Question

The number of degenerate orbitals in an energy level of H-atom having \[{{\text{E}}_{\text{n}}} = \dfrac{{ - {{\text{R}}_{\text{h}}}}}{9}\] , where ${{\text{R}}_{\text{h}}}$ is Rydberg and $1{\text{ }}{{\text{R}}_{\text{h}}} = 2.18 \times {10^{ - 18}}{\text{ J}}$ is:
A.5
B.9
C.6
D.3

Answer 1

Hint: To solve this question the basics of the atomic structure needs to be clear. Firstly, degenerate orbitals in an atom are those which have the same energy. For example, the “p” subshell has three degenerate orbitals, “d” subshell has five, and “f” subshell has seven.

Complete step by step answer:
The energy of the H-atom in terms of Rydberg constant is given by:
\[{{\text{E}}_{\text{n}}} = \dfrac{{ - {{\text{R}}_{\text{h}}}}}{{{{\text{n}}^2}}}\]
 Where, ${{\text{R}}_{\text{h}}}$ is the Rydberg constant here, and ‘n’ is the principal quantum number of the energy level.
According to the question ${{\text{n}}^2} = 9$
Therefore,
\[{\text{n}} = \sqrt 9 = 3\]
So the third orbit with principal quantum number 3 has 9 orbitals, one ‘s’ orbital (3s), three p orbitals $\left( {3{p_x},3{p_y},3{p_z}} \right)$ and five d orbitals \[\left( {3{d_{({x^2} - {y^2})}},{\text{ }}3{d_{xy}},{\text{ }}3{d_{yz}},{\text{ }}3{d_{xy}},{\text{ }}and{\text{ }}3{d_{{z^2}}}} \right)\]

Hence the answer is option B, nine degenerate orbitals.

Additional note:
The energy of an electron in a hydrogen atom is the summation of the kinetic and as well as the potential energy.
The kinetic energy of the electron is \[{\text{K}}{\text{.E}}{\text{.}} = \dfrac{1}{2}{\text{m}}{{\text{v}}^{\text{2}}}\] while the potential energy is \[{\text{P}}{\text{.E}}{\text{.}} = - \dfrac{{{\text{kZ}}{{\text{q}}_{\text{e}}}}}{{\text{r}}}\] where, m is the mass of the electron, v is the velocity, k is a constant, Z is the atomic number, ${{\text{q}}_{\text{e}}}$ is the electronic charge, and r is the radius of the atom.
The final equation of atomic energy that is obtained by adding these two is:
\[{\text{E}} = \dfrac{1}{2}{\text{m}}{{\text{v}}^2} - \dfrac{{{\text{kZ}}{{\text{q}}_{\text{e}}}}}{{\text{r}}}\]

Note:
The formula to find out the energy of an electron of a hydrogen-like atom, where Z is the atomic number and n in the excited state of the electron is:
${{\text{E}}_{\text{n}}}{\text{ = }} - \dfrac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}{{\text{E}}_{\text{0}}}$ where ${{\text{E}}_{\text{0}}}$ is a constant whose value is:
 ${{\text{E}}_{\text{0}}} = \dfrac{{{\text{2}}{{\text{\pi }}^{\text{2}}}{{\text{e}}^{\text{4}}}{\text{m}}{{\text{k}}^{\text{2}}}}}{{{{\text{h}}^{\text{2}}}}} = 13.6{\text{eV}}$
So, the formula can also be written as:
${{\text{E}}_{\text{n}}}{\text{ = }} - \dfrac{{{\text{13}}{\text{.6}}{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}$