| ⇦ |
| ⇨ |
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: - A radioactive substance has a half-life of four months. Three-fourth of substance
- A transistor-oscillator using a resonant circuit with an inductor L
- Three masses are placed on the x-axis : 300 g at origin, 500 g at x = 40 cm
- A jar is filled with two non-mixing liquids 1 and 2 having densities
- Astigmatism is corrected by using
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A radioactive substance has a half-life of four months. Three-fourth of substance
- A transistor-oscillator using a resonant circuit with an inductor L
- Three masses are placed on the x-axis : 300 g at origin, 500 g at x = 40 cm
- A jar is filled with two non-mixing liquids 1 and 2 having densities
- Astigmatism is corrected by using
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