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JEE Advanced 2025 Paper 1 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

Monochromatic light of wavelength 300 nm falls on a metal surface of work function 2.0 eV. The fastest photoelectrons enter a region of uniform…

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Monochromatic light of wavelength 300 nm300\ \mathrm{nm} falls on a metal surface of work function 2.0 eV2.0\ \mathrm{eV}. The fastest photoelectrons enter a region of uniform magnetic field B=1.0×10−3 TB=1.0\times10^{-3}\ \mathrm{T} perpendicular to their velocity. Use hc=1240 eV nmhc=1240\ \mathrm{eV\,nm}, me=9.11×10−31 kgm_e=9.11\times10^{-31}\ \mathrm{kg} and e=1.6×10−19 Ce=1.6\times10^{-19}\ \mathrm{C}. The radius of the circular path of these electrons is closest to
  1. A. 3.5 mm3.5\ \mathrm{mm}
  2. B. 9.9 mm9.9\ \mathrm{mm}
  3. C. 2.5 mm2.5\ \mathrm{mm}
  4. D. 4.9 mm4.9\ \mathrm{mm}
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Answer: (D) 4.9 mm4.9\ \mathrm{mm}

Einstein's equation gives Kmax=hc/λ−ϕ=1240/300−2.0=2.133 eVK_{max}=hc/\lambda-\phi=1240/300-2.0=2.133\ \mathrm{eV}. The momentum of the electron is p=2mK=2(9.11×10−31)(2.133×1.6×10−19)=7.89×10−25 kg m s−1p=\sqrt{2mK}=\sqrt{2(9.11\times10^{-31})(2.133\times1.6\times10^{-19})}=7.89\times10^{-25}\ \mathrm{kg\,m\,s^{-1}}. In the magnetic field the magnetic force supplies the centripetal force, evB=mv2/revB=mv^2/r, so r=peB=7.89×10−25(1.6×10−19)(10−3)=4.93×10−3 m≈4.9 mmr=\dfrac{p}{eB}=\dfrac{7.89\times10^{-25}}{(1.6\times10^{-19})(10^{-3})}=4.93\times10^{-3}\ \mathrm{m}\approx4.9\ \mathrm{mm}. The other values come from dividing or multiplying the correct radius by 2\sqrt2 or 22, which is the typical slip when KK is confused with 2K2K or when p=mKp=\sqrt{mK} is used.
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Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake