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If a radiation corresponds to second line of Balmer series of $L i^{+ 2}$ ion knocked out electron from first excited state of H – atom then K.E of ejected electron in ev is___

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answer is 19.55.

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Detailed Solution

$E = 13.6 \times 3^{2} [\frac{1}{2^{2}} - \frac{1}{4^{2}}] = 13.6 \times \frac{27}{16}$
Energy required to remove $e^{-}$ from 2^nd orbit of H is $= 13.6 \times 1^{2} (\frac{1}{4} - \frac{1}{α^{2}}) = \frac{13.6}{4}$
$K . E_{o f e^{-}} = 13.6 \times \frac{27}{16} - \frac{13.6}{4} = 19.55 e v$

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If a radiation corresponds to second line of Balmer series of Li+2 ion knocked out electron from first excited state of H – atom then K.E of ejected electron in ev is___