Frequency: [tex]1.27\cdot 10^8 Hz[/tex]
Explanation:
The force experienced by an electron in a magnetic field is
[tex]F=qvB[/tex]
where
[tex]q=1.6\cdot 10^{-19}C[/tex] is the electron charge
v is the speed of the electron
B is the strength of the magnetic field
Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:
[tex]qvB=m\frac{v^2}{r}[/tex]
where
r is the radius of the orbit
[tex]m=9.31\cdot 10^{-31}kg[/tex] is the mass of the electron
Re-arranging the equation,
[tex]\frac{v}{r}=\frac{qB}{m}[/tex] (1)
We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):
[tex]v=\frac{2\pi r}{T}[/tex]
Substituting into (1),
[tex]\frac{2\pi}{T}=\frac{qB}{m}[/tex]
We also know that 1/T is equal to the frequency f, so
[tex]f=\frac{qB}{2\pi m}[/tex]
In this problem,
[tex]B=4.62\cdot 10^{-3}T[/tex]
Therefore, the frequency of the electrons is
[tex]f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz[/tex]
