Question

The energy of an electron in the first Bohr orbit of \[{\text{H}}\] atom is \[{\text{ - 13}}{\text{.6eV}}\]. The possible energy value of the excited state for the electron in the Bohr orbits of hydrogen is:
(A) \[{\text{ - 3}}{\text{.4eV}}\]
(B) \[{\text{4}}{\text{.2eV}}\]
(C) \[{\text{ - 6}}{\text{.8eV}}\]
(D) \[{\text{ + 6}}{\text{.8eV}}\]

Answer 1

Hint: In the Bohr model of an atom, the electrons travel in the circular orbits around the nucleus that are defined. These orbits are labelled as the quantum number n. The electrons jump from one orbit to another by emitting energy.
The formula for calculating the energy of electron is –
\[{E_n}{\text{ }} = {\text{ }} - \dfrac{{13.6}}{{{n^2}}}eV\]
where,
\[{{\text{E}}_{\text{n}}}{\text{ = }}\]energy of the electron
\[{{\text{n}}^{\text{2}}}{\text{ = }}\]orbits in which the electrons travel

Complete step by step answer:
Given:
Energy of an electron in the first Bohr orbit of \[{\text{H}}\] atom = \[ - 13.6eV\].
To find: energy value of the excited state for the electron
The formula to calculate the energy of an electron is
\[{E_n}{\text{ }} = {\text{ }} - \dfrac{{13.6}}{{{n^2}}}eV\]
\[ \Rightarrow {E_2}{\text{ }} = {\text{ }} - \dfrac{{13.6}}{{{2^2}}}eV\]
\[ \Rightarrow {E_2}{\text{ }} = {\text{ }} - \dfrac{{13.6}}{4}eV\]
\[ \Rightarrow {E_2}{\text{ }} = {\text{ }} - 3.4{\text{ }}eV\]

Hence, the correct option is (A) i.e.,\[ - 3.4eV\]

Note: A student can get confused between ground state and excited state of an electron.
Ground state – The ground state of an electron is the energy level of the electron that it usually occupies. The ground state is the lowest energy state of the electron.
Excited state – The excited state of an electron is the energy state which the electron temporarily acquires. This state is greater than the ground state.