A) - 3.4 eV
B) + 6.8 eV
C) + 13.6 eV
D) - 6.8 eV
Correct Answer: A
Solution :
\[{{E}_{n}}=-13.6\frac{{{Z}^{2}}}{{{n}^{2}}}eV\] \[=-\frac{13.6\times {{1}^{2}}}{4}eV\] \[=-3.4eV\]
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