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

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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. The potential of a large liquid drop when eight liquid drops are combined is 20 V.
2. The kinetic energy of an electron in the first excited state is
3. Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons
4. A particle moving in a straight line with uniform acceleration is observed
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Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)

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