A radioactive substance emits n beta particles in the first 2 seconds and 0.5 n beta particles in the next 2 seconds. The mean life of the sample is

Question

A radioactive substance emits n beta particles in the first 2 seconds and 0.5 n beta particles in the next 2 seconds. The mean life of the sample is (A) 4 s (B) 2 s (C)  (2/( ln 2))s  (D)  2 ( ln 2) s

Tardigrade Admin · Accepted Answer

We have λ=(0.693/t1 / 2) Mean life τ=(1/λ)=(t1 / 2/0.693) =(2/2.303 log 2)=(2/ ln 2)

Q. A radioactive substance emits $n$ beta particles in the first $2$ seconds and $0.5 n$ beta particles in the next $2$ seconds. The mean life of the sample is

$4 s$

$2 s$

$\frac{2}{( l n 2 )} s$

$2 (ln 2) s$

We have $λ = \frac{0.693}{t _{1/2}}$ Mean life
$τ = \frac{1}{λ} = \frac{t _{1/2}}{0.693}$
$= \frac{2}{2.303 l o g _{2}} = \frac{2}{l n 2}$

Q. A radioactive substance emits n beta particles in the first 2 seconds and 0.5n beta particles in the next 2 seconds. The mean life of the sample is

4s

2s

(ln2)2​s

2(ln2)s