Question

Electron with de Broglie wavelength $\lambda$fall on the target in and X ray tube. The cut off wavelength ($\lambda_{0}$) 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. )$\lambda_{0} = \lambda$

Answer 1

Answer: (A)

$\lambda_{0} = \frac{2mc\lambda^{2}}{h}$

Answer 2

: Let K be the kinetic energy of the incident electron.

Its linear momentum, $p = \sqrt{2mK}$

The de Broglie wavelength is related to the linear momentum as

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mk}}orK = \frac{h^{2}}{2m\lambda^{2}}$

The cut off wavelength of the emitted X rays is related to the kinetic energy of the incident electrons as

$\frac{hc}{\lambda_{0}} = K = \frac{h^{2}}{2m\lambda^{2}}or\lambda_{0} = \frac{2mc\lambda^{2}}{h}$