| ⇦ | 
| ⇨ |
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: - The wavelength λₑ of an electron and λₚ of a photon are of same energy E are related by
- A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle
- A magnet of length 10 cm and magnetic moment 1 A-m² is placed
- In Moseley’s law √ν=a(z-b), the values of the screening constant for K-series
- Two poles of same strength attract each other with a force of magnitude F
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The wavelength λₑ of an electron and λₚ of a photon are of same energy E are related by
- A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle
- A magnet of length 10 cm and magnetic moment 1 A-m² is placed
- In Moseley’s law √ν=a(z-b), the values of the screening constant for K-series
- Two poles of same strength attract each other with a force of magnitude F
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