The half life of `C^(14)(lambda=2.31xx10^(-4)` per year) is

Options

A. `2xx10^(2)` yr
B. `3xx10^(3)` yr
C. `3.3xx10^(4)` yr
D. `4xx10^(3)` yr

Answer

B

Explanation

Radio active decay is first order reaction
So `lambda =(0.69)/(t_(1//2)`
`t_(1//2)=(0.693)/(2.31xx10^(-4))yr`
`=0.3xx10^(4)`
`=3xx10^(3)yr`