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JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Rate of decay and Half-life

  • question_answer
    \[{{T}_{1/2}}\] of \[{{C}^{14}}\] isotope is 5770 years. time after which 72% of isotope left is        [Orissa JEE 2005]

    A)                 2740 years          

    B)                 274 years

    C)                 2780 years          

    D)                 278 years

    Correct Answer: A

    Solution :

               \[K=\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{5770}\]                    \[\therefore \]  \[t=\frac{2.303}{K}\log \frac{100}{72}=\frac{2.303\times 5770}{0.693}\log \frac{100}{72}\]                    \[=19175.05\times (\log 100-\log 72)\]                                 \[19175.05\times 0.143=2742.03\,\]years.


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