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:

  • 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 . $



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