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Determine the lowest energy of photon emitted in Balmer series of hydrogen atom.

Option 1 -

10.02 eV

Option 2 -

1.88 eV

Option 3 -

1.65 eV

Option 4 -

 2.02 eV

0 1 View | Posted 2 weeks ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 weeks ago
    Correct Option - 2


    Detailed Solution:

    For 3 ->2 transitions

                DE = 13.6   ( 1 4 ? 1 9 )

                = 13.6 *   5 3 6

                = 1.88 eV

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