Answer:
Frequency, [tex]f=1.2\times 10^{10}\ Hz[/tex]
Explanation:
It is given that,
Magnetic field, B = 0.43 T
We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
[tex]f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}[/tex]
[tex]f=1.2\times 10^{10}\ Hz[/tex]
Hence, this is the required solution.
