Question

Electron with de-Broglie wavelength l 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. ${\lambda }_0=\frac{2m^2{c}^2{\lambda }^3}{h^2}$
• D. ${\lambda }_0=\lambda$

Answer 1

The correct option is A

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

Let “K” be the kinetic energy of the incident electron its linear momentum $p=\sqrt{2mk}$ De - broglie wavelength

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mk}}ork=\frac{h^2}{2m{\lambda }^2}$ …..(1)

Cut off wavelength of the emmited x-ray

$k=\frac{hc}{{\lambda }_0}$ .....(2) $[valueofk=\frac{h^2}{2m{\lambda }^2}]$

$\frac{h^2}{2m{\lambda }^2}=\frac{hc}{{\lambda }_0}$

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