A radioactive material decays by simultaneous | vedclass

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(A) The rate of decay is given by $\frac{-dN}{dt} = \lambda_1 N + \lambda_2 N = (\lambda_1 + \lambda_2)N$.Integrating this, we get $\ln(\frac{N}{N_0})

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(A) The rate of decay is given by $\frac{-dN}{dt} = \lambda_1 N + \lambda_2 N = (\lambda_1 + \lambda_2)N$.Integrating this, we get $\ln(\frac{N}{N_0}) = -(\lambda_1 + \lambda_2)t$.Given half-lives $T_1 = 1620 \, years$ and $T_2 = 810 \, years$, the decay constants are $\lambda_1 = \frac{\ln 2}{1620}$ and $\lambda_2 = \frac{\ln 2}{810}$.We want to find $t$ when $\frac{N}{N_0} = \frac{1}{4}$, so $\ln(\frac{1}{4}) = -(\lambda_1 + \lambda_2)t$.$-2 \ln 2 = -(\frac{\ln 2}{1620} + \frac{\ln 2}{810})t$.Dividing by $-\ln 2$, we get $2 = (\frac{1}{1620} + \frac{1}{810})t$.$2 = (\frac{1 + 2}{1620})t = \frac{3}{1620}t = \frac{1}{540}t$.Therefore, $t = 2 	imes 540 = 1080 \, years$.

$A$ 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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