Consider a hydrogen atom with its electron in the n t h orbital. An electromagnetic radiation of wavelength…

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 $\mathrm{n}=2$

Asked in: JEE Advanced 2015 (Paper 1)

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