Question

Consider a hydrogen atom with its electron in the $n^{th}$ orbital. An electromagnetic radiation of wavelength $90nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4eV$, then the value of $n$ is ($hc=1242eVnm$)

Answer 1

Given, wavelenght of electromagnetic radiation= $90nm$
energy of the incident photon (E)$=\frac{hc}{\lambda }\Rightarrow \frac{hc}{90}$
And, energy of electron on $n^{th}$ orbital for Hydrogen atom =$\frac{13.6}{n^2}$

So,
The kinetic energy $\left(KE\right)$ of ejected electron= (Energy of incident photon)-(Energy of electron on $n^{th}$ orbital for Hydrogen atom)

$KE=\frac{hc}{90nm}-\frac{13.6}{n^2}$ .....(1)
Given, $KE=10.4eV$ and $hc=1242eVnm$

on putting values on eq.(1).
$10=\frac{1242}{90}-\frac{13.6}{n^2}$
$\frac{13.6}{n^2}=3.8n=2$