Question

An electron with energy $01\, keV$ moves at right angle to the earth's magnetic field of $1 \times 10^{-4} Wbm ^{-2}$ The frequency of revolution of the electron will be (Take mass of electron $=90 \times 10^{-31} kg$ )

• A. $1.6 \times 10^5 Hz$
• B. $5.6 \times 10^5 Hz$
• C. $2.8 \times 10^6 Hz$
• D. $1.8 \times 10^6 Hz$

Answer 1

The Correct Option is C

To find the frequency of revolution of an electron in a magnetic field, we will use the formula for the cyclotron frequency. The cyclotron frequency or gyrofrequency \(\nu\) is given by:

\(\nu = \frac{qB}{2\pi m}\)

where:

• \(q\) is the charge of the electron, \(1.6 \times 10^{-19} C\)
• \(B\) is the magnetic field strength, \(1 \times 10^{-4} Wb/m^2\)
• \(m\) is the mass of the electron, \(9 \times 10^{-31} kg\)

Plugging in these values, we get:

\(\nu = \frac{(1.6 \times 10^{-19}) \times (1 \times 10^{-4})}{2\pi \times (9 \times 10^{-31})}\)

\(\nu = \frac{1.6 \times 10^{-23}}{2\pi \times 9 \times 10^{-31}}\)

\(\nu = \frac{1.6}{2\pi \times 9} \times 10^{7}\)

Calculating further,

\(\nu \approx \frac{1.6}{56.52} \times 10^{7}\)

\(\nu \approx 2.83 \times 10^6 Hz\)

Thus, the frequency of revolution of the electron is approximately \(2.8 \times 10^6 Hz\). This matches the given correct answer.

Therefore, the correct option is:

• \(2.8 \times 10^6 Hz\)