Question

An electron having the de Broglie wavelength $\lambda$ falls on an X-ray tube. The cut-off wavelength of emitted X-ray is

• A. $\frac{2mc{\lambda }^2}{h}$
• B. $\frac{2h}{mc}$
• C. $\frac{h}{mc}$
• D. $\frac{2mc{\lambda }^2}{3h}$

Answer 1

The correct option is A

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

de Broglie wavelength:

de Broglie wavelength is given by, $\lambda =\frac{h}{\sqrt{2mE}}$

where $h$ is Planck constant, $m$ is mass of electron, $E$ is the energy of electron

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

For cutoff wavelength of emitted X-ray: ${\lambda }_0=\frac{hc}{E}=\frac{2mc{\lambda }^2}{h}$

here $c$is the speed of light

Hence, option A is correct.