InterviewSolution
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InterviewSolution
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A small quantity of solution containing `Ne^(24)` radio nucliode (half life `= 15 hour`) of activity `1.0` microcurlar is injected into the blood of a person A sample of the blood of volume `1 cm^(3)` taken a after `5` hour shown an activity of the blood in the body of the person . Assume that redicative solution mixed uniformly in the blood of the person `(1 curie = 3.7 xx 10^(10)` disntegrations per sound) |
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Answer» `t_(1//2) = 15 hour` Actually `A_(0) = 10^(-6)` curie (in small quentyty of solution of `^(24) Na) = 3.7 xx 10^(4) dps` Observation of blood of volume `1 cm^(3)` After `5 hour .,A = 296 dps` The initial activity can be found by the formula ` t = (2.303)/(lambda) log_(10) (A_(0))/(A) rArr 5 = (2.303)/(0.693//15) xx log_(10) (A_(0))/(296)` `rArr log_(10) (A_(0))/(296)= ( 5 xx 0.693)/(2.303 xx 15) = (0.33010)/(3) = 0.10033` `rArr (A_(0))/(296)= 1.26 rArr A_(0) = 373 dpm = (373)/(60) dps` This is the activity level in `1 cm^(3)` Comparing it with the initial activity level of `3.7 xx 10^(4) dps `we find the volume of blood `V = (3.7 xx 10^(4))/(373//60) = 5951.7 cm^(3) = 5.951 litre ` |
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