JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Modern Physics

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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

lo g_{10} 1.88 = 0.274

)

(A) $0.02 min^{- 1}$
(B) $2.7 min^{- 1}$
(C) $0.063 min^{- 1}$ ✓ Correct
(D) $6.3 min^{- 1}$

Answer: (C)

Solution

The activity follows

A = A_{0} e^{- λ t}

.

Given

A_{0} = 4250

dpm,

A = 2250

dpm at

t = 10

min:

2250 = 4250 e^{- λ (10)}

e^{- 10 λ} = \frac{2250}{4250} = \frac{9}{17}

10 λ = ln (\frac{4250}{2250}) = ln (1.888...) \approx 0.635

Therefore

λ \approx 0.063 min^{- 1}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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