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NEET2026PhysicsChapterActual

Light of wavelength is incident on a metal surface. The de-Broglie wavelength of the fastest emitted photoelectrons is _d . If m is the mass of an electron and c is the speed of light in vacuum, the threshold wavelength ₀ of the metal is given by:

Options

  1. A2mc _d^2 2mc _d^2 + h
  2. B2mc _d^2 2mc _d^2 - h
  3. C2mc _d^2 2mc _d^2 - h
  4. D2mc _d^2 h - 2mc _d^2

Correct answer

B. 2mc _d^2 2mc _d^2 - h

Step-by-step solution

From Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by: K_ max = hc - hc ₀ The de-Broglie wavelength of the fastest photoelectron is related to its maximum kinetic energy by: _d = h 2m K_ max Squaring both sides and rearranging for K_ max : K_ max = h^2 2m _d^2 Equating the two expressions for K_ max : hc - hc ₀ = h^2 2m _d^2 Dividing the entire equation by hc : 1 - 1 ₀ = h 2mc _d^2 Rearranging to solve for 1 ₀ : 1 ₀ = 1 - h 2mc _d^2 Taking the common denominato

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