A radioactive sample has a half life of `40` seconds. When its activity is measured `80` seconds after the beginning, it is found to be `6.932xx10^(18)` dps. During this time total energy released is `6xx10^(8)` joule `(ln2=0.6932)`:-

Options

A. The initial number of atoms in the simple is `1.6xx10^(20)`
B. The initial number of atoms in the sample is `1.6xx10^(21)`
C. Energy released per fission is `5xx10^(-13) J`
D. Energy released per fission is `(5)/(3)xx10^(-13)J`

Answer

B, C

Explanation

`A = A_(0) e^(-lambda t)`
`6.932xx10^(18) = A_(0)e^(-(ln2)/(40)(80))`
`= A_(0)e^(-ln4) = (A_(0))/(4)`
`A_(0) = 4xx6.932xx10^(18)`
`lambda N_(0)= 4xx6.932xx10^(18)`
`(0.6932)/(40)N_(0) = 4xx6.932xx10^(18)`
`N_(0)=160xx10^(19)`
`N_(0) =1.6xx10^(21)`
Also `(3)/(4) (1.6xx10^(21))E_(0)=6xx10^(8)`
`E_(0)=5xx10^(-13)`