An electron with energy 0 . 1 keV moves at right angle to the earth's magnetic field of 1 × 10 - 4 Wbm - 2 . The frequency of revolution of the electron will be (Take mass of electron = 9 . 0 × 10 - 31 kg )

Step-by-step solution

The frequency of revolution of the electron is f = e B 2 π m , here, e is charge of electron, B is magnetic field and m is mass of electron. So, f = 1 . 6 × 10 - 19 × 10 - 4 2 π × 9 × 10 - 31 = 2 . 8 × 10 6   Hz