Q.
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantisation of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.
It is found that the excitation frequency from ground to the first excited state of rotation for the $C O$ molecule is close to $\frac{4}{\pi} \times 10^{11} Hz$.
- In a $C O$ molecule, the distance between $C$ $( mass =12\, amu )$ and $O( mass =16\, amu )$, where $1 amu$ $=\frac{5}{3} \times 10^{-27} kg$, is close to
Atoms
Solution:
$r_{1}=\frac{m_{2} d}{m_{1}+m_{2}}$ and $r_{2}=\frac{m_{1} d}{m_{1}+m_{2}}$
Now, $I=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}$
$\therefore d=1.3 \times 10^{-10} m$
