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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    Rn decays into Po by emitting an \[\alpha \]-particle with half-life of 4 days. A sample contains \[6.4\times {{10}^{10}}\] atoms of Rn after 12 days, the number of atoms of Rn left in the sample will be

    A) \[3.2\times {{10}^{10}}\]             

    B)        \[0.53\times {{10}^{10}}\]

    C) \[2.1\times {{10}^{10}}\]             

    D)        \[0.8\times {{10}^{10}}\]

    Correct Answer: D

    Solution :

    Remaining amount\[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\]                                 \[N=6.4\times {{10}^{10}}{{\left( \frac{1}{2} \right)}^{12/4}}\]                                 \[=0.8\times {{10}^{10}}\,\,atoms\]


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