IAT IISER2026PhysicsDual Nature of Matter
An experimental study of the photoelectric effect involves a metal of work function ₀ . What is the smallest wavelength of the incident photon to photoemit an electron of mass m which has the same de Broglie wavelength as that of the incident photon? [Given h is the Planck's constant, c is the speed of light, and ₀ mc^2 ]
Options
- Ah mc (1 - 1 - 2 ₀ mc^2 )⁻¹
- Bh mc (1 + 1 - 2 ₀ mc^2 )⁻¹
- Ch mc (1 - 1 - ₀ mc^2 )⁻¹
- Dh mc (1 + 1 - ₀ mc^2 )⁻¹
Correct answer
B. h mc (1 + 1 - 2 ₀ mc^2 )⁻¹
Step-by-step solution
The energy of the incident photon is E = hc . The momentum of the incident photon is p = h . By the photoelectric equation, the maximum kinetic energy of the photoemitted electron is K_ = hc - ₀ . The de Broglie wavelength of the emitted electron is given to be equal to the wavelength of the incident photon, . Thus, the momentum of the electron is p_e = h . Using the non-relativistic expression for kinetic energy, the kinetic energy of the electron is K = p_e^2 2m = h^2 2m ^2 . For the electron to be emitted, its k