An electron and proton.... |MODERN PHYSICS

JEE MAIN 2021

27-07-2021 S2

Question

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electrons and form a hydrogen atoms in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic enrgy of the emitted photoelectron?

Select the correct option:

7.61 eV

1.41 eV

3.3 eV

No photoelectron would be emitted

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Initially, energy of electron $=+3 \mathrm{eV}$
finally, in $2^{\text {nd }}$ excited state,
energy of electron $=-\frac{(13.6 \mathrm{eV})}{3^2}=-1.51 \mathrm{eV}$
Loss in energy is emitted as photon,
So, photon energy $\frac{\mathrm{hc}}{\lambda}=4.51 \mathrm{eV}$
Now, photoelectric effect equation $$ \begin{aligned} & \mathrm{KE}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi=4.51-\left(\frac{\mathrm{hc}}{\lambda_{\mathrm{th}}}\right) \\ & =4.51 \mathrm{eV}-\frac{12400 \mathrm{eVA}}{4000 \ A}=1.41 \mathrm{eV} \end{aligned} $$

Question Tags

JEE Main

Physics

Easy

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