JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    The binding energy of an electron in the ground state of He-atom is equal to \[{{E}_{0}}=24.6\,eV\].The energy required to remove both the electrons from the atom is

    A)  24.6eV          

    B)  79.0 eV          

    C)  54.4 eV                      

    D)  None of these

    Correct Answer: B

    Solution :

     The total energy required to make the electron free from nucleus is the sum of the energy required to separate the electrons from the influence of each other and the energy required to separate the electrons from the influence of nucleus, i.e., Total energy required = BE of electron in \[\phi '\] atom + lonization energy of He-atom \[\tan \,\phi =\frac{V}{{{H}^{.}}}=\frac{V}{H\,\cos \,\theta }\] \[\frac{\tan \,\phi '}{\tan \,\phi }=\frac{V/H\,\cos \,\theta }{\frac{V}{H}}=\frac{1}{\cos \,\theta }\]


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