A free electron of $2.6 \mathrm{eV}$ energy collides with a $\mathrm{H}^{+}$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.
$\left(h=6.6 \times 10^{-34} \mathrm{Js}\right)$
Correct Option: , 3
For every large distance P.E. $=0$
$\&$ total energy $=2.6+0=2.6 \mathrm{eV}$
Finally in first excited state of $\mathrm{H}$ atom total energy $=-3.4 \mathrm{eV}$
$\begin{aligned} \text { Loss in total energy } &=2.6-(-3.4) \\ &=6 \mathrm{eV} \end{aligned}$
It is emitted as photon
$\lambda=\frac{1240}{6}=206 \mathrm{~nm}$
$f=\frac{3 \times 10^{8}}{206 \times 10^{-9}}=1.45 \times 10^{15} \mathrm{~Hz}$
$=1.45 \times 10^{9} \mathrm{~Hz}$
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