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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
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Topics: Atoms and Nuclei
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A thin wire of resistance 4 ohm is bent to form a circle. The resistance across
- Three capacitors each of capacity 4 μF are to be connected in such a way that the effective
- Two parallel long wires carry currents i₁ and i₂ with i₁ > i₂ .
- A uranium nucleus ₉₂ U ²³⁸ emits an α-particle and a β-particle in succession.
- Large transformers, when used for some time, become hot and are cooled by circulating
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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