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Q.

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6×10−16 s. The frequency of revolution of the electron in its first excited state in s−1 is

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a

7.8×1014

b

5.6×1012

c

6.2×1015

d

1.6×1014

answer is A.

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Detailed Solution

We know Time Period , T∝rv∝n2z×nz∝n3z2               ​⇒T1T2=n13n23=18  ⇒T2=8T1 =8×1.6×10−16=12.8×10−16​Now, frequency, f2=1T2=112.8×10−16≈7.8×1014
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