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 ionisation potential = 13-6 eV so, ionisation energy Rhc = 13-6 eV So, \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] In n = 2 orbit \[{{E}_{2}}=-\frac{13.6}{{{2}^{2}}}=-\frac{13.6}{4}=-3.4eV\] In n = 3 orbit \[{{E}_{3}}=-\frac{13.6}{{{3}^{2}}}\] \[=-\frac{13.6}{9}=1.51eV\] Energy released is \[\Delta E={{E}_{3}}-{{E}_{2}}\] \[=-1.51eV-(-3.4eV)\] \[=-1.5eV+3.4eV\] \[=1.89eV\]
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