Nucleus - 4
Units of activity:- s1 units of activity is Becqueral(Bq) other commonly units are curve (c1) and ratherford(rd)
1Bq = 1 decay/s
1Ci =3.7 X 1010 Bq
1rd = 106 Bq
PROBLEMS
Easy
At time = 0, activity of a radioactive substance is 1600Bq. at t=8s activity remains 100 Bq. Find the activity at t = 2s.
     2n = 16
=> n = 4
tan half lives are equivalent to 80. Hence 2s is equal to one half life activity will remain half of 1600 Bq. i.e 800 Bq
We are given the following atomic masses.
hence the symbol Pa is for the element proactinium(z = 91).
(a) Calculate the energy released during the 
-decay of 
(b) Show that cannot spontaneously emit a proton
[solutions]:-
(a) The -decay of is given by
the energy released in this process = 
= (0.00456 u) x c2
= 0.00456 x 931.5 MeV
= 4.25 MeV
(b) if spantaneously emits a proton , the decay process would be
Mass defect for this process = 
m = Mu - MPa - MH = - 0.00825u
Since mass defect is comming out to be -ve, 
if this reaction occurs than it would require an external supply of energy. Hence emission of proton by spantaneously is not feasible
(3) The half-life of against decay is 4.5 x 109y. what is the activity of ig sample of ?
Ans: T1/2 = 4.5 x 109 = 4.5 x 109 x 3.16 x 107
= 1.42 x 1017s.
A mol of an isotope has a mass Equal to the atomic weight of that isotope expressed in grams
Hence 1g of contains contains = 4.2 x 10-3mol
One mol of an isotope has ava gadros number of atom, and so 1g og contains 4.20 x 103x x 6.023 x 1023 atom
The decay rate is
[4] the mean life of radioactive substance are 1620 yrs and 405 yrs. and 
emission respectively. Find out the time during which 3/4 th of a sample will decay. if it is decaying both by and emission simultaneously.
Solution:- Net rate of decay = rate due to decay + rate due to decay
time for decaying 3/4 th of sample = 2t1/2 = 2 x 224.5324 = 449.064y
[5] Uranium is mixture of u - 234, u - 235, u - 238 in ratio 0.006%: 0.71%: 99.284% t1/2 are 2.5 x 10 5yrs, 7.1 x 108yrs, 4.5 x 109yrs respectively. calculate the contribution to activity of each isotope in %
solution:- let 100 gm of sample be taken
for u = 234
for u = 235
for u = 238
Activity = A1 + A2 + A3 = (N0ln 2)(92.7 x 10-12 + 102.6 x 10-12+4.25 x 10-12)
= (N0ln 2)(199.55 x 10-12)
% contribution of u - 234 = 
x 100 = 51.5 %
% contribution of u - 235 = 
x 100 = 2.13 %
% contribution of u - 238 = 
x 100 = 46.37 %