A) \[-3.4eV\]
B) \[-4.2eV\]
C) \[-6.8eV\]
D) \[+6.8eV\]
Correct Answer: A
Solution :
Values of energy in the excited state \[=-\frac{13.6}{{{n}^{2}}}eV\] \[=\frac{-13.6}{4}=-3.4\,eV\]in which \[n=2,\,3,\,4\]etc.You need to login to perform this action.
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