For a radioactive material, its activity A and rate of change of its activity R are defined as $A = - \frac{d N}{d t}$ and $R = - \frac{d A}{d t}$ , where $N (t)$ is the number of nuclei at time t. Two radioactive sources P (mean life $τ$ ) and Q (mean life 2 $τ$ ) have the same activity at $t = 0$ . Their rates of change of activities at $T = 2 τ$ are $R_{P}$ and $R_{Q}$ , respectively. If $\frac{R_{P}}{R_{Q}} = \frac{n}{e}$ , then the value of n is

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Detailed Solution

$N = N_{0} e^{- λ t} A = - \frac{d N}{d t} =λ N_{0} e^{- λ t} R = - \frac{d A}{d t} = λ^{2} N_{0} e^{- λ t}$
P Mean life $τ$

Q Mean life $2 τ$
$\to$ initially At time ‘t’=0

$A_{P} = A_{Q}$
$\Rightarrow λ_{1} N_{1} e^{- λ (0)} = λ_{2} N_{2} e^{- λ (0)}$
$\Rightarrow λ_{1} N_{1} = λ_{2} N_{2}$

$\to$ After $t = 2 τ$
Ratio of rate of change of activities gives,

$\begin{array}{l} \frac{R_{P}}{R_{Q}} = \frac{λ_{1}^{2} N_{1} e^{- λ_{1} t}}{λ_{2}^{2} N_{2} e^{- λ_{2} t}} \\ \Rightarrow \frac{R_{P}}{R_{Q}} = \frac{λ_{1}}{λ_{2}} \times \frac{e^{- \frac{1}{T} \times 2 T}}{e^{- \frac{1}{2 T} \times 2 T}} \\ \Rightarrow \frac{R_{P}}{R_{Q}} = \frac{τ_{2}}{τ_{1}} \frac{e^{- 2}}{e^{- 1}} = \frac{2 τ}{τ} \times e^{- 1} \\ \Rightarrow \frac{R_{P}}{R_{Q}} = \frac{2}{e} \end{array}$