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Topic: Matter waves, de Broglie relation· Dual Nature of Matter and Radiation

Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ_0) of the emitted X-ray is

1. \(\lambda_0=\frac{2h}{mc}\)

2. \(\lambda_0=\frac{2m^2c^2\lambda^3}{h^2}\)

3. \(\lambda_0=\lambda\)

4. \(\lambda_0=\frac{2mc\lambda^2}{h}\)

1 answer

Best answer

Answer: 4. \(\lambda_0=\frac{2mc\lambda^2}{h}\)

The cutoff wavelength corresponds to the whole kinetic energy \(K\) of the electron turning into one photon: \(\lambda_0=\dfrac{hc}{K}\).

  • de Broglie: \(p=\dfrac h\lambda\), so \(K=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}\).
$$\lambda_0=\frac{hc}{K}=\frac{hc\cdot2m\lambda^2}{h^2}=\frac{2mc\lambda^2}{h}$$

Option (4).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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