The oscillating frequency of a cyclotron is 10 MHz. If the radius of its Dees is o.5 m,

⇦

⇨

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:

1. Unit of electrical conductivity is
2. Which of the following phenomena support the wave theory of light?
3. The earth radiates in the infrared region of the spectrum. The spectrum is correctly
4. The name of ions knocked out from the hot surface is
5. Two balls are dropped from heights h and 2h respectively. The ratio of times

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