A radioactive substance of half life 138. 6 days is placed in a box . After n days only 20% of the substance is present then the value of n is
[ in (5)= 1.61]

Options

A. 693
B. 138.6
C. 277.2
D. 322

Answer

D

Explanation

Half life of a radioactive substance `n_(1//2) = 138.6 ` days
If `N_(0)` be the initial amount of radioactive substance then ramaining amount after n days is given by
`N= 20% " of " N_(0) = (20)/( 100) xx N_(0) = (N_(0))/( 5 )`
By radioative decay law
`N= N_(0) ((1)/(2)) rArr (N_(0))/(5 ) = N_(0) ((1)/(2)) rArr (1)/(5) = (1)/(2)`
Taking log on the both sides we get
`In. (1)/(5) = in (1)/(2) rArr " in "(1)/(5) = (n )/(138.6) " in " ((1)/(2))`
In 5 = `(n )/( 138.6) ` in 2
`rArr n = 138.6 xx (in 5 )/( in 2) `
`n = 138.6 xx (1.61 )/( 0.693 ) rArr n = 322 days `