Question

An electron with energy $0.1\, keV$ 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 $=9.0 \times 10^{-31} kg$ )

• 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

Answer: (C)

$f =\frac{1}{ T }=\frac{ eB }{2 \pi m } $
$ =\frac{1.6 \times 10^{-19} \times 10^{-4}}{2 \pi \times 9 \times 10^{-31}}=2.8 \times 10^6 Hz$