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Light of wavelength 200 nm is incident on a metallic surface having a work function of 2.2 eV. The de Broglie wavelength of the fastest emitted photoelectron is closest to: (Take hc = 1240 eV.nm and the de Broglie wavelength of an electron with kinetic energy K (in eV) as 12.27 K )

Options

  1. A4.93
  2. B8.27
  3. C4.23
  4. D6.14

Correct answer

D. 6.14

Step-by-step solution

The energy of the incident photon is given by: E = hc = 1240 eV.nm 200 nm = 6.2 eV According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is: K_ = E - W = 6.2 eV - 2.2 eV = 4.0 eV The fastest photoelectron has a kinetic energy of 4.0 eV. Its de Broglie wavelength is: = 12.27 K_ Substituting the value of K_ : = 12.27 4.0 = 12.27 2.0 = 6.135 Rounding to two decimal places, the wavelength is closest to 6.14 . Answer: 6.14

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