A) \[{{n}^{2}}\]
B) n
C) \[\frac{1}{n}\]
D) \[\frac{1}{{{n}^{2}}}\]
Correct Answer: D
Solution :
In \[{{H}_{2}}\] molecule total energy \[E=\frac{-n{{z}^{2}}{{e}^{4}}}{8{{\varepsilon }_{0}}^{2}{{h}^{2}}}\left( \frac{1}{{{n}^{2}}} \right)\] \[\therefore \] \[E\propto \frac{1}{{{n}^{2}}}\]
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