A radioactive material has half-life of `10` days. What fraction of the material would remain after `30` days ?

Options

A. 0.5
B. 0.25
C. 0.125
D. 0.33

Answer

C

Explanation

`T_(1//2)=10` days, T=30 days
`T=nxxT_(1//2)" "n=(T)/(T_(1//2))=(30)/(10)=3`
`(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^(3)=(1)/(8)=0.125`.