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

54.4 e V

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13.6 e V

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40.8 e V

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27.2 e V

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Solution

The correct option is D $54.4 e V$
Excitation energy $δ E = E_{2} - E_{1} = 13.6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$\Rightarrow 40.8 = 13.6 \times \frac{3}{4} \times Z^{2}$

$\Rightarrow Z = 2$

Now required energy to remove the electron from ground state $= \frac{+ 13.6 Z^{2}}{(1)^{2}} = 13.6 (Z)^{2} = 54.4 e V$

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

The correct option is D 54.4eVExcitation energy δE=E2−E1=13.6Z2[112−122]⇒40.8=13.6×34×Z2⇒Z=2Now required energy to remove the electron from ground state =+13.6Z2(1)2=13.6(Z)2=54.4 eV

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