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Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ₀) of the emitted X-ray is
Options
(a) λ₀= 2mcλ²/h
(b) λ₀= 2h/mc
(c) λ₀= 2m²c²λ³/h²
(d) λ₀= λ
Correct Answer:
λ₀= 2mcλ²/h
Explanation:
λ = h / p ⇒ p = h / λ
KE of electrons = E = p² / 2m = h² / 2mλ²
Also in X-ray, λ₀ = hc / E = 2mcλ² / h
Related Questions: - In the phenomenon of electric discharge through gases at low pressure,
- The resistance of an ammeter is 13Ω and its scale is graduated for a current
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the phenomenon of electric discharge through gases at low pressure,
- The resistance of an ammeter is 13Ω and its scale is graduated for a current
- All components of the electromagnetic spectrum in vacuum have the same
- A circular disc of radius R and thickness R/6 has moment of inertia I about an axis
- A body rolls down a stair case of 5 steps. Each step has height 0.1 m and width 0.1m
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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