A radioactive material decays by simultaneous | vedclass

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(A) The effective decay constant $\lambda$ is the sum of individual decay constants: $\lambda = \lambda_1 + \lambda_2$.Since $\lambda = \frac{\ln 2}{T

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

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