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# Radioactive nuclei P and Q disintegrate into R with half-lives 1 month and 2 months respectively. At time $t = 0$, number of nuclei of each P and Q is $x$. Time at which rate of disintegration of P and Q are equal, number of nuclei of R is:

Last updated date: 11th Jun 2024
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Hint: In this question the rate of reaction for P and Q is the same that means the disintegration constant for nuclei P and Q are equal. For radioactive disintegration the change of number of nuclei is differentiated with respect to time.

Complete step by step answer:
Let us assume that disintegration constant is $\lambda$. Since the rate of disintegration is same for P and Q so we can write,
$\Rightarrow {\lambda _1}{N_1} = {\lambda _2}{N_2}$
We can write the above expression as,
$\Rightarrow {\lambda _1}{N_0}{e^{ - {\lambda _1}t}} = {\lambda _2}{N_0}{e^{ - {\lambda _2}t}}$
Now, we divide both sides of the above equation by ${\lambda _2}$.
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = {e^{{\lambda _1}t}} - {e^{ - {\lambda _2}t}}$
After simplification of the above equation we get,
$\Rightarrow t = \dfrac{{\ln \dfrac{{{\lambda _1}}}{{{\lambda _2}}}}}{{{\lambda _1} - {\lambda _2}}}$
And now in terms of the half life ${T_1}$ and ${T_2}$, we can write the equation as,
$\Rightarrow t = \dfrac{{\ln \dfrac{{{T_2}}}{{{T_1}}}}}{{\ln 2\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)}}$
Now, we substitute the values in the above equation that is ${T_1} = 1\;{\text{month}}$ and ${T_2} = 2\;{\text{months}}$.
$\Rightarrow t = \dfrac{{\ln 2}}{{\ln 2\left( {1 - \dfrac{1}{2}} \right)}}$
After simplification of the above equation we get,
$t = 2\;{\text{month}}$
In two months, P disintegrates 2 half-lives that is $75\%$.
Now we convert $75\%$ Into fraction as
$\Rightarrow 75\% x = \dfrac{{3x}}{4}$
In two months, Q disintegrates to 1 half-life, so it is $50\%$.
Now we convert $50\%$ into fraction as,
$\Rightarrow 50\% x = \dfrac{x}{2}$
As we know that the amount of R formed can be calculated by adding the disintegration of P and the disintegration of Q.
Add disintegration of P and disintegration of Q.
$\Rightarrow {\text{Amount of R}} = \dfrac{{3x}}{4} + \dfrac{x}{4}$
After simplification we get,
$\Rightarrow {\text{Amount of R}} = \dfrac{{5x}}{4}$

Therefore, the number of nuclei of R is $\dfrac{{5x}}{4}$.

Note: In this question one thing should be noted that the rate of disintegration of P and Q is the same. The amount of R formed by disintegration of P and disintegration of Q is added in this case. Don’t forget to convert the percentage disintegration into fractions for getting the answer in terms of $x$.