Question

Half-lives of two radioactive elements A and B are $20\min$ and $40\min$ respectively. Initially, the samples have an equal number of nuclei. After $80\min$, the ratio of decayed numbers of A and B nuclei will be
$\begin{aligned} & A)1:16 \\\ & B)4:1 \\\ & C)1:4 \\\ & D)5:4 \\\ \end{aligned}$

Answer 1

Half-life of a radioactive element is related to the decay constant of that radioactive element. During the process of radioactive decay, the final number of nuclei in a radioactive element is proportional to the initial number of nuclei present in it. Combining both these facts, the final number of nuclei in both A and B are determined. Decayed number of a radioactive element refers to the difference in initial number of nuclei and final number of nuclei in the radioactive element.

Formula used:
$1){{\tau }_{\dfrac{1}{2}}}=\dfrac{0.693}{\lambda }$
$2)N={{N}_{0}}{{e}^{-\lambda t}}$

Complete answer:
We know that the half-life of a radioactive element is related to the decay constant of that radioactive element. Mathematically, half-life of a radioactive element is given by
${{\tau }_{\dfrac{1}{2}}}=\dfrac{0.693}{\lambda }$
where
${{\tau }_{\dfrac{1}{2}}}$ is then half-life of a radioactive element
$\lambda$ is the decay constant of that element
Let this be equation 1.
We also know that during the process of radioactive decay, the final number of nuclei in a radioactive element is proportional to the initial number of nuclei present in it. Mathematically, final number of nuclei in a radioactive element is given by
$N={{N}_{0}}{{e}^{-\lambda t}}$
where
$N$ is the final number of nuclei in a radioactive element, after radioactive decay
${{N}_{0}}$ is the is the initial number of nuclei in a radioactive element, before radioactive decay
$t$ is the time taken for decay process
$\lambda$ is the decay constant of that element
Let this be equation 2.
Substituting equation 1 in equation 2, we have
$N={{N}_{0}}{{e}^{-\lambda t}}\Rightarrow N={{N}_{0}}{{e}^{-\left( \dfrac{0.693}{{{\tau }_{\dfrac{1}{2}}}} \right)t}}$
Let this be equation 3.
Coming to our question, we are given that half-lives of two radioactive elements A and B are $20\min$ and $40\min$ respectively. Also given that initially, the samples have equal number of nuclei. We are required to determine the ratio of decayed numbers of A and B nuclei after $80\min$.
If ${{N}_{A}}$ represents the final number of nuclei in radioactive element A, then, using equation 3, ${{N}_{A}}$ is given by
${{N}_{A}}={{N}_{0}}{{e}^{-\left( \dfrac{0.693}{{{\tau }_{{{A}_{\dfrac{1}{2}}}}}} \right)t}}={{N}_{0}}{{e}^{-\left( \dfrac{0.693}{20} \right)80}}={{N}_{0}}{{e}^{-2.772}}$
where
${{N}_{A}}$ is the final number of nuclei in radioactive element A, after radioactive decay
${{N}_{0}}$ is the is the initial number of nuclei in radioactive element A, before radioactive decay
$t=80\min$ is the time taken for decay process (as provided)
${{\tau }_{{{A}_{\dfrac{1}{2}}}}}=20\min$ is the half-life of radioactive element A (as provided)
Let this be equation 4.
Similarly, if ${{N}_{B}}$ represents the final number of nuclei in radioactive element B, then, using equation 3, ${{N}_{B}}$ is given by
${{N}_{B}}={{N}_{0}}{{e}^{-\left( \dfrac{0.693}{{{\tau }_{{{B}_{\dfrac{1}{2}}}}}} \right)t}}={{N}_{0}}{{e}^{-\left( \dfrac{0.693}{40} \right)80}}={{N}_{0}}{{e}^{-1.386}}$
where
${{N}_{B}}$ is the final number of nuclei in radioactive element B, after radioactive decay
${{N}_{0}}$ is the is the initial number of nuclei in radioactive element B, before radioactive decay
$t=80\min$ is the time taken for decay process (as provided)
${{\tau }_{{{B}_{\dfrac{1}{2}}}}}=40\min$ is the half-life of radioactive element B (as provided)
Let this be equation 5.
Now, we know that the decayed number of a radioactive element is equal to the difference in initial number of nuclei and final number of nuclei in the radioactive element. Therefore, decayed number of element A is given by
${{N}_{A}}-{{N}_{0}}={{N}_{0}}{{e}^{-2.772}}-{{N}_{0}}={{N}_{0}}\left( 1-{{e}^{-2.772}} \right)$
where
${{N}_{A}}$ is the final number of nuclei in radioactive element A, after radioactive decay
${{N}_{0}}$ is the is the initial number of nuclei in radioactive element A, before radioactive decay
Let this be expression 6.
Similarly, decayed number of element B is given by
${{N}_{B}}-{{N}_{0}}={{N}_{0}}{{e}^{-1.386}}-{{N}_{0}}={{N}_{0}}\left( 1-{{e}^{-1.386}} \right)$
where
${{N}_{B}}$ is the final number of nuclei in radioactive element B, after radioactive decay
${{N}_{0}}$ is the is the initial number of nuclei in radioactive element B, before radioactive decay
Let this be expression 7.
Dividing expression 6 by expression 7, we have
$\dfrac{{{N}_{0}}\left( 1-{{e}^{-2.772}} \right)}{{{N}_{0}}\left( 1-{{e}^{-1.386}} \right)}=\dfrac{1-{{e}^{-2.772}}}{1-{{e}^{-1.386}}}=\dfrac{1-0.06254}{1-0.25007}=\dfrac{\text{0}\text{.93746}}{\text{0}\text{.74993}}=1.25=\dfrac{5}{4}$
Let this be expression 8.
Therefore, from expression 8, it is clear that the ratio of decayed numbers of A and B nuclei is $\dfrac{5}{4}$.
Hence, the correct answer is option D.

Note:
Students can also proceed by taking $0.693$ in equations 3, 4 and 5 as $\ln 2$. In that case, expression 6 turns out to be $\dfrac{15{{N}_{0}}}{16}$ and expression 7 turns out to be $\dfrac{3{{N}_{0}}}{4}$. Further, ratio of decayed numbers of A and B nuclei will turn out to be
$\dfrac{15{{N}_{0}}}{16}\times \dfrac{4}{3{{N}_{0}}}=\dfrac{5}{4}$.