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NEET 2015 (AIPMT) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from NEET 2015 (AIPMT)

Q147NEET 2015 · AIPMTMedium

Photoelectric Effect: Einstein's Equation

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

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Answer: (a)

  1. Photon energy =1240500=2.48= \frac{1240}{500} = 2.48 eV, so Kmax=2.48−2.28=0.20K_{max} = 2.48 - 2.28 = 0.20 eV.
  2. λ=h2mK=1.2270.20\lambda = \frac{h}{\sqrt{2mK}} = \frac{1.227}{\sqrt{0.20}} nm ≈2.8×10−9\approx 2.8 \times 10^{-9} m
  3. Electrons with less than maximum KE have larger λ\lambda, so λ≥2.8×10−9\lambda \geq 2.8 \times 10^{-9} m.
  4. So the answer is (a).
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