JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    The count rate observed from a radioactive source at t second was No and at 4t second it was\[\frac{{{N}_{0}}}{16}\]. The count rate observed, at\[\left( \frac{11}{2} \right)t\] second will be -

    A) \[\frac{{{N}_{0}}}{128}\]                    

    B)        \[\frac{{{N}_{0}}}{64}\]

    C) \[\frac{{{N}_{0}}}{32}\]                      

    D)        None

    Correct Answer: B

    Solution :

    [b] \[{{N}_{0}}\xrightarrow{3t}\frac{{{N}_{0}}}{16}\] \[t=0\]   \[t=t\]       \[t=4t\] \[A={{A}_{0}}{{e}^{-\lambda t}}\] \[\frac{{{N}_{0}}}{16}={{N}_{0}}{{e}^{-\lambda (3t)}}\] \[{{e}^{3\lambda t}}=16\] \[{{N}_{0}}~\xrightarrow{9t/2}N=?\] \[t=t\]   \[t=\frac{11}{2}t\] \[A={{A}_{0}}{{e}^{-\lambda t}}\] \[N={{N}_{0}}{{e}^{-\frac{9t}{2}\lambda }}\] \[N={{N}_{0}}{{\left( {{e}^{-3\lambda t}} \right)}^{3/2}}\] \[N={{N}_{0}}{{\left( \frac{1}{16} \right)}^{3/2}}\] \[N=\frac{{{N}_{0}}}{64}\]


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