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Q.

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

8.8 × 105 ms–1

b

5.5 × 105 ms–1

c

5.5 × 106 ms–1

d

6.8 × 106 ms–1

answer is C.

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Detailed Solution

We know that V=c137Zn;V∝1n
V1V2=n2n1;2.2×106V2=41V2=2.2×1064=0.55×106=5.5×105

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