Atoms And Nuclei Question 377
Question: The activity of a radioactive sample is $ A_{1} $ at time $ t_{1} $ and $ A_{2} $ at time $ t_{2} $ . If $ \tau $ is average life of sample then the number of nuclei decayed in time ( $ t_{2}-t_{1} $ ) is
Options:
A) $ A_{1}t_{1}-A_{2}t_{2} $
B) $ \frac{( A_{2}-A_{1} )}{2}\tau $
C) $ ( A_{1}-A_{2} )( t_{2}-t_{1} ) $
D) $ ( A_{1}-A_{2} )\tau . $
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Answer:
Correct Answer: D
Solution:
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Let $ N_{0} $ be the initial number of nuclei, then
$ N_{1}=N_{0}{{e}^{-\lambda t_{1}}} $
and $ N_{2}=N_{0}{{e}^{-\lambda t_{2}}} $
$ \therefore $Number of nuclei decayed $ =N_{1}-N_{2} $ $ =N_{0}({{e}^{-\lambda t_{1}}}-{{e}^{-\lambda t_{2}}})=\frac{A_{0}}{\lambda }({{e}^{-\lambda t_{1}}}-{{e}^{-\lambda t_{2}}}) $
$ =\frac{A_{1}-A_{2}}{\lambda }=(.A_{1}-A_{2})\tau . $
