Question

Let the potential energy of a hydrogen atom in the ground state be zero. Then, its energy in the first excited state will be:
(A) \[10.2\,{\text{eV}}\]
(B) \[13.6\,{\text{eV}}\]
(C) \[23.8\,{\text{eV}}\]
(D) \[27.2\,{\text{eV}}\]

Answer 1

Hint:First of all, we will find the expression which relates potential energy and total energy of an electron. Then we will find energies in the ground level and the first excited level. We will manipulate the above relation and find the result accordingly.
Complete step by step answer:
In this given question, data are presented as follows:
It is assumed that the potential energy of a hydrogen atom in the ground state to be zero.
For an electron,
Potential energy is related to kinetic energy as:
$P.E = - 2 \times K.E \\$
$K.E = - \dfrac{1}{2} \times P.E \\$
Again, the total energy of an electron is the sum of its potential energy and kinetic energy, which is given by:
\[E = P.E + K.E\] …… (1)
Manipulating equation (1), we get,
$E = P.E + K.E \\$
$E = P.E - \dfrac{1}{2} \times P.E \\$
$E = \dfrac{1}{2} \times P.E \\$
$P.E = 2 \times E \\$
Energy of an electron is given by the equation,
$E = - \dfrac{{13.6}}{{{n^2}}}\\$ …… (2)
Where,
\[E\] indicates the energy of an electron in a particular state.
\[n\] indicates the state level of the electron.
Energy in the first excited state is corresponding to \[n = 2\] .
Total energy in the ground state when \[n = 1\] is given by:
${E_1} = - \dfrac{{13.6}}{{{n^2}}} \\$
$\Rightarrow {E_1} = - \dfrac{{13.6}}{{{1^2}}} \\$
$\Rightarrow {E_1} = - 13.6\,{\text{eV}} \\$
Potential energy in the ground state when \[n = 1\] is given by:
$P.{E_1} = 2 \times {E_1} \\$
$\Rightarrow P.{E_1} = 2 \times \left( { - 13.6} \right) \\$
$\Rightarrow P.{E_1} = - 27.2\,{\text{eV}} \\$
Total energy in the first excited state when \[n = 2\] is given by:
${E_2} = - \dfrac{{13.6}}{{{n^2}}} \\$
$\Rightarrow {E_2} = - \dfrac{{13.6}}{{{4^2}}} \\$
$\Rightarrow {E_2} = - 3.4\,{\text{eV}} \\$
If \[P.{E_1}\] is taken as the reference value, then energy in the first excited state can be written as:
${E_{2'}} = {E_2} - P.{E_1} \\$
$\Rightarrow {E_{2'}} = - 3.4 - \left( { - 27.2} \right) \\$
$\Rightarrow {E_{2'}} = 23.8\,{\text{eV}} \\$
Hence, its energy in the first excited state will be \[23.8\,{\text{eV}}\] .
The correct option is (C).

Note: While trying to solve this problem, you should know the energy of a particular orbit. Always keep in mind that the total energy of an electron is composed of potential energy and kinetic energy. An electron moves from lower to higher when it absorbs energy and vice-versa when it loses energy.