| ⇦ | 
| ⇨ |
The oscillating frequency of a cyclotron is 10 MHz. If the radius of its Dees is o.5 m, the kinetic energy of a proton, which is accelerated by the cyclotron is
Options
(a) 10.2 MeV
(b) 2.55 MeV
(c) 20.4 MeV
(d) 5.1 MeV
Correct Answer:
5.1 MeV
Explanation:
qvB = mv² / r
⇒ (1/2) (mv² / e) = Kinetic energy in electron
v² = r²ω² = r².4π²v² ʋ = 10 × 10⁶ Hz = 10⁷ Hz.
Therefore, K.E. = (1/2) (mv² / e)
= (1/2) [1.673 × (0.5 × 2π × 10⁷)² / (1.6 × 10⁻¹⁹)
= 5.1 Mev
Related Questions: - In a tangent galvanometer, a current of 0.1 A produces a deflection of 30⁰.
- Two parallel metal plates having charges +Q and -Q face each other at a certain
- Which of the following statement about scalar quantity is true?
- The lowest frequency of light that will cause the emission of photoelectrons
- Weight of a body of mass m decreases by 1% when it is raised to height h
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a tangent galvanometer, a current of 0.1 A produces a deflection of 30⁰.
- Two parallel metal plates having charges +Q and -Q face each other at a certain
- Which of the following statement about scalar quantity is true?
- The lowest frequency of light that will cause the emission of photoelectrons
- Weight of a body of mass m decreases by 1% when it is raised to height h
Topics: Magnetic Effects of Current and Magnetism (167)
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