How much energy is required to excite a hydrogen atom from its ground state to the second excited state?

Hint: We know that there are different energy states for atoms. Atoms are most likely to remain in their ground state. But if we provide enough energy for the atoms will jump from its ground state to the excited states. Here we have to find the energy required to excite the hydrogen atom from its ground state to its second excited state. Atoms are in an excited state when the electrons absorb energy and jump to higher energy levels.
Formula used
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
Where ${E_n}$ is the energy of the hydrogen atom in the ${n^{th}}$ energy level and $n$ stands for the required energy level to which we excite the atom. We get the constant value of $13.6$ through calculation.

Complete step by step solution:
We know that the energy of the hydrogen atom at the ${n^{th}}$ energy level is given by,
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
The energy required to excite the hydrogen atom from the ground state to its second excited state will be equal to the difference between the energies of the excited state and the ground state of the hydrogen atom,
i.e.
For ground state,$n = 1$
For the second excited state,$n = 3$
The difference in energy can be written as,
${E_n} = \dfrac{{ - 13.6}}{{{3^2}}} - \dfrac{{ - 13.6}}{{{1^2}}}$
Taking common values outside, we get
${E_n} = - 13.6\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{1^2}}}} \right)$
This can be written as,
${E_n} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right)$
$ \Rightarrow {E_n} = 12.08eV$

The answer is: $12.08eV$

Note:
In the expression for the energy of the ${n^{th}}$energy level, there is a negative sign. This negative sign shows that the electron is bound to the nucleus. If we can provide an energy of $\dfrac{{13.6}}{{{n^2}}}eV$ to the electron located in the ${n^{th}}$orbit, it can escape to a point infinitely far away from the nucleus of the atom. The lowest energy level of an atom is called its ground state.