Question
Question: The oscillating frequency of a cyclotron is \(10MHz.\) If the radius of its Dees is \(0.5m\) , the k...
The oscillating frequency of a cyclotron is 10MHz. If the radius of its Dees is 0.5m , the kinetic energy of a proton, which is accelerated by the cyclotron is
A. 10.2MeV
B. 2.55MeV
C. 20.4MeV
D. 5.1MeV
Solution
A cyclotron is the one of the compact particle accelerators which are used to accelerate the positive charge. Some of the positively charged particles are protons, deuterons and alpha particles. The cyclotron will work in the presence of both electric field and magnetic field. Here in this problem we will find kinetic energy in joules then we will convert to electron volt.
Complete step-by-step solution:
Given:
f=10MHz=10×106Hz
r=0.5m
Kinetic energy, Ek=?
Kinetic energy of charged particle of cyclotron is given by,
Ek=2mq2B2r2 ………(1)
And frequency of cyclotron is given by,
f=2πmqB ………. (2)
On simplifying above equation we get
⇒qB=2πmf ………. (3)
Substituting equation (3) in equation (1)
Ek=2m(2πmf)2r2
On further simplification
⇒Ek=2π2mf2r2 ………….(4)
Substituting the given values in above equation
Ek=2×(3.14)2×1.67×10−27×(10×106)2×(0.5)2
On calculating we get,
Ek=8.23×10−13J
We know that, 1eV=1.6×10−19J or 1J=1.6×10−191eV
Therefore, on converting J to eV we get
EK=1.6×10−198.23×10−13eV
Therefore, Ek=5.1×106eV
⇒ Kinetic energy of charged particle Ek=5.1MeV
Hence, option D is correct. That is 5.1MeV
Note: The cyclotron is a device which works in the presence of both magnetic and electric fields. The presence of an electric field makes the positive charge like protons, deuterons and alpha particles move faster that is to increase in kinetic energy and the presence of a magnetic field makes the positive charge move in the circular path.