The binding energy of an electron in the ground state of $H e$ atom is $E_{0} = 24.6 e V .$ The energy required to remove both the electrons from the atom is

24.6 e V

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79.0 e V

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54.4 e V

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None of these

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Solution

The correct option is B $79.0 e V$
Given, Energy required to remove electron in ground state from a neutral atom is $24.6 e V$ , which makes it ${H e}^{+}$ hydrogen like atom.
Energy required to remove electron from singly ionized helium is $13.6 \times 2^{4}$ .
Therefore, Energy required to remove both the electrons is $24.6 + 54.4 = 79.0 e V$

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Similar questions

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The binding energy of an electron in the ground state of He atom is E0=24.6eV. The energy required to remove both the electrons from the atom is

The binding energy of an electron in the ground state of $H e$ atom is $E_{0} = 24.6 e V .$ The energy required to remove both the electrons from the atom is