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The decay constant of radio isotope is `lambda`. If `A_(1) and A_(2)` are its activities at times `t_(1) and t_(2)` respectively, the number of nuclei which have decayed during the time `(t_(1)-t_(2))`A. `A_(1)t_(1)-A_(2)t_(2)`B. `A_(1)-A_(2)`C. `(A_(1)-A_(2))//lambda`D. `lambda(A_(1)-A_(2))` |
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Answer» Correct Answer - C Activity at time `t_(1), A_(1)=lambdaN_(1) or N_(1)A_(1)//lambda` Activity at time `t_(2), A_(2)=lambdaN_(2) or N_(2)A_(2)//lambda` Therefore, the number of nuclei decayed during time interval `(t_(1)-t_(2))` is `N_(1)-N_(2)=(A_(1))/lambda-(A_(2))/lambda=((A_(1)-A_(2))/lambda)` |
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