An electron with energy $$0.1 \text{ keV}$$ moves at right angle to the earth's magnetic field of $$1 \times 10^{-4} \text{ Wb m}^{-2}$$. The frequency of revolution of the electron will be (Take mass of electron $$= 9.0 \times 10^{-31} \text{ kg}$$)

We need to find the frequency of revolution of an electron moving at a right angle to the Earth's magnetic field.

1. Identify the Given Parameters

• Magnetic field ($$B$$) = $$1 \times 10^{-4}\text{ Wb m}^{-2}$$ (or Tesla, $$\text{T}$$)
• Mass of an electron ($$m$$) = $$9.0 \times 10^{-31}\text{ kg}$$
• Charge of an electron ($$e$$) = $$1.6 \times 10^{-19}\text{ C}$$
• Kinetic energy = $$0.1\text{ keV}$$ (Note: Since the electron moves at a right angle, its path is perfectly circular. The frequency of revolution in a magnetic field is independent of its speed or energy).

2. Formula for Frequency of Revolution

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force provides the necessary centripetal force:

$$evB = \frac{mv^2}{r} \implies \frac{v}{r} = \frac{eB}{m}$$

Since the angular frequency is $$\omega = \frac{v}{r} = 2\pi f$$, the cyclotron frequency ($$f$$) is given by:

$$f = \frac{eB}{2\pi m}$$

3. Calculation

Substitute the given values into the frequency formula:

$$f = \frac{1.6 \times 10^{-19} \times 1 \times 10^{-4}}{2 \times 3.14 \times 9.0 \times 10^{-31}}$$

$$f = \frac{1.6 \times 10^{-23}}{56.52 \times 10^{-31}}$$

$$f \approx 0.0283 \times 10^8\text{ Hz} = 2.83 \times 10^6\text{ Hz}$$

Conclusion

The frequency of revolution of the electron is approximately $$2.8 \times 10^6\text{ Hz}$$, which matches Option C.