BHU PMT BHU PMT (Screening) Solved Paper-2009

  • question_answer
    If elements with principal quantum number \[n>4\]were not allowed in nature, then the number of possible elements would be

    A)  32                                         

    B)  60

    C)  18                                         

    D)  4

    Correct Answer: B

    Solution :

                     If all the elements having\[n>4\]are removed the number of elements that will be present in the periodic table are calculated as: \[n=1\]represents K shell and the number of elements having K shell\[=2[2{{n}^{2}}]\] \[n=2\]represents L shell and the number of elements having L shell = 8 \[n=3\]represents M shell and the number of elements having M shell =18 \[n=4\]represents N shell and the number of elements having N shell =32 So the total number of elements having\[n<5\] are\[2+8+18+32=60\]


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