Question

Which electronic shell is at a higher energy level K or L?

Answer 1

According to Bohr’s Atomic Model, the shell which is closer to the nucleus has lower energy and the shell which is away from the nucleus has higher energy. As we know that K is closer to the nucleus therefore, it has lower energy than L-shell which is comparatively far from the nucleus.

Complete step-by-step answer: First let us discuss about Bohr’s Atomic Model and then further compare the energy levels:-
-The Bohr model explains that the electrons in an atom are in orbits of differing energy around the nucleus like planets orbiting around the sun.
-Bohr used the term energy levels (or shells) to describe these orbits of distinct energy and he said that the energy of an electron is quantized which means that electrons can have one energy level or another but not between them. These energy levels were given mnemonics which are as follows:-
n = 1 = K-shell
n = 2 = L-shell
n = 3 = M-shell
n = 4 = N-shell
and so on. Here n is the principal quantum number or the number of energy shells.
-Bohr once found that the closer an electron is to the nucleus, the less energy it requires whereas the farther away it is, the more energy it requires. So Bohr numbered these electron’s energy levels. The higher the energy level number, the farther away the electron is from the nucleus and thereby higher the energy.
So according to the above data we can conclude that since K-shell (energy level) is closest to the nucleus so it will have the least energy whereas L-shell (energy level) is comparatively far due to which it has more energy.
Therefore ${{E}_{K}}<{{E}_{L}}$ .

Note: -The above answer can also be proven mathematically by the formula of energy of electron shells around an atom.
${{E}_{n}}=-\dfrac{13.6\text{ }\\!\\!~\\!\\!\text{ }{{\text{Z}}^{2}}}{{{n}^{2}}}\text{eV}$
where,
n = number of energy level (shell)
Z = atomic number

Keeping everything constant, let’s calculate energy for K-shell and L-shell:

-Energy for K-shell: ${{E}_{K}}=-\dfrac{13.6\text{ }\\!\\!~\\!\\!\text{ }{{\text{Z}}^{2}}}{{{1}^{2}}}\text{eV}$
-Energy for L-shell: ${{E}_{L}}=-\dfrac{13.6\text{ }\\!\\!~\\!\\!\text{ }{{\text{Z}}^{2}}}{{{2}^{2}}}\text{eV}$
On comparing them we get: $\begin{aligned} & -\dfrac{13.6\text{ }~\text{ }{{\text{Z}}^{2}}}{{{1}^{2}}}\text{eV}-\dfrac{13.6\text{ }~\text{ }{{\text{Z}}^{2}}}{{{2}^{2}}}\text{eV} \\\ & \\\ \end{aligned}$
$\therefore {{\text{E}}_{K}}<{{E}_{L}}$