JEE MainPhysicsDual Nature of Matter
Light of wavelength 310 nm illuminates a metal surface having a work function of 2.0 eV , and the emitted photoelectrons have a certain minimum de Broglie wavelength. When the incident light is changed to a new wavelength ₂ , the minimum de Broglie wavelength of the emitted photoelectrons becomes half of its initial value. The value of ₂ is: (Take hc = 1240 eV nm )
Options
- A207 nm
- B413 nm
- C124 nm
- D155 nm
Correct answer
C. 124 nm
Step-by-step solution
From Einstein's photoelectric equation, the maximum kinetic energy of the photoelectrons for the first incident wavelength is: K₁ = hc ₁ - K₁ = 1240 310 - 2.0 = 4.0 - 2.0 = 2.0 eV The de Broglie wavelength of an electron is related to its kinetic energy by: _d = h 2mK This implies that K 1 _d^2 . Since the minimum de Broglie wavelength is halved ( _ d2 = _ d1 2 ), the new maximum kinetic energy must be four times the initial maximum kinetic energy: K₂ = 4K₁ = 4 2.0 eV = 8.0 eV Applying the photoelectric equation fo