Question

Orbital acceleration of an electron is
A. $\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}$
B. $\dfrac{{{n^2}{h^2}}}{{2{n^2}{r^3}}} $
C. $\dfrac{{4{n^2}{h^2}}}{{{\pi ^2}{m^2}{r^3}}}$
D. $\dfrac{{4{n^2}{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}$

Answer 1

Hint: We will first use Bohr's postulates which state that the angular momentum of a system of electrons is quantized and is a multiple of the Planck’s constant. Using the statement and finding velocity from this equation, we will substitute the velocity in the formula of centripetal acceleration for an orbit.

Complete step by step answer:
According to the Bohr’s postulates, the angular momentum of a system is quantized. An electron can only move in orbit and the angular momentum is an integral multiple of\[\dfrac{h}{{2\pi }}\]. Thus, we have from the above statement, that;
\[mvr = \dfrac{{nh}}{{2\pi }}\]
Here, m is the mass of the particle, v is the velocity of the particle, r is the radius of the orbit of the particle, n is a whole number, that is an integer, and h is the Planck’s constant.

In terms of velocity, the above formula can be expressed as follows:
\[v = \dfrac{{nh}}{{2\pi mr}}\]
Now, as the electron is moving in an orbit, it will possess centripetal acceleration and it is given by:
\[a = \dfrac{{{v^2}}}{r}\]
Where a is the centripetal acceleration.
Thus, substituting the value of velocity obtained through Bohr’s postulate into the equation of centripetal acceleration, we obtain:
\[
a = \dfrac{{{v^2}}}{r} \\
\therefore a = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}
\]
In this way, we find the orbital acceleration of an electron.

Thus, option A is the correct answer.

Note: Here we assume that the orbits of the electrons are perfectly circular, as assumed by the Bohr’s postulate, but it was later found out that the orbits of the electrons can be elliptical and of dumb-bell shaped as well. Only, s shell orbital is spherical and hence Bohr’s postulates are applicable only to the first orbit.