Half life for radioactive C is 5760 yr. In ho many years 200 mg of `^14C` will be reduced to 25 mg?

Options

A. 11520 years
B. 2880 years
C. 1440 years
D. 17128 years

Answer

A

Explanation

We know that `N=((1)/(2))^(n)N_(0)`
`:. 25=((1)/(2))^(n) 100`
`(25)/(100)=((1)/(2))^(n)impliesn=2`
and `T=nxxt_(1//2)`
`:. T=2xx5760=11520years`