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NEET PHYSICSMedium

An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57×10−23.57 \times 10^{-2} T. If the value of e/me/m is 1.76×10111.76 \times 10^{11} C/kg, the frequency of revolution of the electron is:

A

1 GHz

B

100 MHz

C

62.8 MHz

D

6.28 MHz

Step-by-Step Solution

  1. Identify the Physical Relationship: For a charged particle moving in a circular path perpendicular to a uniform magnetic field BB, the magnetic force provides the centripetal force. The angular frequency is ω=qBm\omega = \frac{qB}{m} .
  2. Frequency Formula: The frequency of revolution (cyclotron frequency) is given by ν=ω2π=12π(em)B\nu = \frac{\omega}{2\pi} = \frac{1}{2\pi} \left( \frac{e}{m} \right) B .
  3. Substitute Given Values:
  • B=3.57×10−2B = 3.57 \times 10^{-2} T
  • e/m=1.76×1011e/m = 1.76 \times 10^{11} C/kg
  • ν=12×3.14159×(1.76×1011)×(3.57×10−2)\nu = \frac{1}{2 \times 3.14159} \times (1.76 \times 10^{11}) \times (3.57 \times 10^{-2})
  1. Calculation:
  • ν=6.2832×1096.28318≈1×109\nu = \frac{6.2832 \times 10^{9}}{6.28318} \approx 1 \times 10^{9} Hz
  1. Unit Conversion: 10910^9 Hz is equal to 1 GHz. Therefore, the frequency of revolution is 1 GHz.
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