The disintegration rate of a certain radioactive sample at any instant is $4250$ disintegrations per minute. $10$ minutes later, the rate becomes $2250$ disintegrations per minute. The approximate decay constant is
(Take $\log_{e}1 . 88 = 0 . 63$ )

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The disintegration rate of a certain radioactive sample at any instant is $4250$ disintegrations per minute. $10$ minutes later, the rate becomes $2250$ disintegrations per minute. The approximate decay constant is

(Take $\log_{e}1 . 88 = 0 . 63$ )

Quantrex Academy · Accepted Answer

Correct answer: C. 0 . 063 ^ - 1

At t = 0 disintegration rate A _ 0 = 4250dpm At t = 10min disintegration rate A _ t = 2250dpm The activity of a sample at any time is given by, A _ t = A _ 0 e ^ - ⇒ 2250 = 4250 e ^ - λ(10) ⇒ 10λ = ln(4250)/(2250) = 1 . 88 ⇒ λ = 0 . 063 ^ - 1

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