JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    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

    A)  \[{{A}_{1}}{{t}_{1}}-{{A}_{2}}{{t}_{2}}\]

    B)  \[\frac{\left( {{A}_{2}}-{{A}_{1}} \right)}{2}\tau \]

    C)  \[\left( {{A}_{1}}-{{A}_{2}} \right)\left( {{t}_{2}}-{{t}_{1}} \right)\]

    D)  \[\left( {{A}_{1}}-{{A}_{2}} \right)\tau .\]

    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 \[={{N}_{1}}-{{N}_{2}}\] \[={{N}_{0}}({{e}^{-\lambda {{t}_{1}}}}-{{e}^{-\lambda {{t}_{2}}}})=\frac{{{A}_{0}}}{\lambda }({{e}^{-\lambda {{t}_{1}}}}-{{e}^{-\lambda {{t}_{2}}}})\] \[=\frac{{{A}_{1}}-{{A}_{2}}}{\lambda }=(.{{A}_{1}}-{{A}_{2}})\tau .\]


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