# The actively of a sample of radioactive material is A1 at time t1 and A2 at time t2 (t2 > t1). Its mean life is T such that…

Q: The actively of a sample of radioactive material is A1 at time t1 and A2 at time t2 (t2 > t1). Its mean life is T such that

(a) A1 t1 = A2 t2

(b) $\displaystyle \frac{A_1 – A_2}{t_2 – t_1} = constant$

(c) $\displaystyle A_2 = A_1 e^{(t_1 – t_2)/T}$

(d) $\displaystyle A_2 = A_1 e^{(t_1 /T t_2)}$

Ans: (c)

Sol:

$\displaystyle A_1 = (\frac{dN}{dt})_1 = \lambda N_1 = \lambda N_0 e^{-\lambda t_1}$ …(i)

$\displaystyle A_2 = (\frac{dN}{dt})_2 = \lambda N_2 = \lambda N_0 e^{-\lambda t_2}$ …(ii)

On dividing (ii) by (i)

$\displaystyle \frac{A_2}{A_1} = \frac{e^{-\lambda t_2}}{e^{-\lambda t_1}}$

$\displaystyle \frac{A_2}{A_1} = e^{(t_1 – t_2)\lambda}$

$\displaystyle A_2 = A_1 e^{(t_1 – t_2)\lambda}$

As we know that , λ = 1/T

$\displaystyle A_2 = A_1 e^{(t_1 – t_2)/T}$