Physics - Modern Physics Question with Solution | TestHub

Given,$E_n = \frac{-13.6}{n^2} \mathrm{e}\mathrm{V}$

The energy of photon ejected when an electron jumps from$n=3$state to$n=2$state is given by

$∆E=E_3-E_2$

$\therefore E_3 = - \frac{13.6}{{\left(3\right)}^2} \mathrm{e}\mathrm{V} =- \frac{13.6}{9} \mathrm{e}\mathrm{V}$

$E_2 = - \frac{13.6}{{\left(2\right)}^2} \mathrm{e}\mathrm{V}\mathrm{ }=- \frac{13.6}{4} \mathrm{e}\mathrm{V}$

So,$∆E=E_3-E_2$=$-\frac{13.6}{9}-\left(-\mathrm{ }\frac{13.6}{4}\right)$

$=1.9\mathrm{ }\mathrm{e}\mathrm{V}$(approximately)