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JEE Main 2017 · Chemistry · Atomic Structure · hard

Light emitted when a He^+ ion makes the transition n=3→ n=2 falls on a metal surface of work function 2.5 eV. The de Broglie wavelength of the…

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Light emitted when a He+\mathrm{He^+} ion makes the transition n=3→n=2n=3\to n=2 falls on a metal surface of work function 2.5 eV2.5\ \mathrm{eV}. The de Broglie wavelength of the fastest photoelectron is nearly (use En=−13.6Z2/n2 eVE_n=-13.6Z^2/n^2\ \mathrm{eV} and λ(nm)=1.226/K(eV)\lambda(\mathrm{nm})=1.226/\sqrt{K(\mathrm{eV})} for an electron)
  1. A. 0.55 nm0.55\ \mathrm{nm}
  2. B. 0.35 nm0.35\ \mathrm{nm}
  3. C. 0.42 nm0.42\ \mathrm{nm}
  4. D. 0.77 nm0.77\ \mathrm{nm}
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Answer: (A) 0.55 nm0.55\ \mathrm{nm}

Photon energy =13.6×4 (1/4−1/9)=7.56=13.6\times4\,(1/4-1/9)=7.56 eV. Maximum kinetic energy K=7.56−2.5=5.06K=7.56-2.5=5.06 eV. Then λ=1.226/5.06=0.545\lambda=1.226/\sqrt{5.06}=0.545 nm ≈0.55\approx0.55 nm. Forgetting Z2=4Z^2=4 (giving 1.891.89 eV, below the work function) would predict no emission at all; using 7.567.56 eV directly gives 0.4460.446 nm.
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