1.

The activity of carbon -14 in a piece of an ancient wood is only 12.5 %. If the half-life period of carbon -14 is 5760 years, the age of the piece of wood will be (log 2 = 0.3010)

Answer»

`17.281 xx 10^(2)` years
`172.81 xx 10^(2)` years
`1.7281 xx 10^(2)` years
`1728.1 xx 10^(2)` years

Solution :`t_(1//2) " of " C - 14 = 5760` years, `lamda = (0.693)/(5760)`
Now, `t = (2.303)/(lamda) "log" (.^(14)C " original")/(.^(14)C "after time" t)`
`= (2.303 xx 5760)/(0.693) "log" (100)/(12.5) = (2.303 xx 5760 xx 0.9030)/(0.693)`
` = 17281 = 172.81 xx 10^(2)` years


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