Question

The cyclotron frequency of an electrons gyrating in a magnetic field of $1\,T$ is approximately:
A) $28\,MHz$
B) $280\,MHz$
C) $2.8\,MHz$
D) $28\,GHz$

Answer 1

Hint: To find the value of the frequency of the cyclotron of an electron, use the formula below and substitute the value of the magnetic induction, charge and mass of the electron in it , the obtained result provides the answer for the frequency of cyclotron.

Useful formula:
The formula of the frequency of the cyclotron is given by
$\nu = \dfrac{{qB}}{{2\pi m}}$
Where $\nu $ is the frequency of the cyclotron, $q$ is the charge of the electron, $B$ is the magnetic field and $m$ is the mass.

Complete step by step solution:
It is given that the
Magnetic field that is used in the cyclotron, $B = 1T$
Use the formula of the frequency of the cyclotron,

$\nu = \dfrac{{qB}}{{2\pi m}}$
Substitute the values of the charge $q = 1.6 \times {10^{ - 19}}$ , the magnetic induction from the given, $m = 9.1 \times {10^{ - 31}}$ as the mass of the electron in the above used formula.
$\nu = \dfrac{{1.6 \times {{10}^{ - 19}} \times 1}}{{2 \times 3.14 \times 9.1 \times {{10}^{ - 31}}}}$
Simplifying the above equation, we get
$\nu = \dfrac{{0.8 \times {{10}^{12}}}}{{3.14 \times 9.1}}$
By the further simplification of the above step,
$\nu = 28 \times {10^9}\,Hz$
It is known that the $1\,GHz = {10^9}\,Hz$ , substituting this in the above step
$\nu = 28 \times {10^9}\,GHz$
Hence the cyclotron frequency of the electron is obtained as $\nu = 28\,GHz$. It is not $28\,MHz$ since $1\,MHz = {10^6}\,HZ$ .

Thus the option (D) is correct.

Note: The cyclotron frequency mainly depends upon the specific charge and the magnetic induction. The specific charge is calculated by dividing the charge of the particle and the mass of it. The use of the cyclotron is to produce the radioactive isotopes which are used in the process of the imaging procedures.