Question

The activity of a sample of a radioactive material is A, at time $t_{1}$ and $A_{2}$ at time $t_{2}$ $(t_{2} > t_{1}).$ If its mean life T, then.

• A. $A_{1}t_{1} = A_{2}t_{2}$
• B. $A_{1} - A_{2} = t_{2} - t_{1}$
• C. $A_{2} = A_{1}e^{(t_{1} - t_{2})/T}$
• D. $A_{2} = A_{1}e^{(t_{1}/t_{2})T}$

Answer 1

Answer: (C)

$A_{2} = A_{1}e^{(t_{1} - t_{2})/T}$

Answer 2

$A = A_{0}e^{- \lambda t} = A_{0}e^{- t/\tau};$

where $\tau =$mean life

So $\Rightarrow \Delta L = \frac{h}{2\pi}(n_{2} - n_{1})$ ⇒$A_{0} = \frac{A_{1}}{e^{- t_{1}/T}} = A_{1}e^{t_{1}/T}$

$\therefore A_{2} = A_{0}e^{- t/T} = (A_{1}e^{t_{1}/T})e^{- t_{2}/T} \Rightarrow A_{2} = A_{1}e^{(t_{1} - t_{2})/T}$.