An electron of a hydrogen like atom, having Z = 4, jumps in from energy state to energy state. The energy released in this process, will be: (Given ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant [202

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Solution

Q.
An electron of a hydrogen like atom, having Z = 4, jumps in from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be: (Given $R_{c h} = 13.6 eV$ ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant [2023]

1 13.6 eV

2 10.5 eV

3 3.4 eV

4 40.8 eV

Ans.
(4)

$Δ E = 13.6 Z^{2} [\frac{1}{2^{2}} - \frac{1}{4^{2}}] eV$

$= 13.6 \times {(4)}^{2} (\frac{1}{4} - \frac{1}{16}) eV$

$= 13.6 [4 - 1] eV = 13.6 \times 3 = 40.8 eV$

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