Question

Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes respectively. If initially a sample has equal number of nuclei, then after 6 minutes, the ratio of the decayed numbers of nuclei A and B will be:
A. 9:8
B. 1:8
C. 8:1
D. 3:8

Answer 1

Hint: The decay of radioactive nuclei is governed by the decay equation. The number of decayed nuclei is equal to the initial number of nuclei minus the number of nuclei left after a certain time t. By obtaining this ratio using the decay equation, we can get the required answer.

Formula used:
The decay equation for a radioactive material is given as
$N = {N_0}{e^{ - \lambda t}}$
The decay constant is given as
$\lambda = \dfrac{{0.693}}{{{T_{1/2}}}}$

Complete step by step answer:
We are given two radioactive nuclei A and B in a sample whose half-lives are given as follows:
\[
  {T_{1/2A}} = 10\min \\
  {T_{1/2B}} = 20\min \\
 \]
The decay equation for a radioactive material is given as
$N = {N_0}{e^{ - \lambda t}} = \dfrac{{{N_0}}}{{{2^{\dfrac{t}{{{t_{1/2}}}}}}}}$
Here N represents the number of nuclei of the radioactive material in a given sample at some time t while ${N_0}$ represents the number of nuclei of the radioactive material in a given sample initially at t = 0.
$\lambda $ is known as the decay constant while ${T_{\dfrac{1}{2}}}$ is the half-life of the given nuclei.
We are given that the initial concentration of these two species is the same. Therefore, we have
${N_{0A}} = {N_{0B}} = {N_0}$
We need to find out their concentrations after $t = 60\min $. So, we can use the decay equation in the following way.
$
  {N_A} = {N_{0A}}{e^{ - {\lambda _A}t}} = \dfrac{{{N_0}}}{{{2^{\dfrac{t}{{{t_{1/2A}}}}}}}} = \dfrac{{{N_0}}}{{{2^{\dfrac{{60}}{{10}}}}}} = \dfrac{{{N_0}}}{{{2^6}}} \\
  {N_B} = {N_{0B}}{e^{ - {\lambda _B}t}} = \dfrac{{{N_0}}}{{{2^{\dfrac{t}{{{t_{1/2B}}}}}}}} = \dfrac{{{N_0}}}{{{2^{\dfrac{{60}}{{20}}}}}} = \dfrac{{{N_0}}}{{{2^3}}} \\
 $
For A, the number of decayed nuclei after time t is $ = {N_0} - {N_A} = {N_0} - \dfrac{{{N_0}}}{{{2^6}}} = \dfrac{{63{N_0}}}{{64}}$
For B, the number of decayed nuclei after time t is $ = {N_0} - {N_B} = {N_0} - \dfrac{{{N_0}}}{{{2^3}}} = \dfrac{{7{N_0}}}{8}$
The ratio of the decayed numbers of nuclei A and B will be
$\dfrac{{{N_0} - {N_A}}}{{{N_0} - {N_B}}} = \dfrac{{63{N_0}}}{{64}} \times \dfrac{8}{{7{N_0}}} = \dfrac{9}{8}$
This is the required answer.

So, the correct answer is “Option A”.

Note:
The decay of radioactive nuclei in a sample is a statistical phenomenon which means that the rate of decay depends on the number of initial nuclei in a given sample. The graph follows an exponential behaviour which can be seen in the decay equation used for studying the decay of nuclei in a given radioactive sample.