If `10%` of a radioactive material decays in `5` days, then the amount of original material left after `20` days is approximately.

Options

A. `60%`
B. `65%`
C. `70%`
D. `75%`

Answer

B

Explanation

(b) we have `N=N_(0e^(-lamdat))`
`therefore 0.9 N_(0)=N_(0)e^(-lamda xx5)implies 5lamda =log _(e) "" (1)/(0.9)`
and ` xN_(0)e^(-lamda xx20)implies 20 lamda =log_(e) ((1)/(x)) `
Dividing Eq. (i) by Eq (ii) ,we get
`(1)/(4)=(log_(e)(1//0.9))/(log _(e) (1//x))=(log_(10)(1//0.9))/(log _(10)(1//x))=(log_(10).9)/(log_(10)x)`
`implies log _(10)x=4log_(10)0.9implies x=0.658=65.8% ~=65%`