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JEE Advanced 2018 Paper 1 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

JEE Advanced 2018 Paper 1 Physics: Paragraph 2

In a photoelectric experiment on a metal surface, light of wavelength 400 nm400\ \mathrm{nm} gives a stopping potential of 1.10 V1.10\ \mathrm{V}, while light of wavelength 248 nm248\ \mathrm{nm} gives a stopping potential of 3.00 V3.00\ \mathrm{V}. The value of hchc is not supplied, so it must be deduced from these two readings, and the metal's work function can then be found. Photoelectrons of the highest energy emitted by the 248 nm248\ \mathrm{nm} light are then made to enter a region of uniform magnetic field of 1.0 mT1.0\ \mathrm{mT} perpendicular to their velocity. Take e=1.602×10−19 Ce=1.602\times10^{-19}\ \mathrm{C} and me=9.109×10−31 kgm_e=9.109\times10^{-31}\ \mathrm{kg}.

Question 1

Single correct option

The threshold wavelength of the metal is
  1. A. 400 nm400\ \mathrm{nm}
  2. B. 496 nm496\ \mathrm{nm}
  3. C. 620 nm620\ \mathrm{nm}
  4. D. 775 nm775\ \mathrm{nm}
Show answer and solution

Answer: (C) 620 nm620\ \mathrm{nm}

Einstein's equation gives eVs=hcλ−ϕeV_s=\dfrac{hc}{\lambda}-\phi. Subtracting the two readings eliminates ϕ\phi: e(3.00−1.10)=hc(1248−1400)nm−1e(3.00-1.10)=hc\left(\dfrac1{248}-\dfrac1{400}\right)\mathrm{nm^{-1}}, so hc/e=1.901.5323×10−3=1240 V nmhc/e=\dfrac{1.90}{1.5323\times10^{-3}}=1240\ \mathrm{V\,nm}. Then ϕ=1240400−1.10=2.00 eV\phi=\dfrac{1240}{400}-1.10=2.00\ \mathrm{eV} (the same from the second reading, 5.00−3.005.00-3.00). The threshold wavelength is λ0=hcϕ=12402.00=620 nm\lambda_0=\dfrac{hc}{\phi}=\dfrac{1240}{2.00}=620\ \mathrm{nm}. The value 775 nm775\ \mathrm{nm} would correspond to ϕ=1.6 eV\phi=1.6\ \mathrm{eV}, and 496 nm496\ \mathrm{nm} to 2.5 eV2.5\ \mathrm{eV}.

Question 2

Single correct option

The radius of the circular path of the most energetic photoelectrons from the 248 nm248\ \mathrm{nm} light in the magnetic field is closest to
  1. A. 3.54 mm3.54\ \mathrm{mm}
  2. B. 4.13 mm4.13\ \mathrm{mm}
  3. C. 5.84 mm5.84\ \mathrm{mm}
  4. D. 8.26 mm8.26\ \mathrm{mm}
Show answer and solution

Answer: (C) 5.84 mm5.84\ \mathrm{mm}

The maximum kinetic energy of the photoelectrons is K=eVs=3.00 eV=4.806×10−19 JK=eV_s=3.00\ \mathrm{eV}=4.806\times10^{-19}\ \mathrm{J}. In the perpendicular magnetic field the electron moves on a circle of radius r=mveB=2mKeBr=\dfrac{mv}{eB}=\dfrac{\sqrt{2mK}}{eB}. Numerically 2×9.109×10−31×4.806×10−19=9.358×10−25 kg m s−1\sqrt{2\times9.109\times10^{-31}\times4.806\times10^{-19}}=9.358\times10^{-25}\ \mathrm{kg\,m\,s^{-1}}, so r=9.358×10−251.602×10−19×10−3=5.84×10−3 mr=\dfrac{9.358\times10^{-25}}{1.602\times10^{-19}\times10^{-3}}=5.84\times10^{-3}\ \mathrm{m}. The options 3.54 mm3.54\ \mathrm{mm} and 4.13 mm4.13\ \mathrm{mm} come from using 1.1 eV1.1\ \mathrm{eV} (the 400 nm400\ \mathrm{nm} data) and 1.5 eV1.5\ \mathrm{eV}, and 8.26 mm8.26\ \mathrm{mm} from using 6 eV6\ \mathrm{eV}.
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