Question: 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 :
A) $13.6 \mathrm{eV}$
B) $6.8 \mathrm{eV}$
C) $1.51 \mathrm{eV}$
D) $3.4 \mathrm{eV}$
Answer: (C) $1.51 \mathrm{eV}$
Sol:
The energy levels of a hydrogen atom are given by the formula : $$ E_n=-\frac{13.6 \mathrm{eV}}{n^2} $$ where $E_n$ is the energy of the $n$-th energy level. The ground state of hydrogen $(n=1)$ has an energy of $-13.6 \mathrm{eV}$ as mentioned, which means that it would take $+13.6 \mathrm{eV}$ to ionize it (remove the electron completely) from this state, since ionization implies moving the electron to a state of zero energy.
The second excited state of hydrogen is when $n=3$ (as $n=1$ is the ground state and $n=2$ is the first excited state). Thus, the energy of the second excited state is : $$ E_3=-\frac{13.6 \mathrm{eV}}{3^2}=-\frac{13.6 \mathrm{eV}}{9}=-1.51 \mathrm{eV} $$
Since ionization implies moving the electron from its current energy level to 0 energy, the energy required to ionize the atom from this state is the absolute value of its current energy state. So, it will take $+1.51 \mathrm{eV}$ to ionize a hydrogen atom from its second excited state.
So, the correct answer is Option C : $1.51 \mathrm{eV}$.
