Consider a hydrogen atom with its electron in the ${\mathrm{n}}^{\mathrm{t}\mathrm{h}}$ orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is $\left(\mathrm{h}\mathrm{c}=1242\mathrm{ }\mathrm{e}\mathrm{V}\mathrm{ }\mathrm{n}\mathrm{m}\right)$

Solution

${\mathrm{E}}_{\mathrm{P}\mathrm{h}}=\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}=\frac{1242}{90}=13.8\mathrm{ }\mathrm{e}\mathrm{V}$
${\mathrm{E}}_{\mathrm{P}\mathrm{h}}=∆\mathrm{E}+(\mathrm{K}.\mathrm{E}.)$
$13.8=∆\mathrm{E}+10.4$
$∆\mathrm{E}=3.4\mathrm{ }\mathrm{e}\mathrm{V}$
So electron initially was in $\mathrm{n}=2$

Hence, the correct answer is 2.