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A radioactive element has rate of disintegration `10,000` disintegrations per minute at a particular instant. After four minutes it becomes `2500` disintegrations per minute. The decay constant per minute isA. `0.2 log_(e)^(2)`B. `0.5 log_(e)^(2)`C. `0.6 log_(e)^(2)`D. `0.8 log_(e)^(2)` |
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Answer» Correct Answer - B `(N)/(N_(0))=e^(-lambdat)` `therefore (2500)/(10000)=e^(-lambda xx 4)` `therefore 1/4=e^(-4)` `therefore e^(4lambda)=4` `therefore 4lambda=log_(e)4` `therefore 4lambda=log_(e)2^(2)` `therefore 4lambda=2log_(e)2` `therefore lambda=2/4 log_(e)2` `therefore lambda=0.5 log_(e)2` |
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