JEE Advanced2007PhysicsDual Nature of MatterActual
Electrons with de-Broglie wavelength fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is
Options
- A₀= 2 m c ^2 h
- B₀= 2 h m c
- C₀= 2 m^2 c^2 ^3 h^2
- D₀=
Correct answer
A. ₀= 2 m c ^2 h
Step-by-step solution
Momentum of striking electrons, p= h Kinetic energy of striking electrons, K= p^2 2 m = h^2 2 m ^2 This is also, maximum energy of X -ray photons. Therefore, h c ₀ = h^2 2 m ^2 or ₀= 2 m ^2 c h Correct option is (a).