The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. The energy needed to ionize hydrogen atom…
The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. The energy needed to ionize hydrogen atom from its second excited state will be
$1.51 \mathrm{eV}$
$3.4 \mathrm{eV}$
$13.6 \mathrm{eV}$
$6.8 \mathrm{eV}$
Solution
We know that total energy of electron in hydrogen atom is given by $E_\eta=\frac{-13.6}{n^2}$
Also, total energy of electron in second excited state $=\frac{-13.6}{(3)^2}=-1.51 \mathrm{eV}$
$\Rightarrow$ lonization energy or energy required to ionize hydrogen atom from its second excited state $=1.51 \mathrm{eV}$