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QJEEAdv23Pha3db: JEE Advanced 2023 Physics - Class 12 JEEAdv23Pha3db Solution: In a radioactive decay process, the activity is defined as ,…

In a radioactive decay process, the activity is defined as A=−dNdtA=-\frac{d N}{d t} , where N(t)N(t) is the number of radioactive nuclei at time tt . Two radioactive sources, S1S_1 and S2S_2 have same activity at time t=0t=0 . At a later time, the activities of S1S_1 and S2S_2 are A1A_1 and A2A_2 , respectively. When S1S_1 and S2S_2 have just completed their 3rd 3^{\text {rd }} and 7th 7^{\text {th }} half-lives, respectively, the ratio A1/A2A_1 / A_2 is _________.

Step-by-Step Verified Solution

A1=A0e−λ1t1A_1=A_0 e^{-\lambda_1 t_1}

also A2=A0e−λ2t2A_2=A_0 e^{-\lambda_2 t_2}

At t1=3ln⁡2λ1,t_1=\frac{3 \ln 2}{\lambda_1},

A1=A0e−λ13ln⁡2λ1A_1=A_0 e^{-\lambda_1 \frac{3 \ln 2}{\lambda_1}}

=A0e−3ln⁡2=A_0 e^{-3 \ln 2} ...........(i)

Similarly, at

amp;t2=7ln⁡2λ2,amp;A2=A0e−λ27ln⁡2λ2amp;=A0e−7ln⁡2...........(ii)\begin{aligned}&t_2= \frac{7 \ln 2}{\lambda_2}, \\\\& A_2=A_0 e^{-\lambda_2 \frac{7 \ln 2}{\lambda_2}} \\\\& =A_0 e^{-7 \ln 2} ...........(ii)\end{aligned}

From (i) and (ii)

A1A2=A0e−3ln⁡2A0e−7ln⁡2=2−32−7=12−4=24=16\frac{A_1}{A_2}=\frac{A_0 e^{-3 \ln 2}}{A_0 e^{-7 \ln 2}}=\frac{2^{-3}}{2^{-7}}=\frac{1}{2^{-4}}=2^4=16

∴A1A2=16\therefore \frac{A_1}{A_2}=16

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