Question

Which of the following is the energy of a possible excited state of hydrogen?
A.$+ 13.6eV$
B.$- 6.8eV$
C.$- 3.4eV$
D.$+ 6.8eV$

Answer 1

When an electron absorbs energy in an atom by any source of energy then it leaves its original orbit and shifts to a new orbit of higher energy level, this new position of electron in an atom is known as excited state.

Complete step by step answer:
To study the behavior of extremely small particles like electrons a concept is introduced known as quantum mechanics. According to quantum mechanics, any state of an atom which has more energy than that of ground state is known as excited state. The atom can reach to its ground state by releasing electrons of appropriate energy.
When the electrons present in the atom absorb energy then this is termed as an excited state of atom. At this position the atom and its electron is unstable so to again gain the stability back it releases energy and again moves to its ground state.
The formula used to calculate the energy released when an atom moves from one shell to the another is:
${\text{E = - 13}}{\text{.6}}\left( {\dfrac{{{z^2}}}{{{n^2}}}} \right)$
Where, ${\text{E}}$ is the energy released.
$n$ is the number of orbit and $z$ is the atomic number.
And we all know that the atomic number of hydrogen is one.
So, ${\text{E = - 13}}{\text{.6}}\left( {\dfrac{1}{{{n^2}}}} \right)$
We will calculate the value of energy for different orbits by changing the values of $n$ .

Orbit (n)	Energy (E)
1	${\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{n^2}}}} \right) = {\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{1^2}}}} \right) = - 13.6eV$
2	${\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{n^2}}}} \right) = {\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{2^2}}}} \right) = - 3.4eV$
3	${\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{n^2}}}} \right) = {\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{3^2}}}} \right) = - 1.51eV$
4	${\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{n^2}}}} \right) = {\text{ - 13}}{\text{.6}}\left( {\dfrac{1}{{{4^2}}}} \right) = - 0.85eV$

From the above calculation we can see that the energies in different orbits differ from $- 13.6$ to $- 0.85eV$ .
The ground state means that the value of $n$ is $1$ and all the other values of $n$ except 1 are considered as excited states. So the energy of the first excited state ($n = 2$) is $- 3.4eV$ and this is the only energy value which matches with the options.
Hence, option C is the correct answer.

Note:
When the electron absorbs energy and moves towards the orbits with higher energy levels, it becomes unstable. It tends to come back to its original state. To gain its original position the electron releases the energy and shifts in its parent orbit.