Consider a radioactive nuclide which follow decay rate given by `A (t) = A_(0)2^(-(l//t_(0))),` where A (f) is the fraction of radioactive material remaining after time t from the initial `A_(0)` at zero time. Let `A_(1)` be the fraction of orginal activity which remains after 120 hours. likewise `A_(2)` is the fraction of 200 hours. If `(A_(1))/(A_(2))= 1.6,` then the half-life `(t_(0))` will be

Options

A. 10 hours
B. 20 hours
C. 40 hours
D. 60 hours

Answer

B

Explanation

Given, decay rate, `A = A _(0) 2 ^(-((t)/(t_(o))))`
Let `(t)/(t_(0))=n`
where,` t_(0)=` half life.
As, `(A_(1))/(A_(2)) =16 implies (A _(0) . 2 ^(-n_(1)))/(A_(0) .2 ^(-n _(2)))=16`
`implies 2 ^(n _(2)-n _(1)) =16 = 2 ^(4) implies n_(2) -n _(1) =4`
`(t_(2))/(t _(0)) - (t_(1))/(t _(0)) =4`
Sa, `t _(1) =120h and t _(2) = 200h,`
`implies (200)/(t_(0)) -(120)/(t _(0)) =4 implies (80)/(t _(0)) =4`
`implies t_(0) =(80)/(4) =20h`