An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57×10^(-2) T. If, the value of e/m is 1.76×10^11 C kg^(-1), the frequency of revolution of the electron is

Correct answer: (A) 1 GHz. f=qB/(2πm)=(1.76×10^11×3.57×10^(-2))/(2×3.14) =10^9 Hz=1 GHz.

Q117 / 180 · NEET 2016 (Phase 2) · Physics

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{- 2} T$ . If, the value of $\frac{e}{m}$ is $1 .76 \times 10^{11} C {kg}^{- 1}$ , the frequency of revolution of the electron is

$f = \frac{q B}{2 π m} = \frac{1.76 \times 10^{11} \times 3.57 \times 10^{- 2}}{2 \times 3.14}$

$= 10^{9} Hz = 1 GHz$ .

Hence, the correct answer is option: 'A'.