MHT CET20228 Aug 2022Morning ShiftPhysicsDual Nature of MatterActual
Electrons of mass m with de-Broglie wavelength fall on the target. The cut-off wavelength ₀ is equal to [ h= Planck's constant, C= velocity of light ]
Options
- A2 m c ^2 h
- Bm c h
- C2 h m c ^2
- D2 m c h
Correct answer
A. 2 m c ^2 h
Step-by-step solution
Using de-Broglie equation = h p where p= 2 m E = h 2 m E Energy of the X-ray emitted E= h c ₀ = h 2 m h c ₀ ₀= 2 m c ^2 h