The kinetic energy of an electron in the $3^{r d}$ Bohr orbit of a hydrogen atom is [ $a_{0}$ is Bohr radius]

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$\frac{h^{2}}{2 \times 36 π^{2} m a_{0}^{2}}$

$\frac{36 h^{2}}{π^{2} m a_{0}^{2}}$

$\frac{h^{2}}{36 π^{2} m a_{0}^{2}}$

$\frac{2 h^{2}}{36 π^{2} m a_{0}^{2}}$

answer is B.

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Detailed Solution

$E \propto \frac{Z^{2}}{n^{2}} r_{n} = \frac{r_{1} f o r H \times n^{2}}{Z} r_{n} \propto n^{2} r_{n} \propto 3^{2} a_{0} r_{n} \propto 9 a_{0} m v r \propto \frac{n h}{2 π} = \frac{3 h}{2 π} v = \frac{3 h}{2 π m r} (r = a_{0}) = \frac{3 h}{2 π m (9_{3} a_{0})} = \frac{h}{6 π m a_{0}} K . E = \frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{h}{6 π m a_{0}})}^{2} = \frac{1}{2} m \frac{h^{2}}{36 π^{2} m^{2} {a_{0}}^{2}} = \frac{1}{2} \frac{h^{2}}{36 π^{2} m a_{0}^{2}}$