JEE Challenger
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PhyMedAdv 2015 • Paper 2

Ratio of Rates of Change of Activity for Two Radioactive Sources

For a radioactive material, its activity AA and the rate of change of activity RR are defined as A=−dNdtA = -\frac{dN}{dt} and R=−dAdtR = -\frac{dA}{dt}, where N(t)N(t) is the number of nuclei present at time tt. Two radioactive sources PP (mean life τ\tau) and QQ (mean life 2τ2\tau) have equal activities at t=0t = 0. Their rates of change of activities at t=2τt = 2\tau are RPR_P and RQR_Q, respectively. If RPRQ=ne\frac{R_P}{R_Q} = \frac{n}{e}, then the value of nn is

Official Numerical Answer2

Topics & Concepts

NucleiRadioactivity

Step-by-Step Solution

According to the radioactive decay law, the number of undecayed nuclei at time tt is given by: N(t)=N0e−λtN(t) = N_0 e^{-\lambda t} where λ\lambda is the decay constant, related to the mean life τmean\tau_{\text{mean}} by: λ=1τmean\lambda = \frac{1}{\tau_{\text{mean}}}

The activity A(t)A(t) is defined as: A(t)=−dN(t)dt=λN(t)=λN0e−λt=A0e−λtA(t) = -\frac{dN(t)}{dt} = \lambda N(t) = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t} where A0A_0 is the activity at t=0t = 0.

The rate of change of activity R(t)R(t) is defined as: R(t)=−dA(t)dt=λA(t)=λA0e−λtR(t) = -\frac{dA(t)}{dt} = \lambda A(t) = \lambda A_0 e^{-\lambda t}

For source PP:

  • Mean life τP=τ  ⟹  λP=1τ\tau_P = \tau \implies \lambda_P = \frac{1}{\tau}
  • Initial activity AP(0)=A0A_P(0) = A_0

For source QQ:

  • Mean life τQ=2τ  ⟹  λQ=12τ\tau_Q = 2\tau \implies \lambda_Q = \frac{1}{2\tau}
  • Initial activity AQ(0)=A0A_Q(0) = A_0

At time t=2τt = 2\tau, the rate of change of activity for each source is: RP=λPA0e−λP(2τ)=(1τ)A0e−(1τ)(2τ)=A0τe−2R_P = \lambda_P A_0 e^{-\lambda_P (2\tau)} = \left(\frac{1}{\tau}\right) A_0 e^{-\left(\frac{1}{\tau}\right)(2\tau)} = \frac{A_0}{\tau} e^{-2}

RQ=λQA0e−λQ(2τ)=(12τ)A0e−(12τ)(2τ)=A02τe−1R_Q = \lambda_Q A_0 e^{-\lambda_Q (2\tau)} = \left(\frac{1}{2\tau}\right) A_0 e^{-\left(\frac{1}{2\tau}\right)(2\tau)} = \frac{A_0}{2\tau} e^{-1}

Taking the ratio of RPR_P to RQR_Q: RPRQ=A0τe−2A02τe−1=2e−1=2e\frac{R_P}{R_Q} = \frac{\frac{A_0}{\tau} e^{-2}}{\frac{A_0}{2\tau} e^{-1}} = 2 e^{-1} = \frac{2}{e}

Comparing this with the given relation RPRQ=ne\frac{R_P}{R_Q} = \frac{n}{e}, we get: n=2n = 2

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