If half life of an element is 69.3 hours then how much of its percent will decay in `10^("th")` to `11^("th")` hours.
Initial activity `=50 muCi`

Options

A. `1%`
B. `2%`
C. `3%`
D. `4%`

Answer

A

Explanation

Let active Nuclei at t = 0 is `N_(0)` then,
Active nuclei at t = 10 hour
`N_(1)=N_(0)e^(-10)lambda)`
Active nuclei at t = 11 hour
`N_(2)=N_(0)e^(-11lambda)`
`%" decay"=((N_(1)-N_(2))/(N_(1)))xx100`
`=((N_(0).e^(-10lambda)-N_(0)Ee^(-11lambda))/(N_(1)))xx100`
`=((N_(0).e^(-10lambda)-N_(0)e^(-11lambda))/(N_(0)e^(-10lambda)))xx100`
`=(1-(1)/(e^(lambda)))xx100`
`=(1-(1)/(e^ln 2//T_(1//2)))xx100`
`T_(1//2)="63.3 hrs."`
`=(1-(1)/(e^(0.01)))xx100`
`% " decay "=1%`