A radioactive sample is undergoing α decay. At any time t1, its activity is A and another time t 2, the activity is ( A /5) . What is the average life time for the sample?

Question

A radioactive sample is undergoing α decay. At any time t1, its activity is A and another time t 2, the activity is ( A /5) . What is the average life time for the sample? (A) (ℓ n 5/ t 2- t 1) (B) ( t 1- t 2/ℓ n 5) (C) ( t 2- t 1/ ln 5) (D) (ℓ n ( t 2+ t 1)/2)

Tardigrade Admin · Accepted Answer

Let initial activity be A 0 A=A0 e-λ t1 (i) ( A /5)= A 0 e -λ t 2 ....(ii) (i) ÷ (ii) 5= e λ(t2-t1) λ=(ℓ n 5/ t 2- t 1)=(1/τ) τ=( t 2- t 1/ℓ n 5)

Q. 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?

$\frac{ℓ n 5}{t _{2} - t _{1}}$

$\frac{t _{1} - t _{2}}{ℓ n 5}$

$\frac{t _{2} - t _{1}}{l n 5}$

$\frac{ℓ n ( t _{2} + t _{1} )}{2}$

Let initial activity be $A_{0}$
$A = A_{0} e^{- λ t_{1}} (i)$
$\frac{A}{5} = A_{0} e^{- λ t_{2}} .... (ii)$
(i) $\div$ (ii)
$5 = e^{λ (t_{2} - t_{1})}$
$λ = \frac{ℓ n 5}{t _{2} - t _{1}} = \frac{1}{τ}$
$τ = \frac{t _{2} - t _{1}}{ℓ n 5}$

Q. A radioactive sample is undergoing α decay. At any time t1​, its activity is A and another time t2​, the activity is 5A​. What is the average life time for the sample?

t2​−t1​ℓn5​

ℓn5t1​−t2​​

ln5t2​−t1​​

2ℓn(t2​+t1​)​