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The angular speed of the electron in the...

The angular speed of the electron in the `n^(th)` Bohr orbit of the hydrogen atom is proportional to

A

`n^(2)`

B

`1/n^(2)`

C

`1/(n^(3//2))`

D

`1/n^(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
We know that, angular momentum, `L=mr^(2)omega`
Again, from Bohr's theory,
angular momentum, `L=(nh)/(2pi)`
`therefore momegar^(2)=(nh)/(2pi) rArr omega propto n/r^(2)`
Now, for an electronn in H-atom, radius, `r propto n^(2)`
`therefore omega propto n/(n^(2))^(2) rArr omega propto 1/n^(3)`
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