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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 particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- The kinetic energy of α-particle emitted in the α-dacay of ₈₈Ra²²⁶ is
- A ray of light is incident at an angle of incidence i, on one face of a prism
- The escape velocity from the earth is Ve. The escape velocity from a planet
- A particle has initial velocity(3i⃗ +4j⃗)and has acceleration (0.4i⃗ +0.3j⃗)
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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