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. 13.8 days
B. 16.5 days
C. 33 days
D. 76 days

Answer

B

Explanation

`((1)/(2))^(n) = (N)/(N_(0)) = (1)/(20) ` gives n = 4.32
t = nT = `4.3 xx 3.8 = 16.5` days