Chapter 31 Radioactivity

Example 31.1 Page no 855

In [71]:
#Given
T=138.0                         #days
E=0.693
N=12.5
N0=100

#Calculation
import math
L=E/T
N1=N/N0
t=(2.303*math.log10(8))/L

#Result
print"Time is",round(t,2),"days"

Time is 414.16 days

Example 31.2 Page no 855

In [1]:
#Given
T=1.42*10**17                  #s
A=6.02*10**23                   #mol**-1
E=0.693
L=238.0

#Calculation
N=A/L
L1=E/T
Z=L1*N

#Result
print"Disintegrations per second occur in 1g of U**238 is",round(Z*10**-4,3),"*10**4 s**-1"

Disintegrations per second occur in 1g of U**238 is 1.234 *10**4 s**-1

Example 31.3 Page no 855

In [81]:
#Given
K=-40                             #Cl
M=0.075                           #kg mole**-1
m=1.2*10**-6                      #kg
A=6.0*10**23
D=170                             #s**-1
E=0.693

#Calculation
N=(A/M)*m
L=D/N
T=(E/L)
O=T/31536000.0

#Result
print"Half life of K40 atom is",round(O*10**-9,3),"*10**9 years"

Half life of K40 atom is 1.241 *10**9 years

Example 31.4 Page no 856

In [15]:
#Given
T=6.0                              #hours
E=0.693
A=6.025*10**23
W=99.0
S=10**-12
t=1                                #hours

#Calculation
import math
L=E/T
N0=(A/W)*S
R0=L*N0
N=N0*math.exp(-L)
R=L*N
R1=L*N

#Result
print"Activity at beginning is", round(R0*10**-8,2),"*10**8 /h"
print"Activity at the end is",round(R1*10**-8,3),"*10**8 /h"

Activity at beginning is 7.03 *10**8 /h
Activity at the end is 6.262 *10**8 /h

Example 31.5 Page no 856

In [22]:
#Given
t=30.0                          #years
a=16

#Calculation
import math
l=(2.3026*math.log10(a))/t
T=0.693/l

#Result
print"Half life period is", round(T,1),"years"

Half life period is 7.5 years

Example 31.6 Page no 856

In [25]:
#Given
m1=238.05081
m2=234.04363
m3=4.00260
A=931.5                            #Mev

#Calculation
Ea=(m1-m2-m3)*A

#Result
print"Kinetic energy is",round(Ea,2),"Mev"

Kinetic energy is 4.27 Mev

Example 31.7 Page no 856

In [28]:
#Given
m1=22.994466
m2=22.989770
A=931.5

#Calculation
Eb=(m1-m2)*A

#Result
print"Maximum kinetic energy is",round(Eb,3),"Mev"

Maximum kinetic energy is 4.374 Mev

Example 31.8 Page no 856

In [42]:
#Given
E1=0
E2=0.412*1.6*10**-13
E3=1.088*1.6*10**-13
h=6.62*10**-34
m1=197.968233
m2=197.966760
a=931.5
A=1.088
A1=0.412

#Calculation
v1=(E3-E1)/h
v2=(E2-E1)/h
v3=(E3-E2)/h
Eb1=((m1-m2)*a)-A
Eb2=((m1-m2)*a)-A1

#Result
print"Radiation frequencies are",round(v1*10**-20,2)*10**20,",",round(v2*10**-20,3)*10**20,"and",round(v3*10**-20,2)*10**20
print"Maximum kinetic energies are",round(Eb1,3),"Mev and ",round(Eb2,3),"Mev"

Radiation frequencies are 2.63e+20 , 9.96e+19 and 1.63e+20
Maximum kinetic energies are 0.284 Mev and  0.96 Mev

Example 31.9 Page no 857

In [46]:
#Given
a=15
b=9.0
T=5730

#Calculation
import math
A=a/b
l=2.303*math.log10(A)
t=(l*T)/0.693

#Result
print"Approximation age of the indus valley civilization is",round(t,2),"years"

Approximation age of the indus valley civilization is 4224.47 years