Question
A radioactive clement decays by $$\beta $$ emission. A detector records $$n$$ beta particles in 2 second and in next 2 seconds it records $$0.75\,n$$ beta particles. Find mean life corrected to nearest whole number. Given $$\ell n2 = 0.6931$$ and $$\ell n3 = 1.0986.$$
A.
$$6.9\,\sec $$
B.
$$9.9\,\sec $$
C.
$$10.1\,\sec $$
D.
$$12.2\,\sec $$
Answer :
$$6.9\,\sec $$
Solution :
We know that, $$N = {N_0}{e^{ - \lambda t}}$$
After 2 second, $${N_1} = {N_0}{e^{ - \lambda \times 2}}$$
After $$\left( {2 + 2} \right)$$ second, $${N_2} = {N_0}{e^{ - \lambda \times 4}}$$
According to given conditions,
$$\eqalign{
& {N_0} - {N_1} = n \cr
& {\text{or}}\,\,{N_0} - {N_0}{e^{ - 2\lambda }} = n\,......\left( {\text{i}} \right) \cr
& {\text{and}}\,{N_0}{e^{ - 2\lambda }} - {N_0}{e^{ - 4\lambda }} = 0.75\,n\,......\left( {{\text{ii}}} \right) \cr} $$
After solving above equations, we get
$$\lambda = 0.145\,s$$
Mean life$$T = \frac{1}{\lambda } = 6.9\,s.$$