Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E\ $= $- \frac{13.6}{n^{2}}\ eV$.The energy of a photon ejected when the electron jumps from$\ $n =3 state to $\ n = 2$state of hydrogen , is approximately

Question

Dr. Abubacker · Accepted Answer

(d) Given , $E_{n}\  = \ \frac{13.6}{n^{2}}\ eV$
Energy of photon ejected when electron jumps from $n$=3 state to $n$=2 state is given by
$\mathrm{\Delta}E = E_{3} - E_{2}$
$	herefore\ \ \ \ \ \ \ \ \ \ E_{3}\  = \  - \ \frac{13.6}{(3)^{2}}\ eV\  = - \ \frac{13.6}{9}\ eV$
$\ \ \ \ \ \ \ \ \ \ \ \ E_{2}\  = \  - \ \frac{13.6}{(2)^{2}}\ eV\  = - \ \frac{13.6}{4}\ eV$
So, $\ \ \ \ \ \mathrm{\Delta}E = E_{3} - E_{2}$ = $- \frac{13.6}{9} - \left( - \ \frac{13.6}{4} ight)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 1.9\ eV$ (approximately)

Q. Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E\ $= $- \frac{13.6}{n^{2}}\ eV$.The energy of a photon ejected when the electron jumps from$\ $n =3 state to $\ n = 2$state of hydrogen , is approximately

Explanation

Reviewed by Dr. Abubacker

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