Derivation of Rydberg Formula for Hydrogen Atom Spectrum :

Derivation of Rydberg Formula for Hydrogen Atom Spectrum :

Derivation of Rydberg Formula for Hydrogen Atom Spectrum : 

According to Bohr's model of the atom, energy is radiated in the form of a photon when an electron jumps from a higher energy state to a lower energy state. In other words, energy is radiated in the form of a photon when an electron jumps from an orbit of higher energy ($n_i$) to the lower energy orbit ($n_f$), where $n_i > n_f$. The energy of the emitted radiation or photon is given by:

$h\nu = E_{ni} - E_{nf}$

Using 

$E_n = \frac{-m e^4}{8 \varepsilon_0^2 h^2 n^2}$

we get

$h\nu = \frac{-m e^4}{8 \varepsilon_0^2 h^2 n_i^2}- \left(\frac{-m e^4}{8 \varepsilon_0^2 h^2 n_f^2}\right)$

$h\nu = \frac{m e^4}{8 \varepsilon_0^2 h^2} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

$\nu = \frac{m e^4}{8 \varepsilon_0^2 h^3}\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

Using $c = \lambda \nu$, i.e. $\nu = \frac{c}{\lambda}$ in Eq. (2), we get

$\frac{c}{\lambda} = \frac{m e^4}{8 \varepsilon_0^2 h^3}\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

$\frac{1}{\lambda} = \frac{m e^4}{8 \varepsilon_0^2 c h^3}\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

where $\lambda$ is the wavelength of the spectral line emitted and 

$R = \frac{m e^4}{8 \varepsilon_0^2 c h^3} = 1.0974 \times 10^7 \,\text{m}^{-1}$

is Rydberg's constant.

Equation (3) is known as the $\textbf{Rydberg Formula}$ for the hydrogen atom spectrum.

Wave Number

As $\bar{\nu} = \frac{1}{\lambda}$, i.e. wave number, we can write

$\bar{\nu} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

Using eqn. (3) we can find the wavelengths of the spectral lines emitted by hydrogen atom.

SHORT CUT: Rydberg formula can be used in a convenient form as

$\lambda = \frac{911 \text{Å}}{\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)}$

NUMERICAL EXAMPLE 5:

The wavelength of yellow line of sodium is 5896 Å. What is its wave number?

Solution:

Wave number,

$\bar{\nu} = \frac{1}{\lambda}$

$\bar{\nu} = \frac{1}{5896 \times 10^{-10}}$

$\bar{\nu} = 1696100 \,\text{m}^{-1}$

$\bar{\nu} =16961 \,\text{cm}^{-1}$

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