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

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a
6.8 × 106 ms–1
b
8.8 × 105 ms–1
c
5.5 × 105 ms–1
d
5.5 × 106 ms–1

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detailed solution

Correct option is C

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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