Nuclear Physics And Radioactivity Question 447
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 . $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] 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 . $
BETA
