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 )

Question

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 ) (A) 1.6 × 105 Hz (B) 5.6 × 105 Hz (C) 2.8 × 106 Hz (D) 1.8 × 106 Hz

Tardigrade Admin · Accepted Answer

f =(1/ T )=( eB /2 π m ) =(1.6 × 10-19 × 10-4/2 π × 9 × 10-31)=2.8 × 106 Hz

Q. An electron with energy $0.1 k e V$ moves at right angle to the earth's magnetic field of $1 \times 1 0^{- 4} Wb m^{- 2}$ . The frequency of revolution of the electron will be (Take mass of electron $= 9.0 \times 1 0^{- 31} k g$ )

$1.6 \times 1 0^{5} Hz$

$5.6 \times 1 0^{5} Hz$

$2.8 \times 1 0^{6} Hz$

$1.8 \times 1 0^{6} Hz$

$f = \frac{1}{T} = \frac{e B}{2 πm}$
$= \frac{1.6 \times 1 0 ^{- 19} \times 1 0 ^{- 4}}{2 π \times 9 \times 1 0 ^{- 31}} = 2.8 \times 1 0^{6} Hz$

Q. An electron with energy 0.1keV moves at right angle to the earth's magnetic field of 1×10−4Wbm−2. The frequency of revolution of the electron will be (Take mass of electron =9.0×10−31kg )

1.6×105Hz

5.6×105Hz

2.8×106Hz

1.8×106Hz