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JEE Main 2019 (10 Jan, Shift 2) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2019 (10 Jan, Shift 2)

Q64JEE Main 2019 · 10 JAN · SHIFT 2Difficult

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

A metal plate of area 1 ×\times 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photoelectrons per second and their maximum energy, respectively, will be

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Answer: (c)

  1. Power on plate P=IA=16×10−3×10−4=1.6×10−6P = IA = 16\times10^{-3}\times10^{-4} = 1.6\times10^{-6} W.
  2. Photon energy E=10E = 10 eV =1.6×10−18= 1.6\times10^{-18} J, so photons per second =1.6×10−61.6×10−18=1012= \dfrac{1.6\times10^{-6}}{1.6\times10^{-18}} = 10^{12}.
  3. Only 10% eject electrons: n=1011n = 10^{11} per second.
  4. Kmax=E−ϕ=10−5=5K_{max} = E - \phi = 10 - 5 = 5 eV.
  5. So the answer is (c) 101110^{11} and 5 eV.
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