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} $