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A sample of radioactive ""^(133)I gave with a Geiger counter 3150 counts per minute at a certain time and 3055 counts per minute exactly one hour later. Calculate the half-life period of ""^(133)I |
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Answer» SOLUTION :`N^(0)= 3150, N= 3055, t=1` hours We have, `lamda= (0.6932)/(t_(1//2))` Substituting these VALUES in `lamda= (2.303)/(t) "log" (N^(0))/(N)` `(0.6932)/(t_((1)/(2)))= (2.303)/(1) "log" (3150)/(3055)` `t_((1)/(2))= 22.63` hours. |
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