Question

A radioactive sample is undergoing alpha decay. At any time t1, activity of the sample is A and at a later time t2, the activity is A3. The average life time for the sample is

A) ln2(t2−t1)
B) t2−t1ln3
C) t2−t1ln5
D) ln3(t2+t12)

Answer 1

hello friend

answer (c)

Answer 2

Given info : A radioactive sample is undergoing alpha decay. At any time t₁ , activity of the sample is A and at a later time t₂ , the activity is A/3.

To find : The average life time for the sample is ...

solution : using radioactive decay equation, $A=A_0e^{-\lambda t}$

where A is reactivity at time t, A₀ is initial reactivity and λ is decay constant.

case 1 : at time t₁ , activity of the sample is A.

∴ $A=A_0e^{-\lambda t_1}$ ...(1)

case 2 : at a later time, t₂ , the activity is A/3.

∴ $A/3=A_0e^{-\lambda t_2}$ ...(2)

from equations (1) and (2) we get,

⇒ $\frac{A}{\left(\frac{A}{3}\right)}=\frac{A_0e^{-\lambda t_1}}{A_0e^{-\lambda t_2}}$

⇒3 = $e^{\lambda (t_2-t_1)}$

⇒ln3 = λ(t₂ - t₁)

⇒1/λ = (t₂ - t₁)/ln3

we know average life , τ = 1/λ

∴ average life , τ = (t₂ - t₁)/ln3

Therefore the average life of the sample is (t₂ - t₁)/ln3