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 (a) λ 0 = (2mcλ²)/(h) (b) λ₀ = 2h/mc (c) λ₀ = (2m²c²λ³)/(h²) (d) λ₀ = λ

Answer: (a). Electron kinetic energy: E = p²/2m = (h²)/(2mλ²) Cut-off: λ₀ = hc/E = hc · (2mλ²)/(h²) λ₀ = (2mcλ²)/(h) So the answer is (a).

Electrons of mass m with de-Broglie wavelength λ fall on the target… | NEET 2016 (Phase 2)

Q13NEET 2016 · PHASE 2Medium

Davisson-Germer, Uncertainty, Compton & X-Rays

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

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Answer: (a)

Electron kinetic energy: $E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$
Cut-off: $\lambda_0 = \frac{hc}{E} = hc \cdot \frac{2m\lambda^2}{h^2}$
$\lambda_0 = \frac{2mc\lambda^2}{h}$
So the answer is (a).

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