Q.

When a proton is accelerated in a cyclotron device, maximum energy attained by it is 10 MeV. If radius of ‘Dee’ of the cyclotron is doubled and the magnetic field is doubled, then the maximum energy of on α-particle when accelerated in the cyclotron will be

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

240 MeV

b

80 MeV

c

160 MeV

d

20 Mev

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Maximum kinetic energy attained by a charged particle when accelerated in the cyclotron is given by K=Q2B2R22m∴KαKp=QαQp2.BαBp2.RαRp2.mpmα =(2)2.(1/2)2.(2)2.(4) ⇒Kα=10×16 MeV = 160 MeV.
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
When a proton is accelerated in a cyclotron device, maximum energy attained by it is 10 MeV. If radius of ‘Dee’ of the cyclotron is doubled and the magnetic field is doubled, then the maximum energy of on α-particle when accelerated in the cyclotron will be