The initial activity of a certain radioactive isotope was measured as `16000` counts `min.^(-1)`. Given that the only activity measured was due to this isotope and that its activity after `12 h` was `2100` counts `min.^(-1)`, its half-life, in hours, is nearest to [Given `log_(e) (7.2)=2`].

Options

A. `9.0`
B. `6.0`
C. `4.0`
D. `3.0`

Answer

C

Explanation

`A=A_(0) e^(- lambda t), 2100 =16000 e^(-12 lambda) rArr e^(12 lambda) =7.6`
`rArr 12 lambda =log_(e) 7.6 =2 rArr lambda (2)/(12)=(1)/(6)`
`:. T=(0.6931 xx 6)/(1)=4` .