Q. Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is
A) $\lambda_{0} = \frac{2mc\lambda^{2}}{h}$
B) $\lambda_{0} = \frac{2h}{mc}$
C) $\lambda_{0} = \frac{2m^{2}c^{2}\lambda^{2}}{h^{2}}$
D) $\lambda_{0} = \lambda$
Explanation
(a) Momentum of striking electrons $p = \frac{h}{\ \lambda}$
$\therefore$ Kinetic energy of striking electrons
$K = \frac{p^{2}}{2m} = \frac{h^{2}}{2m\lambda^{2}}$
This is also, maximum energy of X-ray photons.
Therefore, $\frac{hc}{\ \lambda_{0}} = \frac{h^{2}}{2m\lambda^{2}}$
Or $\ \lambda_{0} = \frac{2m\lambda^{2}c}{h}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.