Question

By assuming Bohr’s postulates derive an expression for radius of nth orbit of electron, revolving round the nucleus of hydrogen atom.

Answer 1

Hint: We know that the Bohr’s postulates demands the angular momentum (mvr) of stationary orbits to be the integral multiple of$\dfrac{h}{2\pi }$ , from here we get a relation for velocity. Also, recall that the electrostatic attraction between the nucleus and the electron in its orbit is balanced by centripetal force on the electron. Now, substituting velocity derived from Bohr’s postulates in this relation and then rearranging gives the radius of nth orbit of the electron.
Formula used:
$mvr=\dfrac{nh}{2\pi }$
$\dfrac{m{{v}^{2}}}{r}=\dfrac{K{{Z}^{2}}{{e}^{2}}}{{{r}^{2}}}$

Complete answer:
As a first step, let us revise Bohr’s postulates.
Bohr’s postulates are:
A. Electrons revolve around the nucleus in a fixed circular path termed orbits/shells/energy level.
B. These orbits are the ‘stationary orbits’.
C. Orbital shells are circular orbits with a fixed amount of energy. No radiation of energy takes place if the electrons continue to revolve in the fixed orbital shells.
D. We have numbered each of this energy level (quantum number) as n=1, 2, 3,… and can also be represented by K,L,M,…respectively.
E. The lowest of these energy levels is the ground state.
F. Electrons when it acquires energy gets excited to higher energy levels and emits energy as radiation when moved down to a lower energy level.
There is a very important characteristic of stationary orbits that is worth mentioning here. Their angular momentum (L) should definitely be an integral multiple of $\dfrac{h}{2\pi }$\[\]
 Where ‘h’ is the Planck constant with a value of \[h=6.626\times {{10}^{-34}}J.H{{z}^{-1}}\]
So by Bohr’s postulates we have,
Angular momentum has to be equal to $n\times \dfrac{h}{2\pi }$ , that is,
$mvr=\dfrac{nh}{2\pi }$
Where ‘n’ can be= 1, 2, 3…
m=mass of the electron
v= velocity of the electron
r=radius of the orbit
$v=\dfrac{nh}{2\pi mr}$ ………………… (1)
Also we have, the centripetal force should be equal to the force of attraction between nucleus and electron, for this is the force that balances the centripetal force and hence the electron continues its revolution around the nucleus. So we have,
$\dfrac{m{{v}^{2}}}{r}=\dfrac{K{{Z}^{2}}{{e}^{2}}}{{{r}^{2}}}$ ……………………. (2)
Where, K is the Coulomb constant =$K=9\times {{10}^{9}}N.{{m}^{2}}.{{C}^{-2}}$
e = charge of an electron
Z= atomic of number of the atom
Substituting (1) in (2), we get,
$\begin{align}
  & \dfrac{m{{n}^{2}}{{h}^{2}}}{4r{{\pi }^{2}}{{m}^{2}}{{r}^{2}}}=\dfrac{K{{Z}^{2}}{{e}^{2}}}{{{r}^{2}}} \\
 & \Rightarrow r=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}mKZ{{e}^{2}}} \\
\end{align}$
Here in the question, we have been asked for the radius of nth orbit of a hydrogen atom specifically. As we know the atomic number (Z) of a hydrogen atom is 1, we can substitute this value for Z.
So we get the required relation as,
$r=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}mK{{e}^{2}}}$

Note:
This relation is purely derived on the basis of our understanding of fundamental concepts of Bohr’s model. Any derivation is easy when you are thorough with the concepts. So, we should always avoid by-hearting steps of any derivation and learn concepts perfectly.