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JEE Advanced 2019 Paper 2 · Physics · Atoms, Nuclei and Dual Nature of Radiation · easy

The photon emitted when a He^+ ion makes the transition n=4→ n=2 (Bohr model, En=-54.4/n^2 eV) falls on a metal surface of work function 4.2 eV. Use…

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The photon emitted when a He+\mathrm{He^{+}} ion makes the transition n=4→n=2n=4\to n=2 (Bohr model, En=−54.4/n2 eVE_n=-54.4/n^2\ \mathrm{eV}) falls on a metal surface of work function 4.2 eV4.2\ \mathrm{eV}. Use hc=1240 eV nmhc=1240\ \mathrm{eV\,nm}, and for an electron of kinetic energy KK (in eV) the de Broglie wavelength λ=1.226/K nm\lambda=1.226/\sqrt{K}\ \mathrm{nm}. Which of the following statement(s) is(are) correct?
  1. A. The energy of the emitted photon is 13.6 eV13.6\ \mathrm{eV}.
  2. B. The stopping potential for the photoelectrons is 10.2 V10.2\ \mathrm{V}.
  3. C. The minimum de Broglie wavelength of the photoelectrons is 0.50 nm0.50\ \mathrm{nm}.
  4. D. The threshold wavelength of the metal is 400 nm400\ \mathrm{nm}.
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Answer: C

The photon energy is 54.4(14−116)=10.254.4\left(\tfrac14-\tfrac1{16}\right)=10.2 eV, not 13.613.6 eV, so (A) is wrong. The maximum kinetic energy of the photoelectrons is Kmax⁡=10.2−4.2=6.0K_{\max}=10.2-4.2=6.0 eV, so the stopping potential is 6.06.0 V; the value 10.210.2 V in (B) forgets the work function, and (B) is wrong. The fastest electrons have the shortest wavelength, λmin⁡=1.226/6.0=0.50\lambda_{\min}=1.226/\sqrt{6.0}=0.50 nm, so (C) is correct. The threshold wavelength is hc/ϕ=1240/4.2≈295hc/\phi=1240/4.2\approx295 nm, not 400400 nm, so (D) is wrong. Hence only (C) is correct. The key idea is to chain three steps: Bohr energy difference, Einstein's equation, then the de Broglie relation for the fastest electrons.
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