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

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The angular speed of the electron in the `n^(th)` Bohr orbit of the hydrogen atom is proportional to
A. directly proportional to n
B. inversely proportional to `sqrt(n)`
C. inversely proportional to `n^2`
D. inversely proportional to `n^3`
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Correct Answer - D
`omega=(v)/(r).`Further `vprop(1)/(n)` and `rpropn^2`. Hence `omegaprop(1//n^3)`
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