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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
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A radioactive substance emits n beta particles in the first 2 s and 0.5 n beta
- If a gymnast, sitting on a rotating stool with his arms outstretched, suddenly
- If ₉₂U²³⁸ undergoes successively 8 ∝-decays and 6 β-decays, the resulting nucleus is
- If R is the radius of the earth and g is the acceleration due to gravity
- A calorimeter contains 0.2 kg of water at 30° C. 0.1kg of water at
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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