A) \[{{n}^{2}}\]
B) n
C) \[\frac{1}{n}\]
D) \[\frac{1}{{{n}^{2}}}\]
Correct Answer: D
Solution :
In \[{{H}_{2}}\] molecule total energy \[E=\frac{-n{{z}^{2}}{{e}^{4}}}{8{{\varepsilon }_{0}}^{2}{{h}^{2}}}\left( \frac{1}{{{n}^{2}}} \right)\] \[\therefore \] \[E\propto \frac{1}{{{n}^{2}}}\]You need to login to perform this action.
You will be redirected in
3 sec