Question

What is the potential energy of electrons in the L-shell of the hydrogen atom?
A. $- 13.6{\text{ }}eV$
B. $- 6.8{\text{ }}eV$
C. $- 10.2{\text{ }}eV$
D. $- 1.7{\text{ }}eV$

Answer 1

For solving this we first need to understand the basics of the Bohr’s theory of Hydrogen atoms. As we know that, Neil Bohr’s atomic Hydrogen model was introduced in the year $1913$. In the model, he had explained that the positively charged nucleus consists of protons and neutrons that are surrounded by a negatively charged electron cloud.

Complete step by step answer:

In Bohr’s model, any atom like lithium, fluorine or aluminum the shell lying closest to the nucleus is called the K shell. Then, the next is the L shell and the outer shell is the M shell.
As we know that the principal quantum number for an L-shell in a hydrogen atom is $4$. Now, according to the question we know that the energy of an electron in nth orbit of Bohr's model is denoted by,
${E_n} = - \dfrac{{13.6eV}}{{{n^2}}}$
Here, we have to substitute the value of potential energy of an electron
So, P.E of electron $= - 2() = - \dfrac{{2 \times 13.6eV}}{{{n^2}}} = - \dfrac{{27.2eV}}{{{n^2}}}$
Therefore, for $n = 4\;$
So, we will substitute the value of n in the above formula.
Thus, potential energy of electron $= - \dfrac{{27.2eV}}{{{{(4)}^2}}} = - \dfrac{{27.2eV}}{{16}} = - 1.7eV$
The potential energy of electrons in the L-shell of the hydrogen atom is -1.7 eV (option D).
$\therefore$ Option D is the correct answer.

Note:
We need to keep in mind that the Bohr model violates the Heisenberg’s Uncertainty Principle. In the Bohr’s model it is considered that the electrons are known to have both a known radius and orbit which is impossible according to Heisenberg principle.