Question

Electrons of mass m with de Broglie wavelength lambda fall on the target in an X-ray tube. The cut-off wavelength λ₀ of the emitted X-rays is:

• A. λ₀ = λ
• B. λ₀ = 2mcλ²/h
• C. λ₀ = 2h/mc
• D. λ₀ = 2m²c²λ³/h²

Answer 1

The correct option is B

λ₀ = 2mcλ²/h

The explanation for the correct option:

Let ${\mathrm{E}}_{\mathrm{k}}$be the kinetic energy of the electron

Its linear momentum, $\mathrm{p}=\sqrt{2\mathrm{mE}}$

Using de-Broglie wavelength

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}$

$\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$

The cut-off wavelength of the emitted X-rays is related to the KE of the incident electron as

$\mathrm{E}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_0}$

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{m}}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_0} \\ \mathrm{we}\mathrm{get}, \\ {\mathrm{\lambda }}_0=\frac{2{\mathrm{mc\lambda }}^2}{\mathrm{h}} \end{gathered}$$

Electrons of mass m with de Broglie wavelength lambda fall on the target in an X-rays tube. The cut-off wavelength λ₀ of the emitted X-rays is: ${\mathrm{\lambda }}_0=\frac{2{\mathrm{mc\lambda }}^2}{\mathrm{h}}$

Therefore Option B is the correct option.