| ⇦ | 
| ⇨ |
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: - Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
- The period of oscillation of a mass M suspended from a spring of negligible mass is T
- In Young’s experiment, the third bright band for light of wavelength 6000Å
- A solid sphere of mass M, radius R and having moment of inertia about an axis passing
- A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 cm
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
- The period of oscillation of a mass M suspended from a spring of negligible mass is T
- In Young’s experiment, the third bright band for light of wavelength 6000Å
- A solid sphere of mass M, radius R and having moment of inertia about an axis passing
- A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 cm
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply