Question

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength $(\lambda_0)$ of the emitted X-ray is

• A. $\lambda_0 = \frac{2 mc \lambda^2}{h}$
• B. $\lambda_0 = \frac{2h}{mc}$
• C. $\lambda_0 = \frac{2 m^2c^2 \lambda^3}{h^2}$
• D. $\lambda_0 = \lambda$

Answer 1

Answer: (A)

$\begin{matrix} \lambda_{0} = \frac{hc}{KE_{e}} \\ \lambda_{0} = \frac{hc}{h^{2}/2m \lambda^{2}} \end{matrix} \Bigg| \begin{matrix} \lambda= \frac{h}{\sqrt{2mKE_{e}}} \\ KE_{e} = \frac{h^{2}}{2m \lambda^{2}} \end{matrix} $
$ \lambda_{0} = \frac{2mc}{h} \lambda^{2} $