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. `16.5 days`
C. `33 days`
D. `76 days`

Answer

b

Explanation

`T_(1//2)=3.8 day`
`:. lambda =(0.693)/(t_(1//2))=(0.693)/(3.8)=0.182`
If the initial number of atoms is a =`A_(0)`, then after time t the number of atoms is `a//20`=A. We have to find `t`.
` t=(2.303)/(lambda) log.(A_(0))/(A)=(2.303)/(0.182) log.(a)/(a//20)=(2.303)/(0.182) log.20`
`=16.46 day`.