Two radioactive elements A and B initially ha | vedclass

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Question

(C) The half-life of element $A$ is given by $T_{1/2}(A) = \frac{\ln 2}{\lambda_A}$.The mean life (average life) of element $B$ is given by $	au(B) =

Vedclass · Accepted Answer

$\lambda_A = \lambda_B \ln 2$

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(C) The half-life of element $A$ is given by $T_{1/2}(A) = \frac{\ln 2}{\lambda_A}$.The mean life (average life) of element $B$ is given by $	au(B) = \frac{1}{\lambda_B}$.According to the problem,the half-life of $A$ is equal to the mean life of $B$:$T_{1/2}(A) = 	au(B)$Substituting the expressions:$\frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B}$Rearranging the equation to solve for $\lambda_A$:$\lambda_A = \lambda_B \ln 2$Therefore,the correct relation is $\lambda_A = \lambda_B \ln 2$.

Two radioactive elements $A$ and $B$ initially have the same number of atoms. The half-life of $A$ is equal to the mean life of $B$. If $\lambda_A$ and $\lambda_B$ are the decay constants of $A$ and $B$ respectively,then choose the correct relation from the given options.

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