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# Two radioactive substances A and B have decay constants 5λ and λ respectively. At t=0, they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be (1/e2) after a time: 4λ2λ12λ14λ

A
4λ
B
12λ
C
2λ
D
14λ
Solution
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#### Given : (No)A=(No)B=NoUsing : Kt=lnNNo where, K is the decay constant⟹ Kt=lnN−lnNoFor substance A : λt=lnNA−ln(No)A∴λt=lnNA−lnNo .............(1)For substance B : 5λt=lnNB−ln(No)B∴5λt=lnNB−lnNo .............(2)Subtracting (2) from (1) we get: λt−5λt=lnNANB−4λt=ln1e2−4λt=−2 ⟹t=12λ

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