State the postulates of Bohr atoms model. Obtain an expression for the radius of $n^{t h}$ orbit of an electron of hydrogen atom based on Bohr's theory.

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Solution

The postulates of Bohr's atom model are:
(i) An electron cannot revolve round the nucleus in all possible orbits. The electrons can revolve round the nucleus only in those allowed or permissible orbits for which the angular momentum of the electron is an integral multiple of $\frac{h}{2 π}$ . (where $h$ is Planck's constant $(6.626 \times 10^{- 34} J s)$ . These orbit are called stationary orbits or non-radiating orbits and an electron revolving in these orbits does not radiate any energy. If $m$ and $v$ are the mass and velocity of the electron in a permitted orbit of radius $r$ , then angular momentum of electron $=mvr$ = $\frac{n h}{2 π}$ , where $n$ is called principal quantum number and has the integral values 1, 2, 3..... This is called Bohr's quantization condition.
(ii) An atom radiates energy, only when an electron jumps from a stationary orbit of higher energy to an orbit of lower energy. If the electron jumps from an orbit of energy $E_{2}$ to an orbit of energy, $E_{1}$ , a photon of energy $h v = E_{2} - E_{1}$ is emitted. This condition is called Bohr's frequency condition.
Radius of the nth orbit $(r_{n})$ : Consider an atom whose nucleus has a positive charge $Z e$ , where $Z$ is the atomic number that gives the number of protons in the nucleus and $e$ , the charge of the electron which is numerically equal to that of proton. Let an electron revolve around the nucleus in the $n t h$ orbit of radius $r_{n}$ .
By Coulomb's law, the electrostatic force of attraction between the nucleus and the electron $= \frac{1}{4 π ϵ_{0}} \frac{(Z e) (e)}{r_{a} 2} . . . . . (1)$
where $ϵ o$ is the permittivity of the free space.
Since, the electron revolves in a circular orbit, it experiences a centripetal force,
$\frac{m v^{2}}{r_{n}} = m r_{n} ω_{n}^{2} . . . . (2)$
where $m$ is the mass of the electron, $v_{n}$ and $ω_{n}$ are the linear velocity and angular velocity of the electron in the $n^{t h}$ orbit respectively.
The necessary centripetal force is provided by the electrostatic force of attraction.
For equilibrium, from equation $(1)$ and $(2)$ ,
$\frac{1}{4 π ϵ_{o}} \frac{Z e^{2}}{r_{n}^{2}} = \frac{m v_{n}^{2}}{r_{n}} . . . . (3)$
$\frac{1}{4 π ϵ_{o}} \frac{Z e^{2}}{r_{n}^{2}} = m r_{n} ω_{n}^{2} . . . . . (4)$
From equation $(4)$ ,
$ω_{n}^{2} = \frac{Z e^{2}}{4 π ϵ_{o} m r_{n}^{3}} . . . . . (5)$
The angular momentum of an electron in $n^{t h}$ orbit is,
$L = m v_{n} r_{n} = m r_{n}^{2} ω_{n} . . . . (6)$
By Bohr's first postulate, the angular momentum of the electron
$L = \frac{n h}{2 π} . . . . (7)$
where $n$ is an integer and is called the principal quantum number.
From equations $(6)$ and $(7)$ ,
$m r_{n}^{2} ω_{n} = \frac{n h}{2 π}$
(or) $ω_{n} = \frac{n h}{2 π m r_{n} 2}$
Squaring both sides,
$ω_{n}^{2} = \frac{n^{2} h^{2}}{4 π^{2} m^{2} r_{n}^{2}} . . . . (8)$
From equations $(5)$ and $(8)$ ,
$\frac{Z e^{2}}{4 π ϵ_{o} m r_{n^{3}}} = \frac{n^{2} h^{2}}{4 π^{2} m^{2} r^{2}}$
(or) $r_{n} = \frac{n^{2} h^{2} ϵ_{o}}{π m^{2} Z e^{2}} . . . . (9)$
From equation $(9)$ , it is seen that the radius of the $n^{t h}$ orbit is proportional to the square of the principal quantum number. Therefore, the radii of the orbits are in the ratio $1 : 4 : 9$ .
For hydrogen atom, $Z = 1$
$∴$ from equation $(9)$ ,
$r_{n} = \frac{n^{2} h^{2} ϵ_{o}}{π m e^{2}} . . . . (10)$
Substituting the known values in the above equation we get,
$r_{n} = n^{2} \times 0.53 \circ A$
If $n = 1, r_{1} = 0.53 \circ A$
This is called Bohr radius.

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