A) 3.4 eV
B) 10.2 eV
C) 13.6 eV
D) 1.9 eV
Correct Answer: A
Solution :
lonisgtion energy for hydrogen atom \[E=\frac{13.6}{{{n}^{2}}}\] For first excited level n = 2 \[E=\frac{13.6}{{{2}^{2}}}=3.4\,eV\]You need to login to perform this action.
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