The half-life of a radioactive substance is 2 | vedclass

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(B) The radioactive decay law is given by $N(t) = N_{0} e^{-\lambda t}$ or $N(t) = N_{0} (1/2)^{t/T_{1/2}}$.At time $t_{1}$,$1/3$ of the substance has

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(B) The radioactive decay law is given by $N(t) = N_{0} e^{-\lambda t}$ or $N(t) = N_{0} (1/2)^{t/T_{1/2}}$.At time $t_{1}$,$1/3$ of the substance has decayed,so the remaining amount is $N_{1} = N_{0} - (1/3)N_{0} = (2/3)N_{0}$.At time $t_{2}$,$2/3$ of the substance has decayed,so the remaining amount is $N_{2} = N_{0} - (2/3)N_{0} = (1/3)N_{0}$.We know that $N(t) = N_{0} (1/2)^{t/T_{1/2}}$.Taking the ratio: $N_{2}/N_{1} = [(1/3)N_{0}] / [(2/3)N_{0}] = 1/2$.Since $N_{2}/N_{1} = (1/2)^{(t_{2}-t_{1})/T_{1/2}}$,we have $(1/2)^{1} = (1/2)^{(t_{2}-t_{1})/20}$.Comparing the exponents,$(t_{2}-t_{1})/20 = 1$,which gives $t_{2}-t_{1} = 20 \ min$.

The half-life of a radioactive substance is $20 \ min$. The approximate time interval $(t_{2}-t_{1})$ between the time $t_{2}$,when $2/3$ of it has decayed,and time $t_{1}$,when $1/3$ of it has decayed,is (in $min$):

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