Radioactive nuclei A and B disintegrate into | vedclass

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(B) The rate of disintegration is given by $R = \lambda N$. Given $R_A = R_B$,we have $\lambda_A N_A = \lambda_B N_B$.Since $\lambda = \frac{\ln 2}{T_

Vedclass · Accepted Answer

$1.25x$

Vedclass · Answer

(B) The rate of disintegration is given by $R = \lambda N$. Given $R_A = R_B$,we have $\lambda_A N_A = \lambda_B N_B$.Since $\lambda = \frac{\ln 2}{T_{1/2}}$,we have $\left( \frac{\ln 2}{T} \right) N_A = \left( \frac{\ln 2}{2T} \right) N_B$,which simplifies to $N_B = 2N_A$.Using the decay law $N = N_0 e^{-\lambda t}$,we have $x e^{-\lambda_B t} = 2 x e^{-\lambda_A t}$.Substituting $\lambda_A = \frac{\ln 2}{T}$ and $\lambda_B = \frac{\ln 2}{2T}$,we get $e^{-\frac{\ln 2}{2T} t} = 2 e^{-\frac{\ln 2}{T} t}$.Taking the natural logarithm on both sides: $-\frac{\ln 2}{2T} t = \ln 2 - \frac{\ln 2}{T} t$.Dividing by $\ln 2$: $-\frac{t}{2T} = 1 - \frac{t}{T} \Rightarrow \frac{t}{2T} = 1 \Rightarrow t = 2T$.At $t = 2T$,$A$ has undergone $2$ half-lives,so $N_A = \frac{x}{2^2} = \frac{x}{4}$. Disintegrated $A = x - \frac{x}{4} = 0.75x$.At $t = 2T$,$B$ has undergone $1$ half-life,so $N_B = \frac{x}{2^1} = \frac{x}{2}$. Disintegrated $B = x - \frac{x}{2} = 0.5x$.The total number of nuclei of $C$ formed is the sum of disintegrated $A$ and $B$: $0.75x + 0.5x = 1.25x$.

Radioactive nuclei $A$ and $B$ disintegrate into $C$ with half-lives $T$ and $2T$. At $t = 0$,the number of nuclei of each $A$ and $B$ is $x$. The number of nuclei of $C$ when the rate of disintegration of $A$ and $B$ are equal is:

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