A radioactive material of half-life time of 69.3 days. kept in a container. `(2)/(3)`rd of the substance remains undecayed after (given, In `(3)/(2) =0.4`)

Options

A. A)20 days
B. B)25 days
C. C)35 days
D. D)40 days

Answer

D

Explanation

Given, `t_(1//2) = 69 -3` days and `N= (2)/(3) N_0` and we know,
`(N_0)/(N) = (2)^(t//t_(1//2))`
`(N_0)/((2)/(3) N_0) = (2)^(t//t_(1//2))`
`(3)/(2) = 2^(t//t_(1//2))`
`log (3)/(2) = (t)/(t_(1//2) ) "In" 2`
`0.4 = (t)/(69.3) xx 0.6932`
`0.4= (t)/( 10)`
`t= 0.4 xx 100`
`t= 40` days