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The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay isA. 14 minB. 20 minC. 28 minD. 7 min |
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Answer» Correct Answer - B `therefore N_(1) = N_(0)-1/3N_(0)=2/3N_(0)` and `N_(2)=N_(0)-2/3N_(0)=1/3N_(0)` `therefore N_(2)/N_(1) = (1/2)^(n) rArr n=1` `therefore t_(2)-t_(1)` = one half-life = 20 min |
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