1.

T_(1//2) " of "C^(14) isotope is 5770 years. Time after which 72% of isotope left is

Answer»

2740 YEARS
274 years
2780 years
278 years

Solution :`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


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