Atoms And Nuclei Question 379

Question: Two radioactive substances A and B have decay constants $ 5\lambda $ and $ \lambda $ 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/e)}^{2}} $ after a time interval

Options:

A) $ 4\lambda $

B) $ 2\lambda $

C) $ 1/2\lambda $

D) $ 1/4\lambda $

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Answer:

Correct Answer: C

Solution:

  • $ {\lambda_{A}}=5\lambda $ and $ {\lambda_{B}}=\lambda $

At $ t=0, $

s $ {{(N_0)}_A}={{(N_0)}_B} $

Given, $ \frac{N_{A}}{N_{B}}={{( \frac{1}{e} )}^{2}} $

According to radioactive decay, $ \frac{N}{N_{0}}={{e}^{-\lambda t}} $

$ \therefore ,\frac{N_{A}}{{{(N_{0})}_{A}}}={{e}^{-\lambda A^{t}}} $

and $ \frac{N_{B}}{{{(N_{0})}_{B}}}={{e}^{-\lambda ,B^{t}}} $

From (1) and (2), $ \frac{N_{A}}{N_{B}}={{e}^{-(5\lambda -\lambda )t}} $

$ \Rightarrow ,{{( \frac{1}{e} )}^{2}}={{e}^{-4\lambda t}}={{( \frac{1}{e} )}^{4\lambda t}}\Rightarrow ,4\lambda t=2\therefore ,t=\frac{1}{2\lambda }. $



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