Physics · Ch 13 — Atoms

Electron Orbits

13.2.2

Electron Orbits

The Classical Picture of the Electron's Orbit

In Rutherford's nuclear model, the atom is visualised as a tiny, massive, positively charged nucleus surrounded by electrons that revolve around it in well-defined orbits. This is a purely classical picture — it treats the electron like a planet orbiting the sun, held in place by the electrostatic attraction to the nucleus.

For a hydrogen atom, which has just one electron, the situation is particularly simple. The electron moves in a circular orbit of radius $r$ with a constant speed $v$ . The force that keeps it in this circular path is the centripetal force, and the only force available to provide that is the electrostatic attraction between the electron (charge $-e$ ) and the proton (charge $+e$ ).

The electrostatic force is given by Coulomb's law:

F_e = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}

The centripetal force required for circular motion of an electron of mass $m$ moving with speed $v$ in a circle of radius $r$ is:

F_c = \frac{mv^2}{r}

For a dynamically stable orbit — one that does not collapse — these two forces must be exactly equal. This is the condition that defines a possible classical orbit.

F_e = F_c \quad \Rightarrow \quad \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r}

This single equation is the starting point for everything that follows in this section.

Relating Orbital Radius and Electron Velocity

From the force balance equation, we can rearrange to get a direct relationship between the orbital radius $r$ and the electron's speed $v$ . Multiply both sides by $r^2$ :

\frac{1}{4\pi\varepsilon_0} e^2 = mv^2 r

Then solve for $r$ :

r = \frac{e^2}{4\pi\varepsilon_0 m v^2}

Alternatively, solve for $v$ :

v = \frac{e}{\sqrt{4\pi\varepsilon_0 m r}}

The textbook gives the first form explicitly:

r = \frac{e^2}{4\pi\varepsilon_0 m v^2}

This is equation (12.3) in the NCERT text. It tells us that for a given speed $v$ , there is exactly one radius $r$ that satisfies the classical stability condition. But it does not tell us which speed or which radius actually occurs in nature — that requires the quantum ideas introduced later in the chapter.

Energy of the Electron in a Hydrogen Atom

The electron in its orbit possesses two forms of mechanical energy: kinetic energy due to its motion, and electrostatic potential energy due to its position in the electric field of the nucleus.

Kinetic energy $K$ is straightforward:

K = \frac{1}{2} m v^2

We can express this in terms of $r$ using the force balance relation. From $mv^2 = \frac{e^2}{4\pi\varepsilon_0 r}$ , we get:

K = \frac{1}{2} \left( \frac{e^2}{4\pi\varepsilon_0 r} \right) = \frac{e^2}{8\pi\varepsilon_0 r}

Potential energy $U$ for two point charges $+e$ and $-e$ separated by distance $r$ is:

U = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}

The negative sign is crucial. It arises because the electrostatic force is attractive — work must be done against the field to separate the charges. By convention, potential energy is taken as zero when the charges are infinitely far apart. As they come closer, the potential energy decreases (becomes more negative), meaning the system becomes more tightly bound.

Watch out

The negative sign in $U$ is not optional. It directly reflects the attractive nature of the Coulomb force. Forgetting it will give you the wrong total energy — and the wrong sign for the binding energy.

Total energy $E$ is the sum:

E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r}

Combining the two terms:

E = -\frac{e^2}{8\pi\varepsilon_0 r}

This is equation (12.4) in the textbook. It is a remarkably simple result: the total energy of the electron in a hydrogen atom is negative and inversely proportional to the orbital radius.

E = -\frac{e^2}{8\pi\varepsilon_0 r}

The Meaning of Negative Total Energy

The fact that $E$ is negative is not a mathematical curiosity — it carries deep physical significance.

A negative total energy means the electron is bound to the nucleus. To remove the electron (to ionise the atom), you must supply enough positive energy to bring the total to zero. The amount of energy required is exactly $|E|$ , the magnitude of the total energy. This is called the binding energy or ionisation energy.

If $E$ were positive, the electron would have more kinetic energy than the potential well could contain. It would not follow a closed orbit — it would escape to infinity, and the atom would not exist as a stable entity.

Important

A negative total energy is the signature of a bound system. For the hydrogen atom, $E < 0$ confirms that the electron is confined to the vicinity of the nucleus. The more negative $E$ is, the more tightly the electron is bound.

Worked Example: Determining Orbital Radius and Velocity from the Binding Energy

The textbook provides a concrete example that ties these formulas to experimental data. It is known that 13.6 eV of energy is required to separate a hydrogen atom into a free proton and a free electron. This is the ionisation energy, so the total energy of the electron in the ground state is $E = -13.6$ eV.

Step 1: Convert energy to joules.

E = -13.6 \text{ eV} = -13.6 \times 1.6 \times 10^{-19} \text{ J} = -2.176 \times 10^{-18} \text{ J}

The textbook rounds this to $-2.2 \times 10^{-18}$ J.

Step 2: Use the total energy formula to find the orbital radius.

From $E = -\dfrac{e^2}{8\pi\varepsilon_0 r}$ , solve for $r$ :

r = -\frac{e^2}{8\pi\varepsilon_0 E}

Substitute the known constants. It is convenient to use $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ N m $^2$ /C $^2$ , so $\frac{1}{8\pi\varepsilon_0} = \frac{k}{2}$ :

r = -\frac{k e^2}{2E} = -\frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{2(-2.2 \times 10^{-18})}

The two negatives cancel. Compute the numerator:

(9 \times 10^9) \times (2.56 \times 10^{-38}) = 2.304 \times 10^{-28}

Divide by $2 \times 2.2 \times 10^{-18} = 4.4 \times 10^{-18}$ :

r = \frac{2.304 \times 10^{-28}}{4.4 \times 10^{-18}} = 5.236 \times 10^{-11} \text{ m}

The textbook gives $r = 5.3 \times 10^{-11}$ m. This is the Bohr radius, the radius of the smallest orbit in the hydrogen atom.

Step 3: Use the radius to find the orbital velocity.

From the force balance relation $mv^2 = \frac{e^2}{4\pi\varepsilon_0 r}$ , we have:

v = \frac{e}{\sqrt{4\pi\varepsilon_0 m r}} = \frac{e}{\sqrt{m}} \cdot \frac{1}{\sqrt{4\pi\varepsilon_0 r}}

Again using $k = \frac{1}{4\pi\varepsilon_0}$ :

$v = e \sqrt{\frac{k}{m r}}$ …

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