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The number of revolutions per second mad...

The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of `3`:

A

`6.57xx10^(15)`

B

`6.57xx10^(13)`

C

`1000`

D

`6.57xx10^(14)`

Text Solution

Verified by Experts

The correct Answer is:
A
`v_(1)(t)=n.2pir_(1)Rightarrown=?`
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