Class 12 > Unit # 26: Atomic Physics > Radius and Energy of nth orbit of Hydrogen Atom

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Radius and Energy of the nth Orbit of Hydrogen Atom

Video Lecture: Radius & Energy of Hydrogen Orbit

Watch the complete video lecture deriving the expressions for the radius and energy levels of a hydrogen atom:

Radius of the nth Orbit of Hydrogen Atom

Consider an electron of charge $-e$ revolving in a hydrogen atom around a proton of charge $+e$ with a constant speed $v$ in an orbit of radius $r$.

When the electron revolves around the nucleus, two forces balance its circular motion:

$\text{Coulomb's Force} = \frac{k e^2}{r^2}$     --- (i)

$\text{Centrifugal Force} = \frac{m v^2}{r}$     --- (ii)

Comparing equation (i) and (ii):

$\frac{m v^2}{r} = \frac{k e^2}{r^2}$

$v^2 = \frac{k e^2}{m r}$     --- (iii)

According to Bohr's theory, the orbital angular momentum is an integral multiple of $\frac{h}{2\pi}$:

$m v r = \frac{n h}{2\pi}$

$v = \frac{n h}{2\pi m r}$

Squaring on both sides (S.O.B.S.):

$v^2 = \frac{n^2 h^2}{4 \pi^2 m^2 r^2}$

Substituting this value of $v^2$ into equation (iii):

$\frac{n^2 h^2}{4 \pi^2 m^2 r^2} = \frac{k e^2}{m r}$

$r_n = \frac{n^2 h^2}{4 \pi^2 m k e^2}$     --- (iv)

The above equation gives the general radius of the $n^{\text{th}}$ orbit of the hydrogen atom.

Radii of Various Orbits

The radius of the first orbit ($n = 1$) of a hydrogen atom is calculated by substituting standard fundamental constants:

• $n = 1$
• $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$
• $m = 9.1 \times 10^{-31}\text{ kg}$
• $k = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$
• $e = 1.6 \times 10^{-19}\text{ C}$

$r_1 = \frac{(1)^2 (6.63 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31})(9 \times 10^9)(1.6 \times 10^{-19})^2} \approx 0.53 \times 10^{-10}\text{ m} = 0.53\text{ \AA}$

For other orbits, since $r_n \propto n^2$:

$r_n = r_1 n^2 = (0.53\text{ \AA}) n^2$

Energy of the nth Orbit of Hydrogen Atom

An electron revolving in an orbit of a hydrogen atom possesses both kinetic energy ($\text{K.E.}$) and potential energy ($\text{P.E.}$). Therefore, the total energy ($E$) is given by:

$E = \text{K.E.} + \text{P.E.}$     --- (I)

1. Kinetic Energy

From the force balance equation ($\text{Centrifugal Force} = \text{Coulomb's Force}$):

$\frac{m v^2}{r} = \frac{k e^2}{r^2}$

$m v^2 = \frac{k e^2}{r}$

$\text{K.E.} = \frac{1}{2} m v^2 = \frac{k e^2}{2 r}$

2. Potential Energy

Potential energy is given by the product of charge and electric potential ($V_{\text{pot}}$) due to a point charge:

$\text{P.E.} = q \cdot V_{\text{pot}} = (-e) \left(\frac{k e}{r}\right) = -\frac{k e^2}{r}$

Total Energy

Substituting the values of $\text{K.E.}$ and $\text{P.E.}$ into equation (I):

$E = \frac{k e^2}{2 r} + \left(-\frac{k e^2}{r}\right)$

$E = -\frac{k e^2}{2 r}$

Substituting the expression for radius $r_n$ from equation (iv):

$E_n = -\frac{2 \pi^2 m k^2 e^4}{n^2 h^2}$     --- (ii)

Physical Significance: The negative sign indicates that the electron is bound to the nucleus. When the energy of the electron becomes zero or positive, the electron will escape and leave the nucleus.

Energy Calculation in Electron Volts (eV)

Substituting the numerical values of constants into equation (ii) for the ground state ($n = 1$):

$E_1 = -13.6\text{ eV}$

$E_n = \frac{-13.6}{n^2}\text{ eV}$

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