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JEE Main 2021 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

A hydrogen atom makes a transition from n=3 to n=1 and the emitted photon falls on a metal surface of work function 4.5 eV. Taking En=-13.6/n^2 eV…

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A hydrogen atom makes a transition from n=3n=3 to n=1n=1 and the emitted photon falls on a metal surface of work function 4.5 eV4.5\ \mathrm{eV}. Taking En=−13.6/n2 eVE_n=-13.6/n^2\ \mathrm{eV} and the de Broglie wavelength of an electron of kinetic energy KK (in eV) as λ=1.227K nm\lambda=\dfrac{1.227}{\sqrt K}\ \mathrm{nm}, the de Broglie wavelength of the fastest photoelectron is nearly
  1. A. 0.353 nm0.353\ \mathrm{nm}
  2. B. 0.445 nm0.445\ \mathrm{nm}
  3. C. 0.578 nm0.578\ \mathrm{nm}
  4. D. 0.890 nm0.890\ \mathrm{nm}
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Answer: (B) 0.445 nm0.445\ \mathrm{nm}

Photon energy =13.6(1−19)=12.09 eV=13.6\left(1-\dfrac19\right)=12.09\ \mathrm{eV}. Maximum kinetic energy K=12.09−4.5=7.59 eVK=12.09-4.5=7.59\ \mathrm{eV}. Then λ=1.2277.59=1.2272.755≈0.445 nm\lambda=\dfrac{1.227}{\sqrt{7.59}}=\dfrac{1.227}{2.755}\approx0.445\ \mathrm{nm}. Using the full photon energy gives 0.353 nm0.353\ \mathrm{nm} (a common slip).
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