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JEE Main 2023 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

Light of wavelength 200 nm falls on a metal of work function 2.2 eV. Using hc=1240 eV nm, the de Broglie wavelength of the fastest photoelectrons is…

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Light of wavelength 200 nm200\ \mathrm{nm} falls on a metal of work function 2.2 eV2.2\ \mathrm{eV}. Using hc=1240 eV nmhc=1240\ \mathrm{eV\,nm}, the de Broglie wavelength of the fastest photoelectrons is approximately (take λdB=1.227K nm\lambda_{dB}=\dfrac{1.227}{\sqrt{K}}\ \mathrm{nm} for KK in eV)
  1. A. 0.31 nm0.31\ \mathrm{nm}
  2. B. 0.43 nm0.43\ \mathrm{nm}
  3. C. 0.61 nm0.61\ \mathrm{nm}
  4. D. 1.23 nm1.23\ \mathrm{nm}
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Answer: (C) 0.61 nm0.61\ \mathrm{nm}

Photon energy =1240200=6.2 eV=\dfrac{1240}{200}=6.2\ \mathrm{eV}, so Kmax=6.2−2.2=4.0 eVK_{max}=6.2-2.2=4.0\ \mathrm{eV}. Then λdB=h2mK=1.2274.0 nm=0.61 nm\lambda_{dB}=\dfrac{h}{\sqrt{2mK}}=\dfrac{1.227}{\sqrt{4.0}}\ \mathrm{nm}=0.61\ \mathrm{nm}. Using the full photon energy as KK gives 0.490.49, and using K=1 eVK=1\ \mathrm{eV} gives 1.231.23.
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