A) 3.4 eV
B) 10.2 eV
C) 13.6 eV
D) 1.9 eV
Correct Answer: A
Solution :
lonisation energy \[E=\frac{-13.6}{{{n}^{2}}}eV\] For first excited state \[n=2\] So, \[E=\frac{-13.6}{{{2}^{2}}}=3.4eV\] \[=3.4\text{ }eV\] (numerically)You need to login to perform this action.
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