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JEE Main 2021 (27 Jul, Shift 2) · Other / Mixed: Atoms & Bohr Links

Other / Mixed: Atoms & Bohr Links question from JEE Main 2021 (27 Jul, Shift 2)

Q6JEE Main 2021 · 27 JUL · SHIFT 2Difficult

Other / Mixed: Atoms & Bohr Links

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 4000A˚\text{\AA}. What is the maximum kinetic energy of the emitted photoelectron?

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Answer: (b)

  1. Second excited state is n=3n = 3: E3=−13.69=−1.51 eVE_3 = -\frac{13.6}{9} = -1.51\text{ eV}.
  2. Photon energy =3+1.51=4.51 eV= 3 + 1.51 = 4.51\text{ eV}
  3. Work function ϕ=124004000=3.1 eV\phi = \frac{12400}{4000} = 3.1\text{ eV}
  4. Kmax=4.51−3.1=1.41 eVK_{max} = 4.51 - 3.1 = 1.41\text{ eV}
  5. So the answer is (b) 1.41 eV.
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