Q2 / 60 · JEE Advanced 2012 (Paper 1) · Chemistry

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

  • A
    $\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}$
  • B
    $\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$
  • Correct answer
    C
    $\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$
  • D
    $\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}$

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 $\begin{array}{l} r=\frac{a_{\mathrm{o}} \times 2^{2}}{1}=4 a_{\mathrm{o}} \\ \mathrm{KE}=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{\mathrm{o}}\right)^{2}}\right)=\frac{h^{2}}{32 \pi^{2} m a_{\mathrm{o}}^{2}} \end{array}$

Hence, the correct answer is option: 'C'.

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