Question

Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At the time $ t = 0 $ , the number of P species are $ 4{N_0} $ ​ and that of Q are $ {N_0} $ ​. The half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. Initially, there are no nuclei of R present in the sample. When the number of nuclei of P and Q is equal, the number of nuclei of R present in the sample would be
A) $ 3{N_0} $
B) $ 9{N_0}/2 $ ​​
C) $ 5{N_0}/2 $
D) $ 2{N_0} $ ​

Answer 1

Hint : In this solution, we will first determine the time at which there will be an equal amount of P and Q species in the sample. After we find this time, we will determine the number of nuclei of R present in the sample at this time.

Formula used: In this solution, will use the following formulae,
 $\Rightarrow N = {N_0}{e^{ - kt}} $ where $ N $ is the initial number of nuclei present in the compound at the time $ t $ , $ {N_0} $ is the initial number, $ k = \ln 2/{t_{1/2}} $ decays constant, $ {t_{1/2}} $ is the half-life.

Complete step by step answer
We’ve been given that two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R and that initially there is no nucleus of R present in the compound. Let us start by finding the time where there will be an equal amount of P and Q nuclei present in the compound.
Since $ {N_P} = {N_Q} $ , we can write
 $\Rightarrow 4{N_0}{e^{\dfrac{{ - \ln 2t}}{1}}} = {N_0}{e^{\dfrac{{ - \ln 2t}}{2}}} $
 $ \Rightarrow \dfrac{{4{N_0}}}{{{2^{t/1}}}} = \dfrac{{{N_0}}}{{{2^{t/2}}}} $
Solving for $ t $ , we get
 $\Rightarrow \dfrac{4}{{{2^{t/2}}}} = 1 $ or $ t = 4\,\min $ .
So, at 4 minutes, P and Q will have equal amounts left over. Let us find the amount of R at 4 mins using the equation $ N = {N_0}{e^{ - kt}} $ . Since both P and Q decay into R, the amount of P converted into R at 4 minutes will be
 $\Rightarrow {N_{P \to R}} = 4{N_0} - \dfrac{{4{N_0}}}{{{2^{4/1}}}} $
 $ \Rightarrow {N_{P \to R}} = \dfrac{{15{N_0}}}{4} $
Similarly, the number of R nuclei due to Q will be
 $\Rightarrow {N_{Q \to R}} = {N_0} - \dfrac{{4{N_0}}}{{{2^{4/2}}}} $
 $ \Rightarrow {N_{Q \to R}} = \dfrac{{3{N_0}}}{4} $
So, the total number of R nuclei will be
 $\Rightarrow {N_R} = \dfrac{{15{N_0}}}{4} + \dfrac{{3{N_0}}}{4} $
 $ \Rightarrow {N_R} = \dfrac{{9{N_0}}}{2} $
Which corresponds to option (B).

Note
Here, while calculating the time of decay, we must be careful that the out units will have the same units as the unit of half time of the material. The only P and Q that can have an equal amount of material left over despite P having more nuclei is only if P has a smaller half-life than Q.