For a radioactive material, its activity and rate of change of its activity are defined as and where is the number of nuclei at time . Two radioactive sources (mean life ) and (mean life ) have the same activity at . Their rates of change of activi

Question

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 [2015]

Smart Achievers · Accepted Answer

(2)

$Activity,$

$A = - \frac{d N}{d t}$

$∴$ $R = N_{0} λ^{2} e^{- λ t} = (N_{0} λ) λ e^{- λ t} = A_{0} λ e^{- λ t}$ $[∵ A_{0} = N_{0} λ]$

$∴$ $\frac{R_{P}}{R_{Q}} = \frac{λ_{P} e^{- λ_{P} t}}{λ_{Q} e^{- λ_{Q} t}} = \frac{λ_{P}}{λ_{Q}} \times \frac{e^{λ_{Q} t}}{e^{λ_{P} t}} = \frac{2 τ}{τ} \times \frac{e^{\frac{2 τ}{2 τ}}}{e^{\frac{2 τ}{τ}}} = \frac{2}{e} = \frac{n}{e}$

$∴$ $n = 2$

Solution

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Solution

Ans. (2) Activity, R=-dAdt=-ddt[-dNdt]=d2Ndt2=d2dt2(N0e-λt) ∴ R=N0λ2e-λt=(N0λ)λe-λt=A0λe-λt [∵ A0=N0λ] ∴ RPRQ=λPe-λPtλQe-λQt=λPλQ×eλQteλPt=2ττ×e2τ2τe2ττ=2e=ne ∴ n=2

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