The activity of a radioactive sample falls from $$700 s^{-1}$$ to $$500 s^{-1}$$ in $$30$$ minutes. Its half life is close to :
Correct option is B. $$62 min$$
From equation of radioactivity we know
$$A_t = A_0 e^{-\lambda t}$$
Where $$A_t = $$ Amount of material at time 't'
$$A_0 =$$ Amount of substance at $$t = 0$$
$$\lambda = $$ decay constant
$$t = $$ time.
Taking log we have
$$\ln \left[\dfrac{A_0}{A_t} \right] = \lambda t$$
For half life $$A_t = \dfrac{A_0}{2}$$.
$$\Rightarrow \ln \ 2 = \lambda t_{1/2}$$ __(A) $$t_{1/2} = $$ half life
From given condition.
$$\ln \left[\dfrac{700}{500} \right] = \lambda (30 \ min)$$ __(B)
$$(A) - (B)$$
$$\dfrac{\ln 2}{\ln [7/5]} = \dfrac{t_{1/2}}{30}$$
$$\Rightarrow t_{1/2} = \dfrac{\ln 2 \times 30}{\ln [7/5]} \Rightarrow t_{1/2} = \dfrac{0.693 \times 30}{0.336}$$
$$\Rightarrow t_{1/2} = 61.8 min$$
option (B) is correct.