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Answer :
To find the frequency of revolution of an electron in a magnetic field and the radius of its path, we can use the following physics concepts and formulas.
(a) Frequency of Revolution:
The frequency of revolution (also known as the cyclotron frequency) of a charged particle in a magnetic field is determined by the formula:
[tex]f = \frac{qB}{2\pi m}[/tex]
Where:
- [tex]f[/tex] is the frequency of revolution in Hertz (Hz).
- [tex]q[/tex] is the charge of the electron, approximately [tex]1.6 \times 10^{-19}[/tex] coulombs.
- [tex]B[/tex] is the magnetic field strength, given as [tex]28.9 \ \mu \text{T} = 28.9 \times 10^{-6} \ \text{T}[/tex].
- [tex]m[/tex] is the mass of the electron, approximately [tex]9.11 \times 10^{-31} \ \text{kg}[/tex].
Let's calculate [tex]f[/tex]:
[tex]f = \frac{(1.6 \times 10^{-19}) \times (28.9 \times 10^{-6})}{2\pi (9.11 \times 10^{-31})} \approx 7.96 \times 10^{6} \ \text{Hz}[/tex]
Thus, the frequency of revolution of the electron is approximately [tex]7.96 \times 10^{6} \ \text{Hz}[/tex].
(b) Radius of Path:
The radius of the circular path of an electron moving perpendicular to a magnetic field is given by:
[tex]r = \frac{mv}{qB}[/tex]
We need to first determine the velocity [tex]v[/tex] of the electron using its kinetic energy formula:
[tex]K.E = \frac{1}{2}mv^2 = 97.9 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J/eV}[/tex]
Solving for [tex]v[/tex]:
[tex]v^2 = \frac{2 \times 97.9 \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31}} \\
v = \sqrt{\frac{2 \times 97.9 \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 5.88 \times 10^{6} \ \text{m/s}[/tex]
Now, substituting into the radius formula:
[tex]r = \frac{(9.11 \times 10^{-31}) \times (5.88 \times 10^{6})}{(1.6 \times 10^{-19}) \times (28.9 \times 10^{-6})} \approx 1.18 \times 10^{-2} \ \text{m}[/tex]
Thus, the radius of the path of the electron is approximately [tex]1.18 \times 10^{-2} \ \text{m}[/tex].
In summary:
(a) The frequency of revolution is approximately [tex]7.96 \times 10^{6} \ \text{Hz}[/tex].
(b) The radius of the path is approximately [tex]1.18 \times 10^{-2} \ \text{m}[/tex].
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