The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\left[a_{0}ight.$ is Bohr…

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\left[a_{0}ight.$ is Bohr radius] :
  1. $\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}$
  2. $\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$
  3. $\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$
  4. $\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}$

Solution

As per Bohr s postulate, $m v r=\frac{n h}{2 \pi}$
So, $v=\frac{n h}{2 \pi m r}$
$\mathrm{KE}=\frac{1}{2} m v^{2}$
So, $\mathrm{KE}=\frac{1}{2} m\left(\frac{n h}{2 \pi m r}ight)^{2}$
Since, $r=\frac{a_{\mathrm{o}} \times n^{2}}{z}$
So, for $2^{\text {nd }}$ Bohr orbit
$r=\frac{a_{\mathrm{o}} \times 2^{2}}{1}=4 a_{\mathrm{o}}$
$\mathrm{KE}=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{\mathrm{o}}ight)^{2}}ight)$
$\mathrm{KE}=\frac{h^{2}}{32 \pi^{2} m a_{\mathrm{o}}^{2}}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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