| ⇦ | 
| ⇨ |
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: - “On flowing current in a conducting wire the magnetic field is produced around it”.
- A wire is stretched under a force. If the wire suddenly snaps the temperature
- Two racing cars of masses m₁ and m₂ are moving in circles of radii
- A particle of mass 1mg has the same wavelength as an electron moving with a velocity
- A block of wood weighs 10 N and is resting on an inclined plane. The coefficient
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- “On flowing current in a conducting wire the magnetic field is produced around it”.
- A wire is stretched under a force. If the wire suddenly snaps the temperature
- Two racing cars of masses m₁ and m₂ are moving in circles of radii
- A particle of mass 1mg has the same wavelength as an electron moving with a velocity
- A block of wood weighs 10 N and is resting on an inclined plane. The coefficient
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