Question

Electron of mass m with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength ${\lambda }_0$ of the emitted X-ray 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 KE is the kinetic energy of electron. The relation between kinetic energy and momentum is

$KE=\frac{p^2}{2m}$

$p=\sqrt{2mKE}$

De-Broglie wavelength related to electron is

$\lambda =\frac{h}{p}$

$\lambda =\frac{h}{\sqrt{2mKE}}$

$KE=\frac{h^2}{2m\lambda }...........\left(1\right)$

Relation between cut-off wavelength of emitted X-rays is related to kinetic energy of incident electron as

$\frac{hc}{{\lambda }_0}=KE......\left(2\right)$

From equation (1) and (2)

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

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