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:
4λ
2λ
12λ
14λ
A
2λ
B
4λ
C
12λ
D
14λ
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Solution
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Given : (No)A=(No)B=No
Using : Kt=lnNNo where, K is the decay constant
⟹Kt=lnN−lnNo
For 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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