JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Modern Physics

A radioactive sample is undergoing

α

decay. At any time

t_{1}

, its activity is

A

and another time

t_{2}

, the activity is

\frac{A}{5}

. What is the average life time for the sample?

(A) $\frac{t _{2} - t _{1}}{l n 5}$ ✓ Correct
(B) $\frac{l n ( t _{2} + t _{1} )}{2}$
(C) $\frac{t _{1} - t _{2}}{l n 5}$
(D) $\frac{l n 5}{t _{2} - t _{1}}$

Answer: (A)

Solution

Activity falls off as

A (t) = A_{0} e^{- λ t}

, and the two readings are

A

t_{1}

and

\frac{A}{5}

t_{2}

, so dividing the later reading by the earlier one,

\frac{1}{5} = e^{- λ (t_{2} - t_{1})} \Rightarrow λ (t_{2} - t_{1}) = ln 5

The mean or average life of a sample is the reciprocal of the decay constant,

τ = \frac{1}{λ}

, so the average life time is

\frac{t _{2} - t _{1}}{ln 5}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option

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