`t_(1//2)` of `C^(14)` isotope is 5770 years. time after which 72% of isotope left is:

Options

A. 2740 years
B. 274 years
C. 2780 years
D. 278 years

Answer

A

Explanation

`K = (0.693)/(T_(1//2)) = (0.693)/(5770)`
`:. t = (2.303)/(K)"log"(100)/(72) = (2.303 xx 5770)/(0.693) "log" (100)/(72)`
`= 19175.05 xx (log 100 - log 72)`
`19175.05 xx 0.143 = 2742.03` years