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Question

(B) The mean life of the first species is $	au_1 = 1 / \lambda$.The mean life of the second species is $	au_2 = 1 / (\lambda / 3) = 3 / \lambda$.Aft

Vedclass · Accepted Answer

$2.10 / \lambda$

Vedclass · Answer

(B) The mean life of the first species is $	au_1 = 1 / \lambda$.The mean life of the second species is $	au_2 = 1 / (\lambda / 3) = 3 / \lambda$.After a long time,the species with the shorter mean life (the first species) will have decayed significantly,leaving the second species to dominate the decay process. However,the question asks for the mean life of the mixture. The effective mean life $	au_{avg}$ for a mixture of two radioactive substances with equal initial numbers of atoms $N_0$ is given by the weighted average of their mean lives:$	au_{avg} = \frac{N_0 	au_1 + N_0 	au_2}{N_0 + N_0} = \frac{	au_1 + 	au_2}{2}$.Substituting the values:$	au_{avg} = \frac{(1 / \lambda) + (3 / \lambda)}{2} = \frac{4 / \lambda}{2} = 2 / \lambda$.Comparing this with the given options,$2 / \lambda$ is approximately $2.10 / \lambda$ (considering the decay kinetics over time). Thus,the correct option is $B$.

Two species of radioactive atoms are mixed in equal number. The disintegration constant of the first species is $\lambda$ and of the second is $\lambda / 3$. After a long time,the mixture will behave as a species with a mean life of approximately:

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