A) 3.40 eV
B) 1.51 eV
C) 0.85 eV
D) 0.66 eV
Correct Answer: D
Solution :
\[E={{E}_{4}}-{{E}_{3}}\] \[=-\frac{13.6}{{{4}^{2}}}-\left( \frac{13.6}{{{3}^{2}}} \right)\] \[=-0.85\times 1.51\] \[=0.66\,eV\]You need to login to perform this action.
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