Answer
Given,
Magnetic field, B = 0.0000193 T
speed, v = 121 m/s
mass of electron, m = 9.11 x 10⁻³¹ Kg
charge of electron, q = 1.6 x 10⁻¹⁹ C
radius of the electron path, r = ?
[tex]r = \dfrac{mv}{qB}[/tex]
[tex]r = \dfrac{9.31\times 10^{-31}\times 121}{1.6\times 10^{-19}\times 0.0000193}[/tex]
r = 3.64 x 10⁻⁵ m
We know frequency is inverse of time period
d = v t
[tex]2 \pi r = v \times t[/tex]
[tex]t = \dfrac{2 \pi r}{v}[/tex]
[tex]t = \dfrac{2 \pi \times 3.64\times 10^{-5}}{121}[/tex]
t = 1.889 x 10⁻⁶ s.
now, frequency
[tex]f = \dfrac{1}{t}[/tex]
[tex]f = \dfrac{1}{1.889\times 10^{-6}}[/tex]
[tex]f = 529380\ Hz[/tex]
