Quantization of Energy

1. Introduction

Classical physics failed to explain the stability of the atom and the discrete line spectra of gases. In 1913 Niels Bohr combined Rutherford's nuclear atom with Planck's quantum idea to build a model of the hydrogen atom in which energy is quantized: the electron can occupy only certain allowed orbits with fixed energies. This chapter develops Bohr's model, the hydrogen spectral series, excitation and ionization, the de Broglie hypothesis, and the production and properties of X-rays.

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2. Bohr's Model of the Hydrogen Atom

Postulates:

1. The electron revolves around the nucleus only in certain allowed circular orbits called stationary states, without radiating energy.
2. The allowed orbits are those in which the angular momentum of the electron is an integral multiple of $\dfrac{h}{2\pi}$:

$$m v r = \frac{n h}{2\pi}, \qquad n = 1, 2, 3, \dots$$

where $n$ is the principal quantum number.

3. Energy is emitted or absorbed only when the electron jumps from one allowed orbit to another. If it jumps from an orbit of energy $E_m$ to one of energy $E_n$, the frequency $f$ of the emitted or absorbed photon is:

$$h f = E_m - E_n$$

4. The orbit with $n = 1$ has the lowest energy and is called the ground state; the electron stays here when the atom is undisturbed.

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3. Radius, Velocity and Energy of the nth Orbit

Consider an electron of charge $e$ and mass $m$ moving with speed $v$ in a circular orbit of radius $r$ around a hydrogen nucleus of charge $+e$.

Radius of the nth orbit. The electrostatic (Coulomb) attraction provides the centripetal force:

$$ \begin{aligned} \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} &= \frac{m v^2}{r} \quad \Rightarrow \quad m v^2 \ &= \frac{e^2}{4\pi\varepsilon_0 r} \qquad \text{(1)} \end{aligned} $$

From Bohr's quantization condition, $v = \dfrac{n h}{2\pi m r}$. Substituting into equation (1):

$$m\left(\frac{n h}{2\pi m r}\right)^2 = \frac{e^2}{4\pi\varepsilon_0 r}$$

Solving for $r$:

$$\boxed{r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}}$$

So $r_n \propto n^2$. Putting in the constants gives the ground-state (Bohr) radius:

$$ \begin{aligned} r_1 &= a_0 \ &= 0.529\ \text{Å} \ &= 0.529 \times 10^{-10}\ \text{m} \ \qquad r_n &= n^2 a_0 \end{aligned} $$

Velocity of the electron in the nth orbit. From $v = \dfrac{n h}{2\pi m r_n}$:

$$\boxed{v_n = \frac{e^2}{2\varepsilon_0 n h}}$$

So $v_n \propto \dfrac{1}{n}$: the electron moves slower in higher orbits.

Total energy of the nth orbit. The kinetic energy from equation (1) is $KE = \tfrac12 m v^2 = \dfrac{e^2}{8\pi\varepsilon_0 r}$. The potential energy of the electron in the field of the nucleus is:

$$ \begin{aligned} PE &= -\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r} \ &= -\frac{e^2}{4\pi\varepsilon_0 r} \end{aligned} $$

Total energy:

$$ \begin{aligned} E &= KE + PE \ &= \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} \ &= -\frac{e^2}{8\pi\varepsilon_0 r} \end{aligned} $$

Substituting $r = r_n$:

$$\boxed{E_n = -\frac{m e^4}{8 \varepsilon_0^2 n^2 h^2} = -\frac{13.6}{n^2}\ \text{eV}}$$

The negative sign shows the electron is bound; energy must be supplied to free it. Sample values:

• $n = 1$: $E_1 = -13.6\ \text{eV}$ (ground state)
• $n = 2$: $E_2 = -3.4\ \text{eV}$
• $n = 3$: $E_3 = -1.51\ \text{eV}$
• $n = \infty$: $E_\infty = 0$ (electron free, atom ionized)

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