Q 12-13-089JEE MainJEE Main 2022 (26 Jul, Shift 1)Easy
The disintegration rate of a certain radioactive sample at any instant is $4250$ disintegrations per minute. $10$ minutes later, the rate becomes $2250$ disintegrations per minute. The approximate decay constant is (Take $\log_e1.88 = 0.63$)
Answer: (C) $0.063\ \text{min}^{-1}$
$R = R_0e^{-\lambda t}$:
$$\lambda = \frac1t\ln\frac{R_0}{R} = \frac{1}{10}\ln\frac{4250}{2250} = \frac{\ln1.88}{10} = 0.063\ \text{min}^{-1}$$
Solution by Sreeraj P, M.Sc Physics