A) 13.6 eV
B) 3.4 eV
C) 1.5 eV
D) 0.85 eV
Correct Answer: B
Solution :
\[E=13.6\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right].\]For highest energy in Balmer series \[{{n}_{1}}=2\] and \[{{n}_{2}}=\infty \]Þ \[E=13.6\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(\infty )}^{2}}} \right]=3.4\,eV\]
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