The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\left[a_{0}\right.$ is…
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\left[a_{0}\right.$ is Bohr radius $]$ :
- $\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}$
- $\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$
- $\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$
- $\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}$
Solution
As per Bohr's postulate,
$\begin{array}{l}
m v r=\frac{n h}{2 \pi} \quad \text { So, } v=\frac{n h}{2 \pi m r} \\
\mathrm{KE}=\frac{1}{2} m v^{2} \quad \text { So, } \mathrm{KE}=\frac{1}{2} m\left(\frac{n h}{2 \pi m r}\right)^{2}
\end{array}$
$\text { Since, } r=\frac{a_{\mathrm{o}} \times n^{2}}{z}$
So, for $2^{\text {nd }}$ Bohr orbit
$\begin{array}{l}
r=\frac{a_{\mathrm{o}} \times 2^{2}}{1}=4 a_{\mathrm{o}} \\
\mathrm{KE}=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{\mathrm{o}}\right)^{2}}\right)=\frac{h^{2}}{32 \pi^{2} m a_{\mathrm{o}}^{2}}
\end{array}$
Asked in: JEE Advanced 2012 (Paper 1)
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