Question

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
\[A.\] $10.2MeV$
$B.$ $2.55MeV$
$C.$ $20.4MeV$
$D.$ $5.1MeV$

Answer 1

Hint: 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:
$f = 10MHz = 10 \times {10^6}Hz$
$r = 0.5m$
Kinetic energy, ${E_k} = ?$
Kinetic energy of charged particle of cyclotron is given by,
${E_k} = \dfrac{{{q^2}{B^2}{r^2}}}{{2m}}$ ………$\left( 1 \right)$
And frequency of cyclotron is given by,
$f = \dfrac{{qB}}{{2\pi m}}$ ………. $\left( 2 \right)$
On simplifying above equation we get
$ \Rightarrow qB = 2\pi mf$ ………. $\left( 3 \right)$
Substituting equation $\left( 3 \right)$ in equation $\left( 1 \right)$
${E_k} = \dfrac{{{{\left( {2\pi mf} \right)}^2}{r^2}}}{{2m}}$
On further simplification
$ \Rightarrow {E_k} = 2{\pi ^2}m{f^2}{r^2}$ ………….$\left( 4 \right)$
Substituting the given values in above equation
${E_k} = 2 \times {\left( {3.14} \right)^2} \times 1.67 \times {10^{ - 27}} \times {\left( {10 \times {{10}^6}} \right)^2} \times {\left( {0.5} \right)^2}$
On calculating we get,
\[{E_k} = 8.23 \times {10^{ - 13}}J\]
We know that, \[1eV = 1.6 \times {10^{ - 19}}J\] or $1J = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}eV$
Therefore, on converting $J$ to $eV$ we get
${E_K} = \dfrac{{8.23 \times {{10}^{ - 13}}}}{{1.6 \times {{10}^{ - 19}}}}eV$
Therefore, ${E_k} = 5.1 \times {10^6}eV$
$ \Rightarrow $ Kinetic energy of charged particle ${E_k} = 5.1MeV$
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.