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 $
Show Answer
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 }. $
BETA
