Expression For Radius , Velocity and Total Energy of $n_{th}$ orbit of Hydrogen Atom

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

$F_n = \frac{1}{4 \pi \epsilon_0} \cdot \frac{e \cdot e}{r_n^2} = \frac{e^2}{4 \pi \epsilon_0 r_n^2}.....(1)$

This force provides the necessary centripetal force for the electron to move in a circular orbit of radius $r_n$ with a speed $v_n$.

$F_c=F_n$

$\frac{m v_n^2}{r_n} = \frac{e^2}{4 \pi \epsilon_0 r_n^2}$

$m v_n^2 = \frac{e^2}{4 \pi \epsilon_0 r_n}.......(2)$

According to Bohr’s postulate of quantization of angular momentum,

$L_n = m v_n r_n = \frac{nh}{2\pi}$

$\therefore \quad v_n = \frac{nh}{2 \pi m r_n} ........(3)$

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,

$m \left( \frac{nh}{2 \pi m r_n} \right)^2 = \frac{e^2}{4 \pi \epsilon_0 r_n}$

$\Rightarrow \frac{n^2 h^2}{4 \pi^2 m r_n^2} = \frac{e^2}{4 \pi \epsilon_0 r_n}$

$\therefore \quad r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} .......(4)$

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 $r_0$.

Calculate the value of Bohr’s radius.

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 $r_0 = 0.53$

Speed of an electron in an orbit of hydrogen atom : 

Substituting eqn. (4) in eqn. (3), we get

$v_n = \frac{nh}{2 \pi m r_n} = \frac{nh}{2 \pi m} \times \frac{\pi m e^2}{n^2 h^2 \epsilon_0}$

$\therefore \quad v_n = \frac{e^2}{2 \epsilon_0 n h}........(5)$

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,

$v_n = \alpha \frac{c}{n} = \frac{1}{137}\frac{c}{n}$

$v_n = v_0 . \frac{1}{n}$

When $n=1$, we get

$v_0 = \frac{1}{137}\times c = \frac{c}{137}$

Thus, speed of electron in the first orbit of hydrogen is $\frac{1}{137}$ times the speed of light in vacuum. 

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,

$$K.E_n = \frac{1}{2} m v_n^2$$

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:

$P.E_n$= Electrostatic potential at the $n$th orbit due to the nucleus of hydrogen atom × 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 

$h = 6.63 \times 10^{-34} \, \text{J s}$

$E_n = \frac{-2 \times (3.14)^2 \times 9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^4}{(8.854 \times 10^{-12})^2 (6.63 \times 10^{-34})^2 n^2}$

 $= \frac{-2.17 \times 10^{-18}}{n^2} \, \text{J}$

$E_n= \frac{-2.17 \times 10^{-18}}{1.6 \times 10^{-19} n^2} \, \text{eV}$

$(\because 1 eV = 1.6 \times 10^{-19} \, \text{J})$

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.

Notes : CBSE Class 12 Physics Chapter 12 Atoms - Param Himalaya

1. J.J. Thomson's Model of Atom
2. Rutherford, Geiger and Marsden Experiment (Alpha Particle Scattering Experiment)
3. Alpha-Particle Trajectory and Impact Parameter
4. Rutherford's Model of Atom – Electron Orbits | Limitations
5. Bohr model of hydrogen atom, 
6. Expression for radius of nth possible orbit, velocity and energy of electron in nth orbit,
7. Energy level diagram of hydrogen atom 
8. hydrogen line spectra
9. De-Broglie's Explanation of Bohr's Second Postulate of Quantization of Angular Momentum 
10. Drawbacks or Limitations of Bohr's Atomic Model