MHT CET202616 April 2026Evening ShiftPhysicsDual Nature of MatterActual
A photoemissive substance is illuminated with a radiation of wavelength _i so that it releases electrons with de-Broglie wavelength _e . The longest wavelength of radiation that can emit photoelectron is ₀ . Expression for de-Broglie wavelength is (m = mass of electron, h = Planck's constant, C = Speed of light)
Options
- A( h _i/2 mc )^ 1 2
- B( h ₀/2 mc )^ 1 2
- C[ h /2 mc ( 1 _i - 1 ₀ )]^ 1 2
- D[ h /[2 mc ( 1 _i - 1 ₀ )]^ 1 2
Correct answer
C. [ h /2 mc ( 1 _i - 1 ₀ )]^ 1 2
Step-by-step solution
Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is given by: K_ max = hc _i - hc ₀ = hc ( 1 _i - 1 ₀ ) The de-Broglie wavelength of the emitted electron is: _e = h p = h 2mK_ max Substituting the value of K_ max : _e = h 2mhc ( 1 _i - 1 ₀ ) _e = h^2 2mhc ( 1 _i - 1 ₀ ) _e = [ h 2mc ( 1 _i - 1 ₀ ) ]^ 1 2