
The angular speed (ω) of an electron revolving in nth Bohr orbit and corresponding principal
quantum number (n) are related as_____,
A. $\omega \propto {n^3}$
B. $\omega \propto \dfrac{1}{{{n^3}}}$
C. $\omega \propto {n^2}$
D. $\omega \propto \dfrac{1}{{{n^2}}}$
Answer
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Hint: First, we should know the Bohr model of an atom. In the Bohr model of the atom, the
electrons travel in certain circular orbits around the nucleus. The orbits are labeled by the
quantum number n. Electrons can jump from one orbit to another by emitting or absorbing
energy. In the Bohr model of an atom, the radius of n th orbit is proportional to n 2 . We should know the
relation of angular speed of an electron with the velocity of and electron in the n th orbit and the
radius of the atom. The velocity of an electron in nth orbit has to calculated and also the radius of
the atom.
We know that the velocity of electron in the n th orbit is,
${v_n} = {r_n}{\omega _n}$ …… (1)
Here, ${\omega _n}$ is the angular speed in n th orbit, ${r_n}$ is the radius of the atom from the
center.
But,
${v_n} = \dfrac{{2\pi kz{e^2}}}{{nh}}$ …… (2)
And,
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}kze}}$ …… (3)
Substituting v n and r n from equation (2) and equation (3) in equation (1),
$\begin{array}{l}{\omega _n} = \dfrac{{\dfrac{{2\pi
kz{e^2}}}{{nh}}}}{{\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}kze}}}}\\ \Rightarrow {\omega _n} =
\dfrac{{2\pi kz{e^2}}}{{nh}} \times \dfrac{{4{\pi ^2}kze}}{{{n^2}{h^2}}}\\ \Rightarrow
{\omega _n} \propto \dfrac{1}{{{n^3}}}\end{array}$ …… (4)
From equation (4) it can be observed that the angular speed is inversely proportional to the cube
of quantum numbers (n).
Hence, the correct answer is (B).
Note: In the solution, the students should first understand the Bohr atomic model and its
principles. The students can also study the graphical demonstration of a hydrogen atomic model.
The graphical model helps in understanding the dynamics of the theory. The Bohr atomic model
gives the relation of the angular speed with the quantum number which can be obtained from the
velocity of an electron and its radius.
electrons travel in certain circular orbits around the nucleus. The orbits are labeled by the
quantum number n. Electrons can jump from one orbit to another by emitting or absorbing
energy. In the Bohr model of an atom, the radius of n th orbit is proportional to n 2 . We should know the
relation of angular speed of an electron with the velocity of and electron in the n th orbit and the
radius of the atom. The velocity of an electron in nth orbit has to calculated and also the radius of
the atom.
We know that the velocity of electron in the n th orbit is,
${v_n} = {r_n}{\omega _n}$ …… (1)
Here, ${\omega _n}$ is the angular speed in n th orbit, ${r_n}$ is the radius of the atom from the
center.
But,
${v_n} = \dfrac{{2\pi kz{e^2}}}{{nh}}$ …… (2)
And,
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}kze}}$ …… (3)
Substituting v n and r n from equation (2) and equation (3) in equation (1),
$\begin{array}{l}{\omega _n} = \dfrac{{\dfrac{{2\pi
kz{e^2}}}{{nh}}}}{{\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}kze}}}}\\ \Rightarrow {\omega _n} =
\dfrac{{2\pi kz{e^2}}}{{nh}} \times \dfrac{{4{\pi ^2}kze}}{{{n^2}{h^2}}}\\ \Rightarrow
{\omega _n} \propto \dfrac{1}{{{n^3}}}\end{array}$ …… (4)
From equation (4) it can be observed that the angular speed is inversely proportional to the cube
of quantum numbers (n).
Hence, the correct answer is (B).
Note: In the solution, the students should first understand the Bohr atomic model and its
principles. The students can also study the graphical demonstration of a hydrogen atomic model.
The graphical model helps in understanding the dynamics of the theory. The Bohr atomic model
gives the relation of the angular speed with the quantum number which can be obtained from the
velocity of an electron and its radius.
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