Q.According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.

The key idea is that in classical electromagnetic theory, an accelerating charge radiates energy at the same frequency as its orbital motion. For the hydrogen atom, the electron's orbital frequency is found from the centripetal force condition and the Bohr radius.

Step 1: Orbital frequency from force balance

The centripetal force is provided by the Coulomb attraction:

The orbital frequency is $f = \frac{v}{2\pi r}$.

Step 2: Express $v$ and $r$

From the force equation, $v^2 = \frac{e^2}{4\pi\varepsilon_0 m r}$.

Using the Bohr radius $r = a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2} \approx 5.29 \times 10^{-11} \text{ m}$, we get:

Why Classical Theory Fails — and What It Predicts

The question asks you to step into the shoes of a 19th-century physicist, before quantum mechanics. According to classical electrodynamics, an accelerating charge radiates electromagnetic waves. An electron orbiting a proton is constantly accelerating (centripetal acceleration), so it must continuously lose energy by emitting light.

This is a disaster for the classical model: as the electron loses energy, it spirals into the nucleus, and the frequency of the emitted light changes continuously. But the problem asks for the initial frequency — the frequency of light emitted at the very start, when the electron is in its smallest stable orbit (the Bohr radius).

The key insight: the frequency of the emitted light equals the orbital frequency of the electron, because the electron's circular motion generates a wave at that same frequency.

Step-by-Step Calculation

1. Set up the force balance for a hydrogen atom

The electron (charge $-e$) orbits a proton (charge $+e$) at a distance $r$. The Coulomb force provides the centripetal acceleration:

where $m_e = 9.11 \times 10^{-31} \, \text{kg}$, $e = 1.60 \times 10^{-19} \, \text{C}$, and $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2$.

2. Solve for the orbital speed $v$

From the force equation:

So:

3. Use the Bohr radius for the initial orbit

The smallest stable orbit in the classical sense corresponds to the Bohr radius (the ground state radius in quantum mechanics, which classical theory cannot derive — but we use it as the starting point):

4. Calculate the orbital frequency

First, find $v$:

Let's compute step by step:

• Numerator: $(9.00 \times 10^9)(2.56 \times 10^{-38}) = 2.304 \times 10^{-28}$
• Denominator: $(9.11 \times 10^{-31})(5.29 \times 10^{-11}) = 4.82 \times 10^{-41}$ …

Method: Bohr's Frequency Condition (Quantum Jump Model)

This problem is a trick — classical electromagnetic theory cannot correctly predict the frequency of light emitted by a hydrogen atom. Classical physics says an accelerating electron radiates continuously, spiralling into the nucleus. The actual discrete spectrum comes from quantum mechanics. However, the exam often expects you to use Bohr's frequency condition, which bridges classical orbit ideas with quantum jumps.

Steps

Step 1: Recall Bohr's frequency condition

When an electron jumps from a higher orbit ($n_2$) to a lower orbit ($n_1$), the frequency of emitted light is:

where $h = 6.63 \times 10^{-34} \ \text{J·s}$ is Planck's constant.

Step 2: Write the energy of the electron in the $n$-th orbit of hydrogen

From Bohr's model:

Convert to joules if needed: $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$.

Step 3: Identify the "initial" transition

The phrase "initial frequency" usually means the first emission line of the Lyman series (highest energy jump): from $n=2$ to