Radioactive nuclei P and Q disintegrate into | vedclass

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(B) Given: Half-lives $T_P = 1$ month,$T_Q = 2$ months.Decay constants: $\lambda_P = \frac{\ln 2}{T_P} = \ln 2$ and $\lambda_Q = \frac{\ln 2}{T_Q} = \

Vedclass · Accepted Answer

$1.25$

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(B) Given: Half-lives $T_P = 1$ month,$T_Q = 2$ months.Decay constants: $\lambda_P = \frac{\ln 2}{T_P} = \ln 2$ and $\lambda_Q = \frac{\ln 2}{T_Q} = \frac{\ln 2}{2}$.Rate of disintegration $R = \lambda N$,where $N = N_0 e^{-\lambda t}$.At time $t$,rates are equal: $\lambda_P x e^{-\lambda_P t} = \lambda_Q x e^{-\lambda_Q t}$.$(\ln 2) e^{-(\ln 2) t} = (\frac{\ln 2}{2}) e^{-(\frac{\ln 2}{2}) t}$.$1 = \frac{1}{2} e^{(\ln 2 - \frac{\ln 2}{2}) t} = \frac{1}{2} e^{(\frac{\ln 2}{2}) t}$.$2 = e^{(\frac{\ln 2}{2}) t} \implies \ln 2 = \frac{\ln 2}{2} t \implies t = 2$ months.Number of nuclei remaining: $N_P = x e^{-\lambda_P t} = x e^{-(\ln 2)(2)} = x(2)^{-2} = \frac{x}{4} = 0.25x$.$N_Q = x e^{-\lambda_Q t} = x e^{-(\frac{\ln 2}{2})(2)} = x(2)^{-1} = \frac{x}{2} = 0.5x$.Total nuclei disintegrated into $R$: $N_R = (x - 0.25x) + (x - 0.5x) = 0.75x + 0.5x = 1.25x$.

Radioactive nuclei $P$ and $Q$ disintegrate into $R$ with half-lives $1$ month and $2$ months respectively. At time $t=0$,the number of nuclei of each $P$ and $Q$ is $x$. At the time when the rates of disintegration of $P$ and $Q$ are equal,the number of nuclei of $R$ is ........ $x$.

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