Excitation energy of a hydrogen like ion in its first excitation state is $40.8 e V$ . Energy needed to remove the electron from the ion in ground state is:

10.8 e V

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27.2 e V

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54.4 e V

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13.6 e V

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Solution

The correct option is C $54.4 e V$
Energy of excitation in first excitation state from ground state = $E_{0} (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$⟹ 40.8 e V = E_{0} \times \frac{3}{4}$
$⟹ E_{0} =$ Ground state energy=Energy required to remove electron from gound state $= 54.4 e V$

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Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is:

The correct option is C 54.4 eVEnergy of excitation in first excitation state from ground state = E0(112−122)⟹40.8eV=E0×34⟹E0=Ground state energy=Energy required to remove electron from gound state=54.4 eV

Excitation energy of a hydrogen like ion in its first excitation state is $40.8 e V$ . Energy needed to remove the electron from the ion in ground state is:

The correct option is C $54.4 e V$
Energy of excitation in first excitation state from ground state = $E_{0} (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$⟹ 40.8 e V = E_{0} \times \frac{3}{4}$
$⟹ E_{0} =$ Ground state energy=Energy required to remove electron from gound state $= 54.4 e V$