If `10%` of a radioactive substance decays in every 5 year, then the percentage of the substance that will have decaed in `20 years` will be .

Options

A. `40%`
B. `50%`
C. `65.6%`
D. `34.4%`

Answer

d

Explanation

`N=N_(0) e^(-lambda t)`
` rArr (N_(0))/(N)= e^(lambda t)`
`lambda t=log_(e).(N_(0))/(N)`
`t =(1)/(lambda) log_(e).(N_(0))/(N)`
` t prop log_(e). (N_(0))/(N)`
`5 prop log_(e). (100)/(90)`
and` 20 prop log_(e). (N_(0))/(N)`
Dividing,
`(5)/(20)=(log_(10).(100)/(90))/(log_(10). (N_(0))/(N))`
or `log_(10) .(N_(0))/(N)=4log_(10).(10)/(9)`
or `(N_(0))/(N)=((10)/(9))^(4) rArr (N)/(N_(0)) =0.6561`
Percentage of substance decayes is
`(1-0.6561)xx100 =34.39` .