1. Tardigrade
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  4. An electron of a hydrogen like atom, having Z=4, jumps from 4 text th energy state to 2 text 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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Q. An electron of a hydrogen like atom, having $Z=4$, jumps from $4^{\text {th }}$ energy state to $2^{\text {nd }}$ 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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Solution:

$\Delta E =13.6 Z ^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right] eV $
$ =13.6 \times(4)^2\left(\frac{1}{4}-\frac{1}{16}\right) eV$
$ =13.6[4-1] eV$
$ =13.6 \times 3=40.8\, eV $