The velocity of electron in the 3rd excited state of a hydrogen atom is [ $a_{0}$ is Bohr radius] :

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$\frac{h}{8 π m a_{0}}$

$\frac{h^{2}}{3 π^{2} m a_{0}}$

$\frac{h^{2}}{3 π^{2} m a_{0}^{2}}$

$\frac{8 h}{π m a_{0}}$

answer is D.

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Detailed Solution

Here;

The velocity can be calculated as;
$\begin{array}{l} n = 4 \\ m v r = \frac{n h}{2 π} \Rightarrow v = \frac{n h}{2 π m r} \Rightarrow \frac{4 h}{2 π m [a_{0} \times (4)^{2}]} = \frac{h}{8 π m a_{0}} \end{array}$

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