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Q.

If the first excitation energy of a hydrogen like atom is 27.3 eV, then the ionization energy of this atom will be

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answer is 36.4.

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Detailed Solution

First excitation energy $= Z^{2} Rhc (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = Z^{2} Rhc \frac{3}{4}$

$∴ \frac{3}{4} Rhc = 27 .3 eV$
$∴ Z^{2} Rhc = \frac{4}{3} \times 27 .3 eV = 36 .4 eV$

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