Question

The value of energy for first excited state hydrogen is:
A. $- 13.6{\text{eV}}$
B. $- 3.40{\text{eV}}$
C. $- 1.51{\text{eV}}$
D. $- 0.85{\text{eV}}$

Answer 1

Energy level in a hydrogen atom is defined as a fixed energy corresponding to the orbits in which its electrons move around the nucleus. The energy levels are quantized. The total energy level of the hydrogen atom is the sum of kinetic energy and potential energy.

Complete step by step answer:
Consider one electron of charge ${\text{ - e}}$ and mass ${\text{m}}$ moves in a circular orbit of radius ${\text{r}}$ around a positively charged nucleus with a velocity ${\text{v}}$.
In general, the total energy level ${\text{E}}$ for the atom is:
${{\text{E}}_{\text{n}}} = - \dfrac{{{\text{k}}{{\text{e}}^2}}}{{2{{\text{a}}_0}}}\left( {\dfrac{{{{\text{Z}}^2}}}{{{{\text{n}}^2}}}} \right)$, where ${\text{k}} = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}$, ${\text{e}}$ is the charge of electron $\left( {1.6 \times {{10}^{ - 19}}} \right)$, ${{\text{a}}_0}$ is Bohr radius $\left( {5.31 \times {{10}^{ - 11}}} \right)$, ${\text{Z}}$ is the atomic number, ${\text{n}}$ is the orbit number.
Substituting the values, we get the energy in electron volts.
i.e. ${{\text{E}}_{\text{n}}} = \dfrac{{ - 13.6}}{{{{\text{n}}^2}}}{\text{eV}}$
Thus ${{\text{E}}_{\text{n}}}$ is the energy level of nth orbit $\left( {{\text{n}} = 1,2,3,..} \right)$
We know that ${\text{n = 1}}$ for the ground state (the state of the lowest energy level). It has least potential energy. Electrons have the most potential energy when they are in an excited state.
Excited state is defined as the energy state which is higher than the ground state.
For the first excited state, ${\text{n = 2}}$.
For this value of ${\text{n}}$, energy ${{\text{E}}_{\text{n}}} = \dfrac{{ - 13.6}}{{{2^2}}}{\text{eV = }}\dfrac{{ - 13.6}}{4}{\text{eV = - 3}}{\text{.40eV}}$
Hence the value of energy for the first excited energy of hydrogen atom is $- 3.40{\text{eV}}$.

Hence the correct option is B.

Note:
The negative sign indicates that a negative work has to be done to remove the electron from the bound of the atom to infinity, where it is considered to have zero energy. At infinity, the electron is completely removed from the atom. Excitation energy is the energy required by an electron that raises it to an excited state from its ground state.