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Dual Nature of Radiation question

2019 · 9 Apr · Shift 1 · Q60
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Dual Nature of Radiation question

2019 · 9 Apr · Shift 1 · Q60

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The electric field of light wave is given as E→=10−3cos⁡(2πx5×10−7−2π×6×1014t)x∧NC\overrightarrow E = {10^{ - 3}}\cos \left( {{{2\pi x} \over {5 \times {{10}^{ - 7}}}} - 2\pi \times 6 \times {{10}^{14}}t} \right)\mathop x\limits^ \wedge {{\rm N} \over C}E=10−3cos(5×10−72πx​−2π×6×1014t)x∧CN​ This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is : Given, E (in eV) = 12375/λ\lambdaλ(inÅ)
  1. A
    2.48 V
  2. B
    0.48 V
  3. C
    0.72 V
  4. D
    2.0 V
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