Question

Electrons of mass m with de-Broglie wavelength $\lambda$ fall on the target in an X-rays tube. The cutoff wavelength $\left({\lambda }_0\right)$ of the emitted X-rays is

• A. ${\lambda }_0=\lambda$
• B. ${\lambda }_0=\frac{2mc{\lambda }^2}{h}$
• C. ${\lambda }_0=\frac{2h}{mc}$
• D. ${\lambda }_0=\frac{2m^2{c}^2{\lambda }^3}{h^2}$

Answer 1

Answer: (B)

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

Using de-Broglie equation $\lambda =\frac{h}{p}$ where $p=\sqrt{2mE}$

$⟹$ $\lambda =\frac{h}{\sqrt{2mE}}$

Energy of the X-ray emitted $E=\frac{hc}{{\lambda }_o}$

$\therefore$ $\lambda =\frac{h}{\sqrt{2m\times \frac{hc}{{\lambda }_o}}}$ $⟹{\lambda }_o=\frac{2mc{\lambda }^2}{h}$