Q: The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20th of the radon sample will remain undecayed, is (given log10 e = 0.4343)
(A)3.8 days
(B) 16.5 days
(C) 33 days
(D) 76 days
Ans: (B)
Sol: $\large N = N_0 e^{-\lambda t}$
$\large \lambda = \frac{ln 2}{t_{1/2}} = \frac{ln2}{3.8}$
$\large \frac{N_0}{20} = N_0 e^{-\frac{ln2}{3.8}t}$
on solving we get,
t = 16.5 days