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A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 × 10-34 Js)

Option 1 -

0.19 * 1015 MHz

Option 2 -

1.45 * 109 MHz

Option 3 -

1.45 * 1016 MHz

Option 4 -

9.0 * 1027 MHz

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

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 2


    Detailed Solution:

    Energy of electron in first existed state will be -3.4 eV.

    So total energy difference will be (2.6 + 3.4) eV.

    Wavelength ( λ ) = 1 2 4 2 e V − n m 6 e V = 2 0 7 n m

    F r e q u e n c y = c λ = 3 × 1 0 8 2 0 7 × 1 0 − 9 = 1 . 4 5 × 1 0 9 M H z

     

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