An electron and proton are separated by a large distance.

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 forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic energy of the emitted photoelectron?

  1. 7.61 eV

  2. 1.41 eV

  3. 3.3 eV

  4. No photoelectron would be emitted

Correct Option: 2,

Solution:

Initially, energy of electron = +3eV

finally, in $2^{\text {nd }}$ excited state,

 energy of electron $=-\frac{(13.6 \mathrm{eV})}{3^{2}}$

$=-1.51 \mathrm{lV}$

Loss in energy is emitted as photon,

So, photon energy $\frac{\mathrm{hc}}{\lambda}=4.51 \mathrm{eV}$

Now, photoelectric effect equation

$\mathrm{KE}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi=4.51-\left(\frac{\mathrm{hc}}{\lambda_{\text {th }}}\right)$

$=4.51 \mathrm{eV}-\frac{12400 \mathrm{eV A}}{4000 A}$

$=1.41 \mathrm{eV}$ 

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