Question

A radioactive sample decays by the three modes simultaneously. Half-lives corresponding to these modes are in G.P and the half-life of the sample is 10 years. When the sample decays by the mode having the largest half-life it takes 70 years for the sample to become half. If the sample decays exclusively by the other modes possible value of the half-life is:
A. 28 years
B. 35 years
C. 14 years
D. 17.5 years

Answer 1

Hint: In order to solve this question we have to consider how much nuclei decayed in the three different modes of the three cycles which is given in the question.

Formula used:
$N=\dfrac{{{N}_{o}}}{{{2}^{n}}}$

Complete step-by-step answer:
First we see the given data provided in the question.
${{\left( {{I}_{\dfrac{1}{2}}} \right)}_{total}}=10\text{ years}.....\left( 1 \right)$
The total half-life of the sample is given in the question is 10 years as shown in the equation (1) .Now it is given in the question that the half-life of the sample is done in the three modes. This means sample decay in three different cycles.
Now the sample decay in the first cycle is 70 years. Now we know the formula for the half cycle is
$N=\dfrac{{{N}_{o}}}{{{2}^{n}}}$
Where, N = number of nuclei undecayed
${{N}_{o}}$ = initial number of the nuclei
n = number of the cycles.
For the first cycle number of the nuclei undecayed is,
$\begin{align}
  & \Rightarrow 70=\dfrac{{{N}_{o}}}{{{2}^{1}}} \\
 & \Rightarrow \dfrac{{{N}_{o}}}{2}=70\text{ years}....\left( 1 \right) \\
\end{align}$
For the second cycle number of nuclei undecayed is,
$\begin{align}
  & \Rightarrow N=\dfrac{{{N}_{o}}}{{{2}^{2}}} \\
 & \Rightarrow N=\dfrac{{{N}_{o}}}{2\times 2} \\
 & \therefore N=\left( \dfrac{{{N}_{o}}}{2} \right)\dfrac{1}{2}.....\left( 2 \right) \\
\end{align}$
Now substitute value of equation (1) in equation (2)
$\begin{align}
  & \Rightarrow N=70\times \dfrac{1}{2} \\
 & \therefore N=35years \\
\end{align}$
Now for the third cycle number of nuclei that is undecayed is,
$\begin{align}
  & \Rightarrow N=\dfrac{{{N}_{o}}}{{{2}^{3}}} \\
 & \Rightarrow N=\dfrac{{{N}_{o}}}{2\times 2\times 2} \\
 & \therefore N=\left( \dfrac{{{N}_{o}}}{2} \right)\times \dfrac{1}{4}.....\left( 3 \right) \\
\end{align}$
Now substituting value of equation (1) in equation (3)
$\begin{align}
  & \Rightarrow N=\dfrac{70}{4} \\
 & \therefore N=17.5\text{ }\operatorname{years} \\
\end{align}$
Therefore from the cycle:-2 and the cycle:-3 our correct answer is option (B) and the option (D).

So, the correct answers are “Option B and D”.

Note: While solving this question we have to consider two things.
(i) We have to find half-life of two other modes which means that there will be two options that are correct.
(ii) If we use the direct formula $\lambda ={{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}$ it won’t get solved because we cannot find the values of ${{\lambda }_{2}}\text{ and }{{\lambda }_{3}}.$