A) \[-3E\]
B) \[-\frac{E}{3}\]
C) \[-\frac{E}{9}\]
D) \[-9E\]
Correct Answer: D
Solution :
\[{{E}_{n}}=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}}\] \[{{E}_{n}}\propto \frac{1}{{{n}^{2}}}\] \[\therefore \] \[\frac{{{E}_{3}}}{{{E}_{1}}}={{\left( \frac{{{n}_{1}}}{{{n}_{3}}} \right)}^{2}}\] \[\frac{-E}{{{E}_{1}}}={{\left( \frac{1}{3} \right)}^{2}}=\frac{1}{9}\] \[\therefore \] \[{{E}_{1}}=-9E\]
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