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t(1//2) of C^(14) isotope is 5770 years...

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

A

2740 years

B

274 years

C

2780 years

D

278 years

Text Solution

Verified by Experts

The correct Answer is:
A
`K = (0.693)/(T_(1//2)) = (0.693)/(5770)`
`therefore 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
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