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JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer
    The radioactivity of a sample is \[{{R}_{1}}\] at a time \[{{T}_{1}}\] and \[{{R}_{2}}\]at a time\[{{T}_{2}}\]. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time \[({{T}_{2}}-{{T}_{1}})\] is proportional to

    A)  \[({{R}_{1}}{{T}_{1}}-{{R}_{2}}{{T}_{2}})\]       

    B)  \[({{R}_{1}}-{{R}_{2}})\]

    C)  \[({{R}_{1}}-{{R}_{2}})/T\]         

    D)  \[({{R}_{1}}-{{R}_{2}})\times T\]

    Correct Answer: D

    Solution :

     \[1.\,\lambda =\frac{0.693}{{{t}^{1/2}}}\]                 2. \[R=\lambda {{N}_{t}}\] Radioactivity at \[{{T}_{1}}\] is \[{{R}_{1}}=\lambda {{N}_{1}},\] Radioactivity at \[{{T}_{2}}\] is \[{{R}_{2}}=\lambda {{N}_{2}}\] \[\therefore \]Number of atoms decayed in time \[({{T}_{1}}-{{T}_{2}})=({{N}_{1}}-{{N}_{2}})\] or \[\frac{{{R}_{1}}-{{R}_{2}}}{\lambda }=\frac{({{R}_{1}}-{{R}_{2}})T}{0.693}\]i.e., \[\alpha ({{R}_{1}}-{{R}_{2}})T\]


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