Question

Two Radioactive nuclei P and Q.In a given sample decay into a stable nucleus R. At time t=0, number of P species are 4 and that of Q are. Half-life of P (for conversion to R) is minute whereas that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of the nuclei of R present in the sample would be:
(A) $ 3{{N}_{0}} $
(B) $ \dfrac{9{{N}_{0}}}{2} $
(C) $ \dfrac{5{{N}_{0}}}{2} $
(D) $ 2{{N}_{0}} $

Answer 1

Hint: Using the decay formula i.e.
 $ \begin{align}
 &\Rightarrow N=\dfrac{{{N}_{0}}}{{{2}^{{}^{T}/{}_{{{t}_{1/2}}}}}}\text{ } \\
 & \text{where N is the number of nuclei left after decay} \\
 & {{\text{N}}_{0}}\text{ is the number of nuclei present before decay} \\
 & T\text{ is the time after which the nuclei number is to be calculated} \\
 & {{\text{t}}_{1/2}}\text{ is the half life of the nuclear particle} \\
\end{align} $
From this formula we can find the no. of nuclei left and then subtract the left nuclei from the total to get the new population of R nuclei formed.

Complete step by step answer:
Initially,No. of nuclei of P = 4N0
No. of nuclei of Q = N0
 $ \begin{align}
 &\Rightarrow {{t}_{1/2}}\text{ of P = 1 min} \\
 &\Rightarrow {{t}_{1/2}}\text{ of Q = 2 min} \\
\end{align} $
After time T, the number of nuclei left of P and Q are equal. So,
 $ \begin{align}
 &\Rightarrow \dfrac{4{{N}_{0}}}{{{2}^{{}^{T}/{}_{1}}}}=\dfrac{{{N}_{0}}}{{{2}^{{}^{T}/{}_{2}}}} \\
 &\Rightarrow \dfrac{4}{{{2}^{T}}}=\dfrac{1}{{{2}^{{}^{T}/{}_{2}}}} \\
 &\Rightarrow {{2}^{{}^{T}/{}_{2}}}=4\text{T} \\
\end{align} $
Taking squares on both sides we get;
 $ \begin{align}
 &\Rightarrow {{\text{2}}^{T}}=16 \\
 &\therefore,T=4\min \\
\end{align} $
Now after time T = 4 min, the number of nuclei of P left are:
 $ \begin{align}
 &\Rightarrow {{N}_{P}}=\dfrac{4{{N}_{0}}}{{{2}^{{}^{4}/{}_{1}}}} \\
 &\Rightarrow\dfrac{4{{N}_{0}}}{16} \\
 &\Rightarrow {{N}_{P}}=\dfrac{{{N}_{0}}}{4} \\
\end{align} $
Similarly, the number of nuclei left of Q can be calculated as:
 $ \begin{align}
 &\Rightarrow {{\text{N}}_{Q}}=\dfrac{{{N}_{0}}}{{{2}^{{}^{4}/{}_{2}}}} \\
 &\Rightarrow\dfrac{{{N}_{0}}}{{{2}^{2}}} \\
 &\Rightarrow {{N}_{Q}}=\dfrac{{{N}_{0}}}{4} \\
\end{align} $
Population of nuclei of R formed is the sum of no. of nuclei decayed of P and Q i.e.
 $ \begin{align}
 &\Rightarrow {{\text{N}}_{R}}=\left( 4{{N}_{0}}-{{N}_{P}} \right)+\left( {{N}_{0}}-{{N}_{Q}} \right) \\
 &\Rightarrow\left( 4{{N}_{0}}-\dfrac{{{N}_{0}}}{4} \right)+\left( {{N}_{0}}-\dfrac{{{N}_{0}}}{4} \right) \\
 &\Rightarrow\left( \dfrac{16{{N}_{0}}-{{N}_{0}}}{4} \right)+\left( \dfrac{4{{N}_{0}}-{{N}_{0}}}{4} \right) \\
 &\Rightarrow\dfrac{15{{N}_{0}}}{4}+\dfrac{3{{N}_{0}}}{4} \\
 &\Rightarrow\dfrac{18{{N}_{0}}}{4} \\
 &\Rightarrow {{N}_{R}}=\dfrac{9{{N}_{0}}}{2} \\
\end{align} $.

Note:
The law of radioactive decay is probably the most important law of radioactivity. Keep in mind that the nucleus of R formed is the sum of the nuclei decayed of P and Q and do not confuse it with the nuclei left of P and Q during calculations. Remember that Radioactive decay is not a reversible process.
Radioactive decay is very important for a wide range of human activities, from medicine to electricity production and beyond, and also to astronomers.