A freshly prepared sample of a radioisotope of half-life 1386 s has activity disintegrations per second. Given that the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparatio

Question

A freshly prepared sample of a radioisotope of half-life 1386 s has activity $10^{3}$ disintegrations per second. Given that $l n 2 = 0.693,$ the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is [2013]

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(4)

$For a radioactive decay$

$N = N_{0} (1 - e^{- λ t})$

$∴$ $\frac{N}{N_{0}} = e^{- λ t}$ $∴$ $1 - \frac{N}{N_{0}} = 1 - e^{- λ t}$

$∴$ $\frac{N_{0} - N}{N_{0}} = 1 - e^{- \frac{0.693}{t_{1 / 2}} t} = 1 - e^{- 0.04} = 1 - (1 - 0.04)$ $[∵ e^{- x} = 1 - x, x ≪ 1]$

$% decayed \approx 0.04 \times 100 = 4 %$

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Q. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln2=0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is [2013]

Ans. (4) For a radioactive decay N=N0(1-e-λt) ∴ NN0=e-λt ∴ 1-NN0=1-e-λt ∴ N0-NN0=1-e-0.693t1/2 t=1-e-0.04=1-(1-0.04) [∵ e-x=1-x, x≪1] % decayed≈0.04×100=4%

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