A free electron of 2.6 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. ( $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$ )
Select the correct option:
A
$1.45 \times 10^{16} \mathrm{MHz}$
B
$0.19 \times 10^{15} \mathrm{MHz}$
C
$1.45 \times 10^9 \mathrm{MHz}$
D
$9.0 \times 10^{27} \mathrm{MHz}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
For every large distance P.E. $=0$ \& total energy $=2.6+0=2.6 \mathrm{eV}$
Finally in first excited state of H atom total energy $=-3.4 \mathrm{eV}$
Loss in total energy $=2.6-(-3.4)$
$$
=6 \mathrm{eV}
$$
It is emitted as photon
$$
\begin{aligned}
& \lambda=\frac{1240}{6}=206 \mathrm{~nm} \\
& \begin{aligned}
\mathrm{f}=\frac{3 \times 10^8}{206 \times 10^{-9}} & =1.45 \times 10^{15} \mathrm{~Hz} \\
\quad \quad= & 1.45 \times 10^9 \mathrm{~Hz}
\end{aligned}
\end{aligned}
$$
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