Question

The ionization potential of a hydrogen atom is $13.6\text{eV}$. The energy required to remove an electron from the second orbit of a hydrogen atom is:

• A. $3.4\text{eV}$
• B. $6.8\text{eV}$
• C. $13.6\text{eV}$
• D. $27.2\text{eV}$

Answer 1

The correct option is A $3.4\text{eV}$

Ionization potential = -(Energy of electron in ground state)
i.e $E_1=–13.6\text{eV}$

& we know, $E_n\propto \frac{1}{n^2}$
Energy of electron in second state
$E_2=\frac{-13.6}{(2{)}^2}=-3.4\text{eV}$
Thus, $3.4\text{eV}$ energy is required to remove an electron from second orbit.