Half life of a radio active substance A is two times the half life of another radio active substance B. Initially number of active nuclei of A and B are N1 and N2 respectively. After three half lives of A, number of nuclei of both are equal. Then the ratio N1N2 is
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a
1/4
b
1/8
c
1/3
d
1/6
answer is B.
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Detailed Solution
T1=Half life of A, T2=Half life of B Now, N11=N1(2)tT1=N1(2)3T1T1=N18and N21=N2(2)tT2=N2(2)3T1T2=N2(2)3×2=N264Since N11=N21⇒N18=N264⇒N1N2=18