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(C) The activity of a radioactive sample at time $t$ is given by $A(t) = A_{0} e^{-\lambda t}$.At time $t_{1}$,$A = A_{0} e^{-\lambda t_{1}}$ ... $(i)

Vedclass · Accepted Answer

$\frac{t_{2}-t_{1}}{\ln 5}$

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(C) The activity of a radioactive sample at time $t$ is given by $A(t) = A_{0} e^{-\lambda t}$.At time $t_{1}$,$A = A_{0} e^{-\lambda t_{1}}$ ... $(i)$At time $t_{2}$,$\frac{A}{5} = A_{0} e^{-\lambda t_{2}}$ ... $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{A}{A/5} = \frac{A_{0} e^{-\lambda t_{1}}}{A_{0} e^{-\lambda t_{2}}}$$5 = e^{\lambda(t_{2}-t_{1})}$Taking the natural logarithm on both sides:$\ln 5 = \lambda(t_{2}-t_{1})$$\lambda = \frac{\ln 5}{t_{2}-t_{1}}$The average life time $	au$ is the reciprocal of the decay constant $\lambda$:$	au = \frac{1}{\lambda} = \frac{t_{2}-t_{1}}{\ln 5}$

$A$ radioactive sample is undergoing $\alpha$ decay. At any time $t_{1}$,its activity is $A$ and at another time $t_{2}$,the activity is $\frac{A}{5}$. What is the average life time for the sample?

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