A) \[2.89\,eV\]
B) \[1.89\,eV\]
C) \[3.89\,eV\]
D) \[4.89\,eV\]
Correct Answer: B
Solution :
The energy of electron in nth orbit of H-atom is \[{{E}_{n}}=-\frac{Rhc}{{{n}^{2}}}\] Given ionization potential \[=13.6\,eV\] So, ionization energy \[Rhc=13.6\,eV\] So, \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] In \[n=2\]orbit \[{{E}_{2}}=-\frac{13.6}{4}=-3.4\,eV\] In \[n=3\]orbit \[{{E}_{3}}=-\frac{13.6}{{{3}^{2}}}\] \[=-\frac{13.6}{9}=1.51\,eV\] Energy released is \[\Delta \Epsilon ={{\Epsilon }_{3}}-{{E}_{2}}\] \[=-1.51\,eV-(-3.4\,eV)\] \[=-1.5\,eV+3.4\,eV\] \[=1.89\,eV\]
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