The half life of 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 e = 0.4343 ) `

Options

A. 3.8 days
B. 16.5 days
C. 33 days
D. 76 days

Answer

B

Explanation

(b) By formula `N = N_0 e^(-lamda t)`
Given `(N)/(N_0) = (1)/(20)` and `lamda = (0.6931)/(3.8)`
`rArr 20 = e^((0.6931)/(3.8))`
Taking log of both sides
or `log 20 = (0.6931 xx t)/(3.8) log_10 e`
or `1.3010 = (0.6931 xx t xx 0.4343)/(3.8)`
`rArr t = 16.5 days`.