Atoms And Nuclei Question 308
Question: A diatomic molecule is made of two masses $ m_{1} $ and $ m_{2} $ which are separated by a distance r. If we calculate its rotational energy by applying Bohr’s rule of angular momentum quantization, its energy will be given by: (n is an integer)
Options:
A) $ \frac{{{( m_{1}+m_{2} )}^{2}}n^{2}h^{2}}{2m_{1}^{2}m_{2}^{2}r^{2}} $
B) $ \frac{n^{2}h^{2}}{2( m_{1}+m_{2} )r^{2}} $
C) $ \frac{2n^{2}h^{2}}{( m_{1}+m_{2} )r^{2}} $
D) $ \frac{( m_{1}+m_{2} )n^{2}h^{2}}{2m_{1}m_{2}r^{2}} $
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Answer:
Correct Answer: D
Solution:
- The energy of the system of two atoms of diatomic molecule $ E=\frac{1}{2}I{{\omega }^{2}} $ where I=moment of inertia $ \omega = $ Angular velocity $ =\frac{L}{I}, $ L=Angular momentum $ I=\frac{1}{2}( m_{1}r_{1}^{2}+m_{2}r_{2}^{2} ) $ Thus, $ E=\frac{1}{2}( m_{1}r_{1}^{2}+m_{2}r_{2}^{2} ){{\omega }^{2}} $ ?.(i) $ E=\frac{1}{2}( m_{1}r_{1}^{2}+m_{2}r_{2}^{2} )\frac{L^{2}}{I^{2}} $ ?(ii) $ L=n,\hbar $ (According Bohr’s Hypothesis) $ E=\frac{1}{2}( m_{1}r_{1}^{2}+m_{2}r_{2}^{2} )\frac{L^{2}}{{{( m_{1}r_{1}^{2}+m_{2}r_{2}^{2} )}^{2}}} $
$ E=\frac{1}{2}\frac{L^{2}}{( m_{1}r_{1}^{2}+m_{2}r_{2}^{2} )}=\frac{n^{2}h^{2}}{8{{\pi }^{2}}r^{2}m_{1}m_{2}} $
$ E=\frac{1}{2}\frac{L^{2}}{( m_{1}r_{1}^{2}+m_{2}r_{2}^{2} )}=\frac{n^{2}h^{2}}{8{{\pi }^{2}}r^{2}m_{1}m_{2}} $
$ E=\frac{( m_{1}+m_{2} )n^{2}h^{2}}{8{{\pi }^{2}}r^{2}m_{1}m_{2}} $
