JIPMER2015PhysicsElectromagnetic Waves
The de-Broglie wavelength of electron falling on the target in an X-ray tube is . The cut-off wavelength of the emitted X-ray is
Options
- A₀= (m c )^2 h
- B₀= m^2 c h^2
- C₀= 2 m c ^2 h
- D₀= m c ^2 h^2
Correct answer
C. ₀= 2 m c ^2 h
Step-by-step solution
The de-Broglie wavelength is given by = h p = h 2 m E ...(i) where, E is the energy of the electron. The cut-off wavelength ₀ is given by ₀= h C E ...(ii) From Eq. (i), ^2= h^2 2 m E E= h^2 2 m X^2 ... (iii) Substituting the value of E from Eq. (ii), we get ₀= h c h^2 2 m ^2 = 2 m c ^2 h