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The .(6)C^(14) and .(6)C^(12) ratio in a...

The `._(6)C^(14)` and `._(6)C^(12)` ratio in a piece of woods is `1//16` part of atmosphere. Calculate the age of wood. `t_(1//2) "of" C^(14)` is 5577 years?

Text Solution

Verified by Experts

Given, `((N_(C_(14)))/(N_(C_(12)))) = (1)/(16) (N_(0C_(14))/(N_(0C_(12))))`
Since, only `C^(14)` undergoes decay
`:. N_(C12) = N_(OC12)`
or `(N_(0C_(14)))/(N_(C_(14))) = (16)/(1)`
`:.t = (2.303)/(lambda) log (16)/(1) = (2.303)/(0.693) xx 5577 log 2^(4)`
`t = 5577 xx 4 = 22308` year
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