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

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(A) When a radioactive material decays by two simultaneous processes,the effective decay constant $\lambda$ is given by $\lambda = \lambda_1 + \lambda_2$.Since the half-life $T$ is related to the decay constant by $T = \frac{\ln 2}{\lambda}$,the effective half-life $T$ is given by $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$.Given $T_1 = 1620$ years and $T_2 = 810$ years,the effective half-life is:$T = \frac{T_1 T_2}{T_1 + T_2} = \frac{1620 	imes 810}{1620 + 810} = \frac{1620 	imes 810}{2430} = 540$ years.We want to find the time $t$ after which $\frac{1}{4}$ of the material remains.The law of radioactive decay states that $N(t) = N_0 \left(\frac{1}{2}ight)^{t/T}$.Setting $N(t) = \frac{1}{4} N_0$,we get $\frac{1}{4} = \left(\frac{1}{2}ight)^{t/T}$,which implies $\left(\frac{1}{2}ight)^2 = \left(\frac{1}{2}ight)^{t/T}$.Thus,$\frac{t}{T} = 2$,so $t = 2T = 2 	imes 540 = 1080$ 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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