Q: The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20^{th} of the radon sample will remain undecayed, is (given log_{10} *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