A free electron of 2.6 eV energy collides with a H⁺ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. $\left(\mathrm{h}=6.6\times {10}^{-34} \mathrm{Js}\right)$

Given that, energy of free electron,

       ${\mathrm{E}}_1=2.6 \mathrm{eV}$

We know that, the energy of H-atom in its first excited state (n = 2),

       ${\mathrm{E}}_2=\frac{-13.6}{2^2}=-\frac{13.6}{4}\mathrm{eV}$

Now, the energy of emitted photons will be the difference of these two energies E₁ and E₂ .

$\begin{array}{l} \mathrm{\Delta E}={\mathrm{E}}_1-{\mathrm{E}}_2\\ \therefore \mathrm{hv}=\left[2.6-\left(-\frac{13.6}{4}\right)\right]\mathrm{eV}\\ \Rightarrow \mathrm{v}=\frac{(10.4+13.6)\times 1.6\times {10}^{-19}}{4\times 6.6\times {10}^{-34}}=1.45\times {10}^9\mathrm{MHz}\end{array}$