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WB JEE Medical WB JEE Medical Solved Paper-2007

  • question_answer
    The ionization potential of hydrogen is 13.6 V. The energy required to remove an electron from the second orbit of hydrogen is

    A)  3.4 eV                 

    B)  6.8 eV

    C)  13.6 eV                               

    D)  1.51 eV

    Correct Answer: A

    Solution :

                     Required energy \[=\frac{13.6}{{{2}^{2}}}=3.4\,eV.\]


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