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

The Correct Option is A

Wavelength and Energy Calculation

The given equations represent the relationship between the wavelength and kinetic energy:

\(\begin{matrix} \lambda_{0} = \frac{hc}{KE_{e}} \\ \lambda_{0} = \frac{hc}{\frac{h^{2}}{2m \lambda^{2}}} \end{matrix} \Bigg| \begin{matrix} \lambda = \frac{h}{\sqrt{2m KE_{e}}} \\ KE_{e} = \frac{h^{2}}{2m \lambda^{2}} \end{matrix}\)

By rearranging the equations, we get the final relationship between the wavelength and the kinetic energy:

\(\lambda_{0} = \frac{2mc}{h} \lambda^{2}\)

Explanation:

• $\lambda_0$: This represents the wavelength based on the energy and other parameters.
• $KE_e$: This represents the kinetic energy of the particle.
• $\lambda$: The wavelength related to the system's energy.
• $h$: Planck's constant.
• $m$: The mass of the particle.
• $c$: The speed of light.

Conclusion:

The equation \(\lambda_{0} = \frac{2mc}{h} \lambda^{2}\) shows the relationship between the wavelength and the kinetic energy of a particle, accounting for mass and other physical constants.