Answer
(a) n=2, $E_{2}$= -3.4 eV
(b) n=3, $E_{3}$= -1.51 eV
Work Step by Step
Energies for hydrogen atom $E_{n}= \frac{-13.6 eV}{n^{2}}$
where n= 1,2,3...........are state of electron ( principle Quantum number )
for (a) n=2 $E_{n}= \frac{-13.6 eV}{2^{2}}$= -3.4 eV
(b) n=3 $E_{n}= \frac{-13.6 eV}{3^{2}}$= -1.51 eV
