Question

The orbital frequency of an electron in a hydrogen atom is proportional to
A. ${n^3}$
B. ${n^{ - 3}}$
C. ${n^1}$
D. ${n^0}$

Answer 1

Hint: Equate the force which provides the necessary centripetal force with the equation of the centripetal force. Again equate the expression for quantization of angular momentum. Find out separately the radius and time period in terms of. Lastly calculate the orbital frequency.

Complete step by step answer:
We use the Bohr atomic theory to calculate the answer
Since both electrons and protons are charged particles, there exists an electrostatic force between the electrons and nucleus. This electrostatic force provides the necessary centripetal force.
Hence we can say that
$
\dfrac{{m{v^2}}}{r} = \dfrac{{kZ{e^2}}}{{{r^2}}} \\
   \Rightarrow m{v^2} = \dfrac{{kZ{e^2}}}{r} \\
   \Rightarrow m{v^2} = \dfrac{{kZ{e^2}}}{r} \\
   \therefore r = \dfrac{{kZ{e^2}}}{{m{v^2}}} \\
 $
Here $m$ is the mass of electron, $v$ is the orbital velocity of electron, $r$ is the radius of the ${n^{th}}$ orbit, $k$ is a constant in electrostatics, $Z$ is the atomic number and $e$ is the charge of one electron.
Again we know that in the Bohr Atomic model, the angular momentum is quantized
Therefore
$
 mvr = \dfrac{{nh}}{{2\pi }} \\
   \Rightarrow v = \dfrac{{nh}}{{2\pi mr}} \\
 $
Here $n$ is a whole number corresponding to the ${n^{th}}$ orbit, $h$ is the Plank’s constant
From here we take the value of $v$ and put it in the equation
Thus we get
$
 r = \dfrac{{kZ{e^2}}}{{m{{\left( {\dfrac{{nh}}{{2\pi mr}}} \right)}^2}}} \\
   \Rightarrow r = \dfrac{{4{\pi ^2}{m^2}{r^2}kZ{e^2}}}{{m{n^2}{h^2}}} \\
   \therefore r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mkZ{e^2}}} \\
 $
Again we put the value of $r$ and put in the equation
From here we get
$
 v = \dfrac{{nh}}{{2\pi mr}} \\
   \Rightarrow v = \dfrac{{nh}}{{2\pi m\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mkZ{e^2}}}}} \\
   \Rightarrow v = \dfrac{{2\pi kZ{e^2}}}{{nh}} \\
 $
Finally to find the orbital frequency we need to divide the velocity by the circumference of the orbit
Thus orbital frequency is given by
$
f = \dfrac{v}{{2\pi r}} \\
   \Rightarrow f = \dfrac{{\dfrac{{2\pi kZ{e^2}}}{{nh}}}}{{2\pi \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mkZ{e^2}}}}} \\
   \Rightarrow f = \dfrac{{4{\pi ^2}m{k^2}{Z^2}{e^4}}}{{{n^3}{h^3}}} \\
 $
Therefore,From here we can infer that
$f \propto {n^{ - 3}}$

So, the correct answer is “Option B”.

Note:
Students should also know to use the Bohr’s atomic theory and basic physics and maths confined to the expressions of the radius, orbital velocity and energy of an electron in the orbit. One should also substitute the values of the known constants, simplify the values and learn the result by heart. The drawbacks of this model are also a common question.