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JEE Main 2014 (12 Apr, Shift 1) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2014 (12 Apr, Shift 1)

Q65JEE Main 2014 · 12 APR · SHIFT 1Difficult

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

A beam of light has two wavelengths 4972 A˚\text{\AA} and 6216 A˚\text{\AA} with a total intensity of 3.6 ×\times 10−-3 Wm−2{}^{-2} equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in 2s is approximately:

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Answer: (b)

  1. Photon energies: E=12400λ(A˚)E = \dfrac{12400}{\lambda(\text{Å})} eV gives 124004972≈2.49\dfrac{12400}{4972} \approx 2.49 eV and 124006216≈1.99\dfrac{12400}{6216} \approx 1.99 eV.
  2. Only the 4972 Å photons exceed ϕ=2.3\phi = 2.3 eV, so only they eject electrons.
  3. Power of that half: P=1.8×10−3×10−4=1.8×10−7P = 1.8\times10^{-3}\times10^{-4} = 1.8\times10^{-7} W.
  4. Electrons per second =1.8×10−72.49×1.6×10−19≈4.5×1011= \dfrac{1.8\times10^{-7}}{2.49\times1.6\times10^{-19}} \approx 4.5\times10^{11}; in 2 s ≈9×1011\approx 9\times10^{11}.
  5. So the answer is (b) 9×10119\times10^{11}.
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