A) 10.2ev
B) 12.09ev
C) 1.9ev
D) 0.65ev
Correct Answer: C
Solution :
[c] \[1.9\,eV\] \[{{E}_{n}}=\frac{-\,13.6}{{{n}^{2}}}eV\] (energy of electron in nth orbit) \[{{E}_{3}}-{{E}_{2}}=13.6\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)eV\] \[13.6\left( \frac{1}{4}-\frac{1}{9} \right)=13.6\left( \frac{9-4}{36} \right)1.9eV\]You need to login to perform this action.
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