=math.sqrt((2*K*e)/(4.0*m))

#Result
print"(i) Kinetic energy is", round(K,1),"Mev"
print"(ii) Speed of particle is", round(S*10**-4,1),"*10**7 m/s"

(i) Kinetic energy is 5.3 Mev
(ii) Speed of particle is 1.6 *10**7 m/s


## Example 25.23 Page no 1337¶

In [145]:
#Given
b=238
c=206
d=92
e=82

#Calculation
a=(b-c)/4.0
A=-d+(2*a)+e

#Result
print"(i) The emission of alpha particle will reduce the mass number by 4a  and charge number by 2a"
print"(ii) Number of alpha particle is", a
print"Number of beta particle is",A

(i) The emission of alpha particle will reduce the mass number by 4a  and charge number by 2a
(ii) Number of alpha particle is 8.0
Number of beta particle is 6.0


## Example 25.24 Page no 1338¶

In [148]:
#Given
a=218
b=84

#Calculation
A=a-4
Z=b-2

#Result
print"Atomic number of new element formed is", A
print"Mass number of new element formed is",Z

Atomic number of new element formed is 214
Mass number of new element formed is 82


## Example 25.27 Page no 1340¶

In [154]:
#Given
mp=10.016125
mn=4.003874
mp1=13.007490
mn1=1.008146
a=931.5

#Calculation
Mr=mp+mn
Mp=mp1+mn1
Md=Mr-Mp
A=a*Md

#Result
print"Energy released in the reaction is",round(A,3),"Mev"

Energy released in the reaction is 4.064 Mev


## Example 25.28 Page no 1345¶

In [159]:
#Given
a=10**6                     #J/s
E=200*10**6*1.6*10**-19

#Calculation
N=a/E

#Result
print"Number of fission per second is", round(N*10**-16,2)*10**16

Number of fission per second is 3.13e+16


## Example 25.29 Page no 1345¶

In [166]:
#Given
P=3*10**8                   #W
E=200*10**6*1.6*10**-19
a=235
m=6.023*10**23

#Calculation
E1=P*3600
N=E1/E
M1=(a*N)/m

#Result
print"Mass of uranium fissioned per hour is", round(M1,2),"g"

Mass of uranium fissioned per hour is 13.17 g


## Example 25.30 Page no 1345¶

In [173]:
#Given
m=6.023*10**26
a=235.0
t=30                          #Days
E=200*10**6*1.6*10**-19

#Calculation
N=(2/a)*m
A=N/(t*24*60.0*60.0)
P=E*A

#Result
print"Power output is", round(P*10**-6,1),"Mev"

Power output is 63.3 Mev


## Example 25.31 Page no 1348¶

In [181]:
#Given
m=1.0076
mp=4.0039
a=931.5*10**6                       #ev

#Calculation
Mr=4*m
Md=Mr-mp
E=Md*a*1.6*10**-19

#Result
print"Energy released is", round(E*10**13,2)*10**-13,"J"

Energy released is 3.95e-12 J


## Example 25.32 Page no 1349¶

In [184]:
#Given
a=6*10**-3                        #Kg
c=3*10**8

#Calculation
E=a*c**2

#Result
print"Energy liberated is", E,"J"

Energy liberated is 5.4e+14 J