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Question

# The activity of a radioactive sample falls from $$700 s^{-1}$$ to $$500 s^{-1}$$ in $$30$$ minutes. Its half life is close to :

A
$$52 min$$
B
$$72 min$$
C
$$62 min$$
D
$$66 min$$
Solution
Verified by Toppr

#### 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 lifeFrom 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.

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