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The half life of `C^(14)(lambda=2.31xx10^(-4)` per year) isA. `2xx10^(2)` yrB. `3xx10^(3)` yrC. `3.3xx10^(4)` yrD. `4xx10^(3)` yr |
Answer» Correct Answer - B Radio active decay is first order reaction So `lambda =(0.69)/(t_(1//2)` `t_(1//2)=(0.693)/(2.31xx10^(-4))yr` `=0.3xx10^(4)` `=3xx10^(3)yr` |
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