Question

Electrons with de-Broglie wavelength $\lambda$ fall on the target in a $X$-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. ${\lambda }_0=\frac{2m^2{c}^2{\lambda }^3}{h^2}$
• D. ${\lambda }_0=\lambda$

Answer 1

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

de-Broglie wave length is $\lambda =\frac{h}{\sqrt{2mE}}$ ...(1)

Where $E$ is the kinetic energy of the electrons. The cut-off wave length is

${\lambda }_0=\frac{hc}{E}$

From equation (1) :

$E=\frac{h^2}{2m{\lambda }^2}$

Hence, ${\lambda }_0=\frac{2mc{\lambda }^2}{h}$