Verified Solution • 2025–26 Session moderate

QJEEM21Phf540: JEE Main 2021 Physics - Class 12 JEEM21Phf540 Solution: An electron and proton are separated by a large distance. The…

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Ao\mathop A\limits^o . What is the maximum kinetic energy of the emitted photoelectron?

Multiple Choice Options

  • A 7.61 eV
  • B 1.41 eV
  • C 3.3 eV
  • D No photoelectron would be emitted

Step-by-Step Verified Solution

initially, energy of electron = + 3eV

finally, in 2nd excited state,

energy of electron = −(13.6eV)32- {{(13.6eV)} \over {{3^2}}}

=−1.51eV= - 1.51eV

Loss in energy is emitted as photon,

So, photon energy hcλ=4.51eV{{hc} \over \lambda } = 4.51eV

Now, photoelectric effect equation

KEmax⁡=hcλ−ϕ=4.51−(hcλth)K{E_{\max }} = {{hc} \over \lambda } - \phi = 4.51 - \left( {{{hc} \over {{\lambda _{th}}}}} \right)

=4.51eV−12400eVAo4000Ao= 4.51eV - {{12400eV\mathop A\limits^o } \over {4000\mathop A\limits^o }}

=1.41eV= 1.41eV

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