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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    A \[H{{e}^{+}}\] ion is in its first excited state. Its ionization energy is:- [JEE Main 9-4-2019 Afternoon]

    A) 6.04 eV                       

    B) 13.60 eV

    C) 54.40 eV                     

    D) 48.36 eV

    Correct Answer: B

    Solution :

    Energy levels in Hydrogen like atom is given by \[E=-13.6\frac{{{z}^{2}}}{{{n}^{2}}}eV\] As \[H{{e}^{+}}\] is 1st excited state \[\therefore \]\[z=2,n=2\] \[E=-13.6eV\] As total energy of \[H{{e}^{+}}\]in 1st excited state is \[13.6\text{ }eV,\]ionisation energy should be \[+13.6eV.\]


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