Free electron of 2.6 eV energy|MODERN PHYSICS|JEE-Main 2021

JEE MAIN 2021

31-08-2021 S1

Question

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:

$1.45 \times 10^{16} \mathrm{MHz}$

$0.19 \times 10^{15} \mathrm{MHz}$

$1.45 \times 10^9 \mathrm{MHz}$

$9.0 \times 10^{27} \mathrm{MHz}$

✓ 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 $$ \begin{aligned} & \lambda=\frac{1240}{6}=206 \mathrm{~nm} \\ & \begin{aligned} \mathrm{f}=\frac{3 \times 10^8}{206 \times 10^{-9}} & =1.45 \times 10^{15} \mathrm{~Hz} \\ \quad \quad= & 1.45 \times 10^9 \mathrm{~Hz} \end{aligned} \end{aligned} $$

Question Tags

JEE Main

Physics

Easy

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