Question

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

• A. ${\lambda }_0=\frac{2m{\lambda }^{2}c}{h}$
• B. ${\lambda }_0=\frac{2m^2{c}^2{\lambda }^{2}c}{h^2}$
• C. ${\lambda }_0=\frac{2h}{mc}$
• D. ${\lambda }_0=\lambda$

Answer 1

The correct option is A ${\lambda }_0=\frac{2m{\lambda }^{2}c}{h}$

Momentum of striking electrons $p=\frac{h}{\lambda }$
$\therefore$ Kinetic energy of electrons $k=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$
The energy equal to maximum energy of X-ray photons
Therefore, $\frac{hc}{{\lambda }_0}=\frac{h^2}{2m{\lambda }^2}\Rightarrow {\lambda }_0=\frac{2m{\lambda }^{2}c}{h}$