A nucleus at rest splits into two nuclear parts having radii in the ratio 1 : 2. Their velocities are in the ratio

Options

A. `6:1`
B. `2:1`
C. `8:1`
D. `4:1`

Answer

C

Explanation

Let in time `t_(1) , 50 % ` of the substance decay and in time `t_(2) , 87.5 % ` of the substance decay .
Then in time `t_(1), 50 % ` of the substance left undecayed and in time `t_(2), 12.5 % ` of the substance left undecayed .
According to radioactive decay law
`N = N_(0)e^(-lambdat)` where `lambda` is the decay constant .
or `N/(N_(0)) = e^(-lambdat)`
` :. 50/100 = e^(-lambdat_(1)) or 1/2 = e^(-lambdat_(1))....(i)`
and ` (12.5)/100 = e^(lambdat_(2)) or 1/8 = e^(-lambdat_(2)) " " ...(ii)`
Dividing eqn . (ii) by eqn . (i) , we get
` (1/8)/(1/2) = (e^(-lambdat_(2)))/(e^(-lambdat_(1))) or 1/4 = e^(-lambda (t_(2)-t_(1)))` or `lambda (t_(2)-t_(1)) = 1n 4 `
` t_(2) - t_(1) = (1n4)/lambda = (21n2)/lambda " " ( :' 1n 4 = 2 1n 2)`
` = (21n2)/(((1n2)/(T_(1//2))))= 2T_(1//2) ( :' lambda = (1n2)/(T_(1//2)))`
Here,
`T_(1//2) = 20 ` minutes
` :. t_(2) - t_(1) = 2 ` (20 minutes ) = 40 minutes