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The half life period of `._(53)I^(125)` is 60 days. What % of radioactivity would be present after 240 days. |
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Answer» Correct Answer - `6.25 %` No. of half life`=(240)/(60)=4=("Given time")/(t_(1//2))` `N_(t)=(N_(0))/(2^(n))=(N_(0))/(2^(4))` `100xx(N_(t))/(N_(0))=(1)/(16)xx100rArr 6.25%` |
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