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(A) When a radioactive material decays by two simultaneous processes,the effective decay constant is $\lambda = \lambda_1 + \lambda_2$.Since $\lambda

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$1080$

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(A) When a radioactive material decays by two simultaneous processes,the effective decay constant is $\lambda = \lambda_1 + \lambda_2$.Since $\lambda = \frac{\ln 2}{T}$,the effective half-life $T$ is given by $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$.Substituting the given values $T_1 = 1620 \, years$ and $T_2 = 810 \, years$:$\frac{1}{T} = \frac{1}{1620} + \frac{1}{810} = \frac{1 + 2}{1620} = \frac{3}{1620} = \frac{1}{540}$.Thus,the effective half-life $T = 540 \, years$.The amount of material remaining is given by $N = N_0 (\frac{1}{2})^n$,where $n = \frac{t}{T}$ is the number of half-lives.We want $N = \frac{1}{4} N_0$,so $(\frac{1}{2})^n = \frac{1}{4} = (\frac{1}{2})^2$,which implies $n = 2$.Therefore,$t = n 	imes T = 2 	imes 540 = 1080 \, years$.

$A$ radioactive material decays by simultaneous emission of two particles with respective half-lives $1620 \, years$ and $810 \, years$. The time (in years) after which one-fourth of the material remains is

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