Question

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

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

Answer 1

Answer: (B)

De Broglie wavelength $=\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }$ ......(i)
$K.E$ of electrons $E=\frac{p^2}{2m}$ ....(ii)
From (i) & (ii),
$E=\frac{h^2}{{\lambda }^2\times 2m}$
$\therefore$ The cut-off wave length ${\lambda }_0=\frac{hc}{E}=\frac{hc}{h^2}\times {\lambda }^2\times 2m=\frac{2mc{\lambda }^2}{h}$