Question

The ground state energy of $H$- atom is $-13.6eV$. The energy needed to ionise $H$- atoms from its second excited state is:

• A. $1.51eV$
• B. $3.4eV$
• C. $13.6eV$
• D. $12.1eV$

Answer 1

The correct option is A $1.51eV$

The energy of $n^{th}$ state in the hydrogen state will be $E_n=-\frac{13.6eV}{n^2}$

so this energy for the second $excited$ state i.e. $n=3$ will be $E_3=\frac{-13.6eV}{9}=-1.51ev$

So ionization energy for this state will be $-(-1.51eV)=1.51eV$