The ratio of `C^(14)` to `C^(12)` in a living matter is measured to be `(C_(14))/(C_(12)) =1.3 xx 10^(-12)` at the present time Activity of `12.0gm` carabon sample is 180 dpm. The half life of `C^(14) ` is nearly _______ `xx 10^(12) sec`.
[Given : `N_(A)=6 xx 10^(23)`]

Options

A. `0.18`
B. `1.8`
C. `0.384`
D. 648

Answer

A

Explanation

Ratio of `C^(14) ` to `C^(12)` in living matter `=1.3 xx 10 ^(-12)`
wt. of sample =12 g
Total no. of atoms in sample `=(12)/(12) xx N_(A)=6.02 xx 10^(23)`
(As the ratio of C-40 is quite lesser avg. atomic mass of sample can be taken as 12)
No. of nuclei of C-14 in sample
=No. of atoms of C-14 in sample `=6.02 xx 10^(23) xx 1.3 xx 10^(-12)`
Activity = 180 dpm = 3 dps= `lambda` N
`3=lambda xx 6.02 xx 10 ^(-23) xx 1.3 xx 10^(-12)`
`lambda = (3)/(6.02 xx 10^(23) xx1.3 10^(-12))`
`t_(1//2) = (0.693)/(lambda)=(0.693 xx 6.02 xx 10^(23) xx 1.3 xx 10 ^(-12))/(3) =0.18 xx 10^(12) sec`.