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(B) Let the initial number of radioactive atoms for both elements be $N_0$.The number of atoms remaining at time $t$ for element $X_1$ is $N_1(t) = N_

Vedclass · Accepted Answer

$\frac{1}{9 \lambda}$

Vedclass · Answer

(B) Let the initial number of radioactive atoms for both elements be $N_0$.The number of atoms remaining at time $t$ for element $X_1$ is $N_1(t) = N_0 e^{-10 \lambda t}$.The number of atoms remaining at time $t$ for element $X_2$ is $N_2(t) = N_0 e^{-\lambda t}$.The ratio of their atoms is given by $\frac{N_1}{N_2} = \frac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}} = e^{-9 \lambda t}$.We are given that this ratio becomes $\frac{1}{e} = e^{-1}$.Equating the exponents: $-9 \lambda t = -1$.Therefore,$t = \frac{1}{9 \lambda}$.

At any instant,two elements $X_1$ and $X_2$ have the same number of radioactive atoms. If the decay constants of $X_1$ and $X_2$ are $10 \lambda$ and $\lambda$ respectively,then the time when the ratio of their atoms becomes $\frac{1}{e}$ will be:

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