JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Modern Physics

At a given instant, say

t = 0

, two radioactive substances

A

and

B

have equal activites. The ratio

\frac{R _{B}}{R _{A}}

of their activities after time

t

itself decays with time

t

e^{- 3 t}

. If the half-life of

A

ℓ n 2

, the half-life of

B

is:

(A) $4 ℓ n 2$
(B) $\frac{l n 2}{2}$
(C) $\frac{l n 2}{4}$ ✓ Correct
(D) $2 ℓ n 2$

Answer: (C)

Solution

Activity decays as

R = R_{0} e^{- λ t}

, so the ratio

\frac{R _{B}}{R _{A}} = e^{- (λ_{B} - λ_{A}) t}

. We are told this equals

e^{- 3 t}

, so

λ_{B} - λ_{A} = 3

. Since the half-life of

A

ln 2

, its decay constant is

λ_{A} = \frac{l n 2}{l n 2} = 1

. Hence

λ_{B} = 4

, and the half-life of

B

\frac{l n 2}{λ _{B}} = \frac{l n 2}{4}

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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