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Q.

The angular speed of the electron in the $n^{t h}$ orbit of Bohr's hydrogen atom is:

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a

directly proportional to $n$ .

b

inversely proportional to $n^{2}$ .

c

inversely proportional to $\sqrt{n}$ .

d

inversely proportional to $n^{3}$ .

answer is D.

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Detailed Solution

$ω = \frac{v}{r},$ Further $v \propto \frac{1}{n}$ and $r \propto n^{2}$ ,

Hence $ω \propto (1 / n^{3})$ .

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