Atoms And Nuclei Question 385
Question: If nuclei of a radioactive element is produced at constant rate $ \alpha $ and they decays with decay constant $ \lambda $ . At t=0, number of nuclei is zero than the number of nuclei at time t is
Options:
A) $ \frac{\alpha }{\lambda }( 1-{{e}^{-\lambda t}} ) $
B) $ \alpha -\frac{\alpha }{\lambda }{{e}^{-\lambda t}} $
C) $ \frac{\alpha }{\lambda }{{e}^{-\lambda t}} $
D) $ \alpha ( 1-{{e}^{-\lambda t}} ) $
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Answer:
Correct Answer: A
Solution:
- $ \frac{dN}{dt}=\alpha -\lambda N $ $ \int_{0}^{N}{\frac{dN}{\alpha -\lambda N}},=\int_{0}^{t}{dt} $ Solving $ N=\frac{\alpha }{\lambda }(1-{{e}^{-\lambda t}}). $
