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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: - Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
- Activity of a radioactive sample decreases to (1/3)rd of its original value
- A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm
- The de-Broglie wavelength of an electron moving with a velocity 1.5×10⁸ ms⁻¹ is equal
- We get Balmer series when electron comes from
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
- Activity of a radioactive sample decreases to (1/3)rd of its original value
- A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm
- The de-Broglie wavelength of an electron moving with a velocity 1.5×10⁸ ms⁻¹ is equal
- We get Balmer series when electron comes from
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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