Bohr’s Theory of Hydrogen Atom :
Hydrogen atom consists of a nucleus having charge $+e$ and an electron having charge $-e$.
The electron is assumed to revolve around the nucleus in circular orbit of radius $r$.
Speed of Electron in terms of radius of an orbit of Hydrogen atom
Coulomb’s force of attraction between the nucleus and the electron revolving in an orbit of radius $r_n$ is given by
$F_n = \frac{1}{4 \pi \epsilon_0} \cdot \frac{e \cdot e}{r_n^2} = \frac{e^2}{4 \pi \epsilon_0 r_n^2} \tag{1}$
This force provides the necessary centripetal force for the electron to move in a circular orbit of radius $r_n$ with a speed $v_n$.
$\frac{m v_n^2}{r_n} = \frac{e^2}{4 \pi \epsilon_0 r_n^2}$
$m v_n^2 = \frac{e^2}{4 \pi \epsilon_0 r_n}$
According to Bohr’s postulate of quantization of angular momentum,
$L_n = m v_n r_n = \frac{nh}{2\pi}$
$\therefore \quad v_n = \frac{nh}{2 \pi m r_n} \tag{3}$
Radius of an orbit of Hydrogen atom
On the basis of Bohr’s atomic model, derive expressions for radius of $n^{th}$ stationary orbit of hydrogen atom.
Substituting eqn. (3) in eqn. (2), we get,
$m \left( \frac{nh}{2 \pi m r_n} \right)^2 = \frac{e^2}{4 \pi \epsilon_0 r_n}$
$\Rightarrow \frac{n^2 h^2}{4 \pi^2 m r_n^2} = \frac{e^2}{4 \pi \epsilon_0 r_n}$
$\therefore \quad r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} \tag{4}$
Since
$\frac{h^2 \epsilon_0}{\pi m e^2} =\text{constant}$
$r_n \propto n^2$
Thus, radius of an orbit of an electron is directly proportional to the square of the principal quantum number ($n$) of the orbit}. Therefore, radii of orbits of hydrogen atom increase with increase in the principal quantum number.
Since $n = 1, 2, 3, 4, \ldots$ (an integer), hence the ratio of the radii of the orbits of hydrogen atom is
$r_1 : r_2 : r_3 : r_4 : \ldots = 1 : 4 : 9 : 16 : \ldots$
Variation of the radius $(r)$ of orbit of hydrogen atom with the principal quantum number $(n)$is shown in figure.
Bohr's Radius : The radius of the innermost orbit ($n=1$) of an electron in hydrogen atom is called Bohr’s radius. It is denoted by $r_0$.
Speed of an electron in an orbit of hydrogen atom :
Substituting eqn. (4) in eqn. (3), we get
$v_n = \frac{nh}{2 \pi m r_n} = \frac{nh}{2 \pi m} \times \frac{\pi m e^2}{n^2 h^2 \epsilon_0}$
$\therefore \quad v_n = \frac{e^2}{2 \epsilon_0 n h}\tag{5}$
Since
$\frac{e^2}{2 h \epsilon_0} = \text{constant}$
$\therefore \quad v_n \propto \frac{1}{n}$
This , speed of an electron in an orbit is inversely proportional to the principal quantum number $n$, i.e., the speed of electron in the orbits of hydrogen atom decreases with increase in principal quantum number.
Variation of the speed $(v)$ of an electron in an orbit of hydrogen atom with the principal quantum number $(n)$ is shown in figure.
Fine Structure Constant and speed of an electron in an orbit of hydrogen atom
We know that
$v_n = \frac{e^2}{2 \epsilon_0 h n}$
So, we can rewrite it as
$v_n = \left( \frac{e^2}{2 \epsilon_0 c h} \right) \frac{c}{n} = \frac{\alpha c}{n}$
where, $\alpha$ is called the \textbf{fine structure constant
$\alpha = \frac{e^2}{2 \epsilon_0 c h}$
Using standard values, we get
$\alpha = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.854 \times 10^{-12} \times 3 \times 10^8 \times 6.625 \times 10^{-34}}= \frac{1}{137}$
Then,
$v_n = \alpha \frac{c}{n} = \frac{1}{137}\frac{c}{n}$
When $n=1$, we get
$v_1 = \frac{1}{137}\times c = \frac{c}{137}$
Thus, speed of electron in the first orbit of hydrogen is $\frac{1}{137}$ times the speed of light in vacuum. Similarly, speed of electron in other orbits can be calculated.
