# Chapter 2 Nuclear Structure and Radioactivity¶

## Example 2_1 pgno:54¶

In [5]:
from math import log
N0=3396.;#no. of counts per minute given by radioactive nuclide at a given time#
N=1000.;#no. of counts per minute given by radioactive nuclide one hour later#
thalf=0.693*60/(2.303*log(N0/N));#half life of nuclide in minutes#
t1=2.303*log(100/25)*thalf/0.693;#time required for the activity to decrease to 25% of the initial activity in minutes#
print'\nTime required for the activity to decrease to 25percent of the initial activity=t1=minutes',round(t1)
t2=2.303*log(100/10)*thalf/0.693;#time required for the activity to decrease to 10% of the initial activity in minutes#
print'\nTime required for the activity to decrease to 10percent of the initial activity=t2=minutes',round(t2)

Half life of radioactive nuclide=t1/2=minutes 14.7674928978

Time required for the activity to decrease to 25percent of the initial activity=t1=minutes 68.0

Time required for the activity to decrease to 10percent of the initial activity=t2=minutes 113.0


## Example 2_2 pgno:55¶

In [6]:
R=3.7*10**10;#no. of alpha particles per second emitted by 1g of 226Ra#
N=(6.023*10**23)/226;#no. of atoms of 226Ra#
yr=3.15*10**7;#no of seconds in a year#
thalf=0.693*N/(R*yr);#half life of 226Ra in years#
print'Half life of 226Ra molecule=t1/2=years',round(thalf)#here the answer written in textbook is wrongly printed actual answer will be the one we are getting here#

Half life of 226Ra molecule=t1/2=years 1585.0


## Example 2_3 pgno:57¶

In [7]:
thalf=14.8*60*60;#half life of 24Na atom in seconds#
v=3.7*10**10;#1 Ci of radioactivity in disintegrations per second#
w=(24*10**6*v*thalf)/(0.693*L);#weight of 1 Ci of 24Na in grams#
print'Weight of 1 Ci of 24Na=w=micrograms=1.13*10**-7grams',round(w,4)

Weight of 1 Ci of 24Na=w=micrograms=1.13*10**-7grams 0.1134


## Example 2_4 pgno:60¶

In [8]:
Mp=1.00728;#mass of proton in amu#
Mn=1.00866;#mass of neutronin amu#
MH=2.01355;#isotopic mass of H atom in amu#
dM=((1*Mp)+(1*Mn)-MH);#dM value of H atom in amu#
print'dM value of H atom=dM=amu',dM
BE=dM*931;#binding energy of H atom in MeV#
print'\nBinding energy of H atom=BE=MeV',round(BE,3)

dM value of H atom=dM=amu 0.00239

Binding energy of H atom=BE=MeV 2.225


## Example 2_5 pgno:65¶

In [10]:
from math import log
N0=15.3;#decay rate of Contemporary Carbon in disintegrations/min/gram#
N=2.25;#decay rate of 14C specimen in disintegrtions/min/gram#
thalf=5670.;#half life of nuclide in years#
t=2.303*log(N0/N)*thalf/0.693;#Age of the specimen in years#
print'Age of the specimen=t=years',round(t)#here the answer given in textbook is actually wrong we get twice that of the answer which is shown through execution#

Age of the specimen=t=years 36120.0


## Example 2_6 pgno:65¶

In [11]:
from math import log
thalf=4.5*10**9;#half life of Uranium in years#
print'Here N0 and N must be in terms of Uranium.N is proportional to 1gram og Uranium'
print'\nN0 can be calculated from the given data.0.0453grams of 206Pb corresponds to 238*0.0453/206=0.0523grams of 238U,i.e 0.0453 grams of 206Pb must have been formed by the decaying of 0.523grams of 238U.\nSince N is proportional to 1,N0 is proportional to 1.0523.'
N0=1.0523;
N=1;
t=2.303*log(N0/N)*thalf/0.693;#Age of the mineral in years#
print'\nAge of the mineral=t=years=7.62*10**8years',round(t)#here also the answer given in textbook is wrong the one resulted through execution is the right one#

Here N0 and N must be in terms of Uranium.N is proportional to 1gram og Uranium

N0 can be calculated from the given data.0.0453grams of 206Pb corresponds to 238*0.0453/206=0.0523grams of 238U,i.e 0.0453 grams of 206Pb must have been formed by the decaying of 0.523grams of 238U.
Since N is proportional to 1,N0 is proportional to 1.0523.

Age of the mineral=t=years=7.62*10**8years 762356479.0