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JEE Main 2006 (AIEEE) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2006 (AIEEE)

Q161JEE Main 2006 · AIEEEEasy

Photoelectric Effect: Einstein's Equation

The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV6.2\text{ eV} and the stopping potential for a radiation incident on this surface is 5 V.5\text{ V.} The incident radiation lies in

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Answer: (a)

  1. E=ϕ+eV0=6.2+5=11.2 eVE = \phi + eV_0 = 6.2 + 5 = 11.2\text{ eV}
  2. λ=124011.2≈111 nm\lambda = \frac{1240}{11.2} \approx 111\text{ nm}
  3. This is below 400 nm and far above X-ray wavelengths (below about 10 nm), so it is ultra-violet.
  4. So the answer is (a) ultra-violet region.
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