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