The activity of radioactive sample is measured as number of count per minute at t=0 and N0e2 counts per minute at t=10 minutes. The time in minute at which activity reduce to half its initial value is

N=N0e−λt⇒N0e2=N0e10λ⇒10λ=2⇒λ=15T1/2=ln2λ=5ln2

The activity of radioactive sample is measured as number of count per minute at t=0 and N0e2 counts per minute at t=10 minutes. The time in minute at which activity reduce to half its initial value is

N=N0e−λt⇒N0e2=N0e10λ⇒10λ=2⇒λ=15T1/2=ln2λ=5ln2

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The activity of radioactive sample is measured as number of count per minute at $t = 0$ and $\frac{N_{0}}{e^{2}}$ counts per minute at $t = 10$ minutes. The time in minute at which activity reduce to half its initial value is

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$5 \ln 2$

$\frac{\ln 2}{5}$

$\ln 4$

$\frac{5}{2} \ln 2$

answer is C.

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Detailed Solution

$N = N_{0} e^{- λ t} \Rightarrow \frac{N_{0}}{e^{2}} = \frac{N_{0}}{e^{10 λ}} \Rightarrow 10 λ = 2 \Rightarrow λ = \frac{1}{5}$

$T_{1 / 2} = \frac{\ln 2}{λ} = 5 \ln 2$

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The activity of radioactive sample is measured as number of count per minute at t=0 and N0e2 counts per minute at t=10 minutes. The time in minute at which activity reduce to half its initial value is

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The activity of radioactive sample is measured as number of count per minute at $t = 0$ and $\frac{N_{0}}{e^{2}}$ counts per minute at $t = 10$ minutes. The time in minute at which activity reduce to half its initial value is