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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.10MHz. If the radius of its Dees is 0.5m0.5m , the kinetic energy of a proton, which is accelerated by the cyclotron is
A.A. 10.2MeV10.2MeV
B.B. 2.55MeV2.55MeV
C.C. 20.4MeV20.4MeV
D.D. 5.1MeV5.1MeV

Explanation

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=10×106Hzf = 10MHz = 10 \times {10^6}Hz
r=0.5mr = 0.5m
Kinetic energy, Ek=?{E_k} = ?
Kinetic energy of charged particle of cyclotron is given by,
Ek=q2B2r22m{E_k} = \dfrac{{{q^2}{B^2}{r^2}}}{{2m}} ………(1)\left( 1 \right)
And frequency of cyclotron is given by,
f=qB2πmf = \dfrac{{qB}}{{2\pi m}} ………. (2)\left( 2 \right)
On simplifying above equation we get
qB=2πmf\Rightarrow qB = 2\pi mf ………. (3)\left( 3 \right)
Substituting equation (3)\left( 3 \right) in equation (1)\left( 1 \right)
Ek=(2πmf)2r22m{E_k} = \dfrac{{{{\left( {2\pi mf} \right)}^2}{r^2}}}{{2m}}
On further simplification
Ek=2π2mf2r2\Rightarrow {E_k} = 2{\pi ^2}m{f^2}{r^2} ………….(4)\left( 4 \right)
Substituting the given values in above equation
Ek=2×(3.14)2×1.67×1027×(10×106)2×(0.5)2{E_k} = 2 \times {\left( {3.14} \right)^2} \times 1.67 \times {10^{ - 27}} \times {\left( {10 \times {{10}^6}} \right)^2} \times {\left( {0.5} \right)^2}
On calculating we get,
Ek=8.23×1013J{E_k} = 8.23 \times {10^{ - 13}}J
We know that, 1eV=1.6×1019J1eV = 1.6 \times {10^{ - 19}}J or 1J=11.6×1019eV1J = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}eV
Therefore, on converting JJ to eVeV we get
EK=8.23×10131.6×1019eV{E_K} = \dfrac{{8.23 \times {{10}^{ - 13}}}}{{1.6 \times {{10}^{ - 19}}}}eV
Therefore, Ek=5.1×106eV{E_k} = 5.1 \times {10^6}eV
\Rightarrow Kinetic energy of charged particle Ek=5.1MeV{E_k} = 5.1MeV
Hence, option DD is correct. That is 5.1MeV5.1MeV

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.