Question

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an $X-ray$ tube. The cut-off wavelength of the emitted $X-rays$ is

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

Answer 1

Answer: (A)

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

Answer 2

Momentum of striking electrons
$\, \, \, \, p = \frac{h}{\lambda}$
$\therefore$ Kinetic energy of striking electrons
$\, \, \, K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$
This is also, maximum energy of X-ray photons.
$Therefore, \frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}\, \, or\, \, \lambda_0 = \frac{2m\lambda^2c}{h}$