Question

What is the energy of the first excited state of a hydrogen atom? How high is it above the ground state if the ground state is at $- 13.61eV$ ?

Answer 1

The energy of the hydrogen atom in the ground state can be determined by substituting the value of atomic number and the number of orbits as one. The energy of the excited state of a hydrogen atom can be determined by substituting the value of atomic number and the number of orbits as two. The difference of these two energies gives the value of how high the ground state is to the excited state.
Formula used:
${E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}$
${E_n}$ is energy of hydrogen atom in ${n^{th}}$ state
$Z$ is atomic number
$n$ is number of orbit or state

Complete answer:
Given that the energy of the electron in the hydrogen atom present in ground state has $- 13.61eV$ Ground state means the number of orbits is one.
The energy of electron in ground state is ${E_1} = \dfrac{{ - 13.6{{\left( 1 \right)}^2}}}{{{1^2}}} = - 13.6eV$
The energy of electron in first excited state is ${E_2} = \dfrac{{ - 13.6{{\left( 1 \right)}^2}}}{{{2^2}}} = - 3.4eV$
The difference in the two energy states will be ${E_2} - {E_1} = - 3.40 - \left( { - 13.6} \right) = 10.20eV$
Thus, the energy of the first excited state of a hydrogen atom is $10.20eV$ high to the ground state.

Note:
The Hydrogen atom is the element with atomic number $1$ it has only one electron in its orbital. This one electron will excite from ground state to excited state. When the electron is excited to the first excited state, the number of orbit or state will be two and the atomic number is the same as the hydrogen atom only.