Question

An electron with de-Broglie wavelength $\lambda$ fall on the target in an $\text{X}$-rays tube. The cut-off wavelength ${\lambda }_0$ of the emitted $\text{X}$-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 ${\lambda }_0=\frac{2mc{\lambda }^2}{h}$

Using de- Broglie equation for striking electron,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

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

But, as we know, maximum energy of photon

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

From equations $\left(1\right)\&\left(2\right)$, we get

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

Hence, option $\left(\text{A}\right)$ is correct.