# 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.