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