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The half life of a substance is 20 minutes The time interval between 33% decay and 67% decay |
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Answer» Solution :For 33% decay, `N/(N_0)=1/(2^(n_1))implies 1/(100) = 1/(2^(n_1)) .....(1)"" n_(1) = t_(1)/T` For 67% decay , `N/N_(0) = 1/(2^(n_2)) implies 33/100 = 1/(2^(n_2)).....(2)` `n_(2) = t_(2)/T` `((2))/((1))=33/67=1/(2^((n_(2)-n_(1))))implies1//2^(1)=1/(2((t_(2)-t_(1)))/T)implies(t_(2)-t_(1))/T=1` `t_(2)-t_(1) = T = 20 ` minutes. |
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