Question

The energy of the ground electronic state of the hydrogen atom is $ - 13.4eV$. The energy of the first excited state will be
$\left( A \right) - 54.4eV$
$\left( B \right) - 27.2eV$
$\left( C \right) - 6.8eV$
$\left( D \right) - 3.4eV$

Answer 1

Hint: Energy of the ground state is given, that is when the principal quantum number, n is equal to $1$ and the atomic number Z of the hydrogen atom is equal to $1$. Using the formula of energy level at different excited states and taking their ratio we can calculate the value of energy at the first excited state where the value of n equals to $2$.

Complete step by step answer:
We know the energy level formula which is represented as given below,
$En = - {Z^2} \times \dfrac{{13.6eV}}{{{n^2}}}$
Where,
Atomic number of hydrogen atoms = Z = 1
This will remain the same in both states, hence it is a constant value in this problem.
The principal quantum number is equa;s to n.
This is for ${n^{th}}$ excitation
From this, we can say that,
$En\,\propto \dfrac{1}{{{n^2}}}$
Now for \[n = 1\]that is for ground state,
$E_1 = - {Z^2} \times \dfrac{{13.6eV}}{{{1^2}}} \ldots \ldots \left( 1 \right)$
Or we can say,
$E_1\,\propto \dfrac{1}{{{1^2}}}$
Noe for $n = 2$ that is for the first excited state,
$E_2 = - {Z^2} \times \dfrac{{13.6eV}}{{{2^2}}} \ldots \ldots \left( 2 \right)$
Or we can say,
$E_2\,\alpha \dfrac{1}{{{2^2}}}$
We know the value of energy at ground state that is $ - 13.4eV$.
Taking the ratio of equation $\left( 1 \right)$ and $\left( 2 \right)$ we will get,
$\dfrac{{E_1}}{{E_2}} = \dfrac{{ - {Z^2} \times \dfrac{{13.6eV}}{{{1^2}}}}}{{ - {Z^2} \times \dfrac{{13.6eV}}{{{2^2}}}}}$
Now putting the value of $E_1$ we will get,
$\dfrac{{ - 13.4eV}}{{E_2}} = \dfrac{{ - {Z^2} \times \dfrac{{13.6eV}}{{{1^2}}}}}{{ - {Z^2} \times \dfrac{{13.6eV}}{{{2^2}}}}}$
Cancelling the common terms we will get,
$\dfrac{{ - 13.4eV}}{{E_2}} = \dfrac{{\dfrac{1}{{{1^2}}}}}{{\dfrac{1}{{{2^2}}}}}$
$ \Rightarrow \dfrac{{ - 13.4eV}}{{E_2}} = \dfrac{{{2^2}}}{{{1^2}}}$
Rearranging the above equation we will get,
$E_2 = \dfrac{{ - 13.4eV}}{{{2^2}}}$
On further solving we will get,
$E_2 = \dfrac{{ - 13.4eV}}{4}$
Hence the value of energy at the first excitation state is $E_2 = - 3.4eV$. Therefore the correct option is $\left( D \right)$.

Note:
We can also solve the problem if the energy of the ground state is not given because we know the atomic number of hydrogen atoms and principal quantum number is also known to us. Here the formula that we used in this problem is the energy level for only hydrogen-like atoms. For example $L{i^{2 + }},H{e^ + },etc$.