An electron of a hydrogen like atom, having $Z = 4$ , jumps from 4^th energy state to 2^nd energy state. The energy released in this process, will be:
(Given, $Rch = 13.6 eV$ )
Where, R = Rydberg constant , c = Speed of light in vacuum , h = Planck’s constant

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10.5 eV

40.8 eV

3.4 eV

13.6 eV

answer is A.

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Detailed Solution

Energy of any orbit in hydrogen like atom is given $E_{n} = \frac{13.6 Z^{2}}{n^{2}} eV$
$\begin{array}{l} E_{4} - E_{2} = 13.6 \times 4^{2} (\frac{1}{4} - \frac{1}{16}) = 13.6 \times 16 \times \frac{3}{16} \\ = 40.8 eV \end{array}$

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