Numerical Example 3 : Calculate the radius of the innermost orbit of an electron in hydrogen atom.
Or
Calculate the value of Bohr’s radius.
Solution: We know, radius of $n$th orbit of hydrogen atom is given by
$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$
Here $h = 6.625 \times 10^{-34}\, \text{J·s}$
$\epsilon_0 = 8.854 \times 10^{-12}\text{m}^{-2}\text{C}^2$
$m = 9.1 \times 10^{-31}\, \text{kg}$
$e = 1.6 \times 10^{-19}\, \text{C}$.
Taking $n=1$ (for innermost orbit), we get
$r_0 = \frac{(1)^2 \times (6.625 \times 10^{-34})^2 \times (8.854 \times 10^{-12})}{(3.14) \times 9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^2}$
$r_0= 5.29 \times 10^{-11} \text{m}$
$r_0=0.529 \text{A} \approx 0.53\text{A}$
Thus, Bohr’s radius = $0.53 \text{A}$
$\therefore r_n = n^2 r_0$ where $r_n$ is the radius of $n$th orbit and $r_0 = 0.53$
Total Energy of an Electron in an Orbit of Hydrogen Atom :
The total energy of an electron in $n^{\text{th}}$ orbit of hydrogen atom is the sum of its kinetic energy and potential energy in that orbit.
$E_n = K.E_n + P.E_n \tag{6}$
Now,
$K.E_n = \tfrac{1}{2} m v_n^2$
Using eqn. (2), we get
$K.E_n = \frac{1}{2} m v_n^2 = \frac{1}{2}\,\frac{e^2}{(4\pi\epsilon_0) r_n} \tag{7}$
Potential energy:
$P.E_n = \text{Electrostatic potential at the $n$th orbit due to the nucleus of hydrogen atom} \times \text{charge on the electron}$
i.e.,
$P.E_n = \frac{1}{4\pi\epsilon_0}\frac{e}{r_n}\times (-e)$
or
$P.E_n = -\frac{1}{4\pi\epsilon_0}\frac{e^2}{r_n} \tag{8}$
Substituting values of eqn. (7) and (8) in eqn. (6), we get
$E_n = \frac{e^2}{8\pi\epsilon_0 r_n} - \frac{e^2}{4\pi\epsilon_0 r_n}$
$E_n = \frac{-e^2}{8 \pi \epsilon_0 r_n}$
Using eqn. (4), we get,
$E_n = \frac{-m e^4}{8 h^2 \epsilon_0^2 n^2}$
Since, $\epsilon_0 = 8.854 \times 10^{-12} \, \text{N}^{-1} \text{m}^{-2} \text{C}^2$, $m = 9.1 \times 10^{-31} \, \text{kg}$, $e = 1.6 \times 10^{-19} \, \text{C}$ and $h = 6.63 \times 10^{-34} \, \text{J s}$
$E_n = \frac{-2 \times (3.14)^2 \times 9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^4}{(8.854 \times 10^{-12})^2 (6.63 \times 10^{-34})^2 n^2}$
$= \frac{-2.17 \times 10^{-18}}{n^2} \, \text{J} = \frac{-2.17 \times 10^{-18}}{1.6 \times 10^{-19} n^2} \, \text{eV} \quad (\because 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J})$
$\ldots$ or
$E_n = \frac{-13.6}{n^2} \, \text{eV}$
Thus, total energy of the electron in an orbit is negative. Negative energy tells that electron and nucleus in a hydrogen atom form an attractive system i.e., a bound system. In other words, the electron can be taken away from its orbit if external energy = $(13.6/n^2) \, \text{eV}$ is supplied to it.
Since $n$ has discrete values, so energy of an electron in hydrogen atom is quantized. This is called Bohr's energy quantization.