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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2013

  • question_answer
    The ionisation energy of an electron in the ground state of helium atom is 24.6 eV. The energy required to remove both the electron is

    A)  \[51.8eV\]          

    B)  \[79\text{ }eV\]

    C)  \[38.2eV\]          

    D)  \[49.2\text{ }eV\]

    Correct Answer: B

    Solution :

    : First ionization energy of helium atom is \[{{I}_{1}}=24.6eV\] Second ionization energy of helium atom is \[{{I}_{2}}=\frac{13.6{{(2)}^{2}}}{{{(1)}^{2}}}eV=54.4eV\] \[\therefore \] Energy required to remove both the electrons from helium atom is \[={{I}_{1}}+{{I}_{2}}=24.6eV+54.4eV=79eV\]


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