Atomic Structure and Spectrum Analysis

Absorption Spectra

An absorption spectrum occurs when white light is passed through a solution or the vapors of a chemical substance before being analyzed by a spectroscope. In this process, the substance absorbs specific wavelengths of light corresponding to its unique chemical nature. When viewed through a spectroscope, this appears as dark lines superimposed on a continuous spectrum.

Key Distinction: While molecular substances often produce broad band spectra, individual atoms produce sharp, distinct line spectra.

The Emission Spectrum of Hydrogen

One of the most significant discoveries in atomic theory is the line spectrum of hydrogen. When hydrogen gas at low pressure is subjected to an electric discharge, the light emitted can be examined to reveal a specific pattern of lines.

These lines are not random; they are the result of electrons moving between "stationary orbits" within the atom. When an electron transitions from a higher energy orbit ($n_2$) to a lower energy orbit ($n_1$), it must release energy in the form of a photon.

Mathematical Modeling of Transitions

To calculate the energy of the emitted photon during a transition from $n_2$ to $n_1$ (where $n_2 > n_1$) in a hydrogen-like sample, we use the energy levels of the atom. The energy of an electron in a specific orbit is given by:

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

The energy of the emitted photon ($\Delta E$) is the difference between these two levels:

$\begin{aligned} \Delta E &= E_{n_2} - E_{n_1} \\ &= \left( -13.6 \frac{Z^2}{n_2^2} \right) - \left( -13.6 \frac{Z^2}{n_1^2} \right) \\ &= 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV} \end{aligned}$

Where $Z$ is the atomic number.

Wavelength and Wave Number

Once the energy of the photon is known, we can determine its wavelength ($\lambda$) using the relationship between energy and wavelength:

$\lambda = \frac{hc}{\Delta E}$

Substituting the energy equation into this relationship, we get:

$\frac{1}{\lambda} = \frac{2.13.6 Z^2}{hc} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

In spectroscopy, it is often more convenient to work with the wave number ($\nu$), which is the reciprocal of the wavelength ($1/\lambda$). This is expressed using the Rydberg constant ($R$):

$\nu = \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

The value of the Rydberg constant is approximately: $R \approx 1.09678 \times 10^7 \text{ m}^{-1}$ Or more simply for quick calculations: $R \approx 1.1 \times 10^7 \text{ m}^{-1}$

Exam Tip: Remember that the Rydberg formula applies to "H-like" atoms (atoms with only one electron, like $\text{He}^+$ or $\text{Li}^{2+}$). Always ensure you include the $Z^2$ term when dealing with ions other than pure Hydrogen.

Summary Table of Spectra Types