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For ""^(24)Na,t_((1)/(2))=14.8 hours. In what period of time will a sample of this substance lose 90% of its radioactive intensity ? |
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Answer» Solution :Let the initial radioactive intensity be 100 which corresponds to `N^(0)`. The radioactive intensity after a time period, say t HOURS, will be 10 (CORRESPONDING to N) as the SUBSTANCE has lost 90% of its radioactive intensity. `lamda= (0.6932)/(t_((1)/(2))) = (0.6932)/(14.8)` We have, `lamda= (2.303)/(t) "log" (N^(0))/(N)` `(0.6932)/(14.8) =(2.303)/(t) "log" (100)/(10)` t= 49.17 hours. |
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