The half-life of a radioactive element X is e | vedclass

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(B) Given: Half-life of $X$ $(T_{1/2, X})$ = Mean life of $Y$ $(	au_Y)$.We know that $T_{1/2, X} = \frac{\ln 2}{\lambda_X} \approx 0.693 	au_X$.Also

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$Y$ will decay faster than $X$.

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(B) Given: Half-life of $X$ $(T_{1/2, X})$ = Mean life of $Y$ $(	au_Y)$.We know that $T_{1/2, X} = \frac{\ln 2}{\lambda_X} \approx 0.693 	au_X$.Also, $	au_Y = \frac{1}{\lambda_Y}$.Since $T_{1/2, X} = 	au_Y$, we have $0.693 	au_X = 	au_Y$, which implies $	au_X > 	au_Y$.Since the decay constant $\lambda = \frac{1}{	au}$, it follows that $\lambda_Y > \lambda_X$.The rate of decay is given by $R = \lambda N$. Since the initial number of atoms $N_0$ is the same for both, the element with the larger decay constant will decay faster.Therefore, $Y$ decays faster than $X$.

The half-life of a radioactive element $X$ is equal to the mean life of another radioactive element $Y$. Initially, the number of atoms for both is the same. Then:

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