Bohr's Model of the Atom: A Quantum Leap in Atomic Theory

Introduction

In 1913, Niels Bohr revolutionized our understanding of atomic structure by proposing a model that combined classical physics with the emerging field of quantum mechanics. His model addressed the limitations of Rutherford's atomic model and explained the mysterious spectral lines of hydrogen that had puzzled scientists for decades.

The Three Fundamental Postulates

Bohr's atomic theory rests on three fundamental postulates:

1. Quantized Energy Levels

Electrons orbit the nucleus in circular paths with specific, discrete energy levels. These orbits, called stationary states, have quantized energies given by:

$E_n = -\frac{E_1}{n^2}$

where $E_1$ is the ground state energy and $n$ is the principal quantum number.

2. Energy Transitions

Electrons can jump between energy levels by absorbing or emitting specific amounts of energy. This energy exchange follows:

$\Delta E_{orbit} = E_{final} - E_{initial} = -E_{photon}$

The negative sign indicates that energy is released during emission and absorbed during excitation.

3. Angular Momentum Quantization

The angular momentum of electrons is quantized according to:

$L_n = n\frac{h}{2\pi}$

where $h$ is Planck's constant and $n$ is the principal quantum number.

The Hydrogen Spectrum

The hydrogen emission spectrum provides compelling evidence for Bohr's model. When hydrogen gas is heated or electrically excited, it emits light at specific wavelengths, creating a distinctive pattern of spectral lines.

The Balmer Series

The Balmer series represents transitions to the n=2 energy level, producing visible spectral lines. The wavelengths are given by Rydberg's equation:

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

where:

Limitations of Bohr's Model

Despite its success in explaining the hydrogen spectrum, Bohr's model has several limitations:

Sample Calculation

Let's calculate the frequency of a photon emitted during a transition from n=3 to n=2:

Using Rydberg's equation:

$\frac{1}{\lambda} = (1.097 × 10^7)(\frac{1}{2^2} - \frac{1}{3^2})$

$\frac{1}{\lambda} = (1.097 × 10^7)(0.25 - 0.111)$

$\lambda = 656.3 \text{ nm}$

The frequency can be calculated using:

$f = \frac{c}{\lambda} = \frac{3.0 × 10^8}{656.3 × 10^{-9}} \\= 4.57 × 10^{14} \text{ Hz}$

The energy of this photon is:

$E = hf = (6.626 × 10^{-34})(4.57 × 10^{14}) \\= 3.03 × 10^{-19} \text{ J}$