Answer
binding energy
(a) n=3, $B_{3}=1.51 eV$
(b) n=5, $B_{5}=0.544 eV$
(c) n=7, $B_{7}=0.2775 eV$
Work Step by Step
Energy of an electron in $n^{th}$ state $E_{n}=\frac{-13.6 eV}{n^{2}}$.
Energy needed to free the electrons from this state = $-E_{n}$, and this energy is called binding energy.
so binding energy of electron in $n^{th}$ state $B_{n}= -E_{n}=\frac{13.6 eV}{n^{2}}$.
binding energy
(a) n=3, $B_{3}=\frac{13.6 eV}{3^{2}}$=$1.51 eV$
(b) n=5, $B_{5}=\frac{13.6 eV}{5^{2}}$=$0.544 eV$
(c) n=7, $B_{7}=\frac{13.6 eV}{7^{2}}$=$0.2775 eV$
