If speed of electron in ground state energy level is 2.2× 106 ms–1, then its speed in fourth excited state will be

If speed of electron in ground state energy level is 2.2× 106 ms–1, then its speed in fourth excited state will be

  1. A

    6.8 × 106 ms–1

  2. B

    8.8 × 105 ms–1

  3. C

    5.5 × 105 ms–1

  4. D

    5.5 × 106 ms–1

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    Solution:

    We know that V=c137Zn;V1n
    V1V2=n2n1;2.2×106V2=41V2=2.2×1064=0.55×106=5.5×105

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