Radioactivity - Nuclei

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Question

The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. 5 min later, the rate becomes 2700 min^-1. Then, the half-life of the sample (in min) is…….. $(\ln (\frac{475}{200}) = 0.56)$

Moderate

Solution

From the relation,
$R = R_{0} e^{- λt}$ (Here, R= activity of sample)
Substituting the values, we have
$\begin{array}{l} 2700 = 4750 e^{- 5 λ} \\ ∴ 5 λ = \ln (\frac{4750}{2700}) = 0 .56 \\ ∴ λ = 0 .113 \min^{- 1} \end{array}$
Half-life of the sample,
$t_{1 / 2} = \frac{\ln 2}{λ} = \frac{0 .693}{0 .113} = 6 .132 \min$

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Radioactivity - Nuclei

Remember concepts with our Masterclasses.

The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. 5 min later, the rate becomes 2700 min-1. Then, the half-life of the sample (in min) is…….. ln475200=0.56

From the relation,R=R0e−λt (Here, R= activity of sample)Substituting the values, we have 2700=4750e−5λ∴5λ=ln⁡47502700=0.56∴λ=0.113min−1Half-life of the sample,t1/2=ln⁡2λ=0.6930.113=6.132min

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The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. 5 min later, the rate becomes 2700 min^-1. Then, the half-life of the sample (in min) is…….. $(\ln (\frac{475}{200}) = 0.56)$