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. $\lambda _{0}=\frac{2 h}{m c}$
• B. $\lambda _{0}=\frac{2 m c \lambda ^{2}}{h}$
• C. $\lambda _{0}=\frac{2 m^{2} c^{2} \lambda ^{2}}{h^{2}}$
• D. $\lambda _{0}=\lambda $

Answer 1

Answer: (B)

De Broglie wavelength $=\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }$ ......(i)
$K.E$ of electrons $E=\frac{p^{2}}{2 m}$ ....(ii)
From (i) & (ii),
$E=\frac{h^{2}}{\lambda ^{2} \times 2 m}$
$\therefore $ The cut-off wave length $\lambda _{0}=\frac{h c}{E}=\frac{h c}{h^{2}}\times \lambda ^{2}\times 2m=\frac{2 mc \lambda ^{2}}{h}$