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A radioactive sample at any instant has its disintegration ratye `5000` disintegrations per minute After `5` minutes , the rate is `1250` disintegration per Then , the decay constant (per minute)A. `0.4 in 2`B. `0.2 in 2`C. `0.1 in 2`D. `0.8 in 2` |
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Answer» Correct Answer - A `lambda = (1)/(t) log_(e) (A_(0))/(A) = (1)/(5) log _(e) (5000)/(1250) = 0.4 log _(e) 2 ` |
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