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(A) For a radioactive nucleus with multiple decay modes,the total decay constant $\lambda_B$ is the sum of the individual decay constants: $\lambda_B

Vedclass · Accepted Answer

$3 / 7$

Vedclass · Answer

(A) For a radioactive nucleus with multiple decay modes,the total decay constant $\lambda_B$ is the sum of the individual decay constants: $\lambda_B = \lambda_1 + \lambda_2$.Since the decay constant $\lambda$ is related to the half-life $	au$ by $\lambda = \ln(2) / 	au$,we can write the relationship as $1 / 	au_B = 1 / 	au_1 + 1 / 	au_2$.Given that if decay mode $2$ were absent,the half-life is $	au_1 = 	au_A / 2$.Given that if decay mode $1$ were absent,the half-life is $	au_2 = 3 	au_A$.Substituting these into the formula for the actual half-life $	au_B$:$1 / 	au_B = 1 / (	au_A / 2) + 1 / (3 	au_A)$$1 / 	au_B = 2 / 	au_A + 1 / (3 	au_A)$$1 / 	au_B = (6 + 1) / (3 	au_A) = 7 / (3 	au_A)$Therefore,$	au_B = (3 / 7) 	au_A$,which implies $	au_B / 	au_A = 3 / 7$.

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