An electron placed in a magnetic field of $3.14 × 10^{-4} T$ moves in a circle. The frequency of the revolution of the electron is

$8.79 × 10^6 Hz$

The correct answer is Option (1) → $8.79 × 10^6 Hz$

Given:

Magnetic field, B = 3.14 × 10⁻⁴ T

Charge of electron, e = 1.6 × 10⁻¹⁹ C

Mass of electron, m = 9.11 × 10⁻³¹ kg

Frequency of revolution of a charged particle in a magnetic field:

$f = \frac{e B}{2 \pi m}$

Substitute values:

$f = \frac{1.6 \times 10^{-19} \cdot 3.14 \times 10^{-4}}{2 \pi \cdot 9.11 \times 10^{-31}}$

$f = \frac{5.024 \times 10^{-23}}{2 \pi \cdot 9.11 \times 10^{-31}}$

$f = \frac{5.024 \times 10^{-23}}{5.723 \times 10^{-30}} \approx 8.78 \times 10^6\ \text{Hz}$

Frequency of revolution = 8.78 × 10⁶ Hz