The activity of a radioactive sample is measures as `N_0` counts per minute at `t = 0` and `N_0//e` counts per minute at `t = 5 min`. The time (in minute) at which the activity reduces to half its value is.

Options

A. `log_(2)2//5`
B. `(5)/(log_(e)2)`
C. `5log_(10)2`
D. `5log_(2)2`

Answer

C

Explanation

Given that the half-life of radioactive substnace is 20 min. so, `t_(1//2)=20`min,
For 20% decay, we have 80% of the substance left, hence
`(80N_(0))/(100)=N_(0)e^(lamdat_(20))` . .. (i)
where, `N_(0)`=initial undecayed substance and `t_(20)` is the time taken for 20% decay.
for 80% decay, we have 20% of the substance left, hence
`(20N_(0))/(100)=N_(0)e^(-lamdat_(80))` . . (ii)
Dividing Eq. (i) and Eq. (ii), we get
`4=e^(lamda(t_(80)-t_(20)))`
`implies"ln "4=lamda(t_(80)-t_(20))` (taking log on both sides)
`implies" 2 ln 2"=(0.693)/(t_(1//2))(t_(80)-t_(20))`
`implies t_(80)-t_(20)=2xxt_(1//2)=40` min