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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 × 10-31kg)

Option 1 -

1.6 * 105Hz    

Option 2 -

5.6 * 105Hz

Option 3 -

2.8 * 106 Hz

Option 4 -

1.8 * 106 Hz

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago
    Correct Option - 3


    Detailed Solution:

    f = qB2πm=1.6×10−19×1.0×10−42×3.14×9×10−31=0.028×108Hz=2.8×106Hz

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