20% of a radioactive substances decay in 10 days . Calculate the amount of the original material left after 30 days.

Options

A. `78%`
B. `62%`
C. `51%`
D. `48%`

Answer

C

Explanation

( c) amount of element left undecayed after 10 days
`=1-(20)/(100)=(8)/(10)`
now ,`N =(8)/(10)N_(0) and t= 10` days
we have ,`N=N_(0) e^(-lamda t)`
`(8)/( 10) N_(0)=N_(0) e^(-lamda t) implies =(10)/(8)implies lamda =(log_(e) (10//8))/(10)`
` therefore lamda = 0.022`
if `T_(1//2)` is half -life period
` T_(1//2)=(0.693)/(lamda ) =(0.693)/(0.022)=31.5 `days
Again ,` N=N_(0) ((1)/(2))^(n)=N_(0) ((1)/(2))^((31.5)/(30))`
`implies % "of" N//N_(0) = 51.2%`