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 `log_10 e=0.4343`)

Options

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

Answer

B

Explanation

By the formula `N=N_(0)e^(-lamdat)`
Given, `(N)/(N_(0))=(1)/(20)and lamda=(0.6931)/(3.8)implies20=e""^((0.6931xxt)/3.8)`
`" "(becauset_(1//2)=3.8days)`
On taking log both side, we get
or `log 20=(0.6931xxt)/(3.8)log_(10)e`
or `1.3010=(0.6931xxtxx0.4343)/(3.8)impliest=16.5days`