| ⇦ | 
| ⇨ |
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 mixture consists of two radioactive materials A₁ and A₂ with half lives of 20s
- One way in which the operation of a n-p-n transistor differs from that of a p-n-p
- The heart of a man pumps 5 litres of blood through the arteries per minute at
- Maximum kinetic energy of electrons emitted in photoelectric effect increases when
- If the ratio of amplitude of two superposed waves is 2:1, then the ratio of maximum
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20s
- One way in which the operation of a n-p-n transistor differs from that of a p-n-p
- The heart of a man pumps 5 litres of blood through the arteries per minute at
- Maximum kinetic energy of electrons emitted in photoelectric effect increases when
- If the ratio of amplitude of two superposed waves is 2:1, then the ratio of maximum
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