Question
In a radioactive material the activity at time $${t_1}$$ is $${R_1}$$ and at a later time $${t_2}$$ it is $${R_2}.$$ If the decay constant of the material is $$\lambda ,$$ then
A.
$${R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}}$$
B.
$${R_1} = {R_2}{e^{\lambda \left( {{t_1} - {t_2}} \right)}}$$
C.
$${R_1} = {R_2}\left( {\frac{{{t_2}}}{{{t_1}}}} \right)$$
D.
$${R_1} = {R_2}$$
Answer :
$${R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}}$$
Solution :
The decay rate $$R$$ of a radioactive material is the number of decays per second.
From radioactive decay law, $$ - \frac{{dN}}{{dt}} \propto N$$
$${\text{or}}\,\, - \frac{{dN}}{{dt}} = \lambda N$$
i.e. Rate of reaction is directly proportional to the initial concentration of reactants.
$$\eqalign{
& {\text{Thus,}}\,\,R = - \frac{{dN}}{{dt}}\,\,{\text{or}}\,\,R \propto N \cr
& {\text{or}}\,\,R = \lambda N\,\,{\text{or}}\,\,R = \lambda {N_0}{e^{ - \lambda t}}\,......\left( {\text{i}} \right) \cr} $$
where $${R_0} = \lambda {N_0}$$ is the activity of the radioactive material at time $$t = 0.$$
$$\eqalign{
& {\text{At}}\,{\text{time}}\,{t_1},\,\,{R_1} = {R_0}{e^{ - \lambda {t_1}}}\,......\left( {{\text{ii}}} \right) \cr
& {\text{At}}\,{\text{time}}\,{t_2},\,\,{R_2} = {R_0}{e^{ - \lambda {t_2}}}\,......\left( {{\text{iii}}} \right) \cr} $$
Dividing Eq. (ii) by Eq. (iii), we have
$$\eqalign{
& \frac{{{R_1}}}{{{R_2}}} = \frac{{{e^{ - \lambda {t_1}}}}}{{{e^{ - \lambda {t_2}}}}} = {e^{ - \lambda \left( {{t_1} - {t_2}} \right)}} \cr
& {\text{or}}\,\,{R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}} \cr} $$