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

Answer: (C)

$2.8 \times 10^6 Hz$

Answer 2

f=T1​=2πmeB​
=2π×9×10−311.6×10−19×10−4​=2.8×106Hz