Question

A sample of radioactive material decays simultaneously by two processes A and B with half lives $\dfrac{1}{2}h$ and $\dfrac{1}{4}h$, respectively. For the first half hour it decays with the process A, next one hour with the process B, and the further half an hour with both A and B. if, original, there are ${N_0}$ nuclei, find the number of nuclei after 2 h of such decay.

Answer 1

Determine the half life of the sample when both the processes happen simultaneously. Calculate the number of half lives in the given time for each process. Use the formula for the number of nuclei present in the sample after n half lives.

Formula used:
The number of nuclei present in the sample after n half lives is,
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^n}$
Here, ${N_0}$ is the initial number of nuclei present in the sample.

Complete step by step answer:
We have given the half life for the process A is, ${t_1} = \dfrac{1}{2}h$.
Half life for the process B is ${t_2} = \dfrac{1}{4}h$.
We can determine the half life for both the processes happening simultaneously as follows,
$\dfrac{1}{{{t_3}}} = \dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}}$.
$\Rightarrow \dfrac{1}{{{t_3}}} = \dfrac{1}{{\dfrac{1}{2}}} + \dfrac{1}{{\dfrac{1}{4}}}$
$\Rightarrow {t_3} = \dfrac{1}{6}h$
We have the formula for the number of nuclei present in the sample after n half lives is,
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^n}$
Here, ${N_0}$ is the initial number of nuclei present in the sample.
We see in the first half hour, there can be only one half life of $\dfrac{1}{2}h$. Therefore, the number of nuclei present in the sample after first half our is,
${N_{t = 0\,h\,{\text{to }}\dfrac{1}{2}h}} = {N_0}{\left( {\dfrac{1}{2}} \right)^1}$
$\Rightarrow {N_{t = 0\,h\,{\text{to }}\dfrac{1}{2}h}} = {N_0}\left( {\dfrac{1}{2}} \right)$
We have given that for the next hour, the decay is followed by the process B. In the process B, the half life is $\dfrac{1}{4}h$. Therefore, in one hour there will be 4 half lives.
We can calculate the number of nuclei present after next one hour as follows,
${N_{t = 0h\,{\text{to}}\,1\dfrac{1}{2}h}} = {N_{t = \dfrac{1}{2}h}} \times {N_{t = 1\,h}}$
$\Rightarrow {N_{t = 0h\,{\text{to}}\,1\dfrac{1}{2}h}} = {N_0}{\left( {\dfrac{1}{2}} \right)^1} \times {N_0}{\left( {\dfrac{1}{2}} \right)^4}$
$\Rightarrow {N_{t = 0h\,{\text{to}}\,1\dfrac{1}{2}h}} = {N_0}{\left( {\dfrac{1}{2}} \right)^5}$
We have given that for the half hour, the decay is followed by both the processes A and B. In both the processes happening simultaneously, the half life is $\dfrac{1}{6}h$. Therefore, in half an hour there will be 3 half lives.
We can calculate the total number of nuclei present after next half hour as follows,
${N_{t = 0h\,{\text{to}}\,2h}} = {N_{t = 0h\,\,{\text{to }}1\,\dfrac{1}{2}h}} \times {N_{t = \dfrac{1}{2}h}}$
$\Rightarrow {N_{t = 0h\,{\text{to}}\,2h}} = {N_0}{\left( {\dfrac{1}{2}} \right)^5} \times {N_0}{\left( {\dfrac{1}{2}} \right)^3}$
$\Rightarrow {N_{t = 0h\,{\text{to}}\,2h}} = {N_0}{\left( {\dfrac{1}{2}} \right)^8}$

Therefore, the total number of nuclei present after the end of 2 hours is ${N_0}{\left( {\dfrac{1}{2}} \right)^8}$.

Note:
Do not forget to multiply the previous number of nuclei present in the sample while determining the number of nuclei remaining in the sample for the next process. When the different processes of decay happen simultaneously, the resultant half life of the sample decreases as we have shown in the above solution. To solve such types of questions, students should remember all the formulae relating to the decay equation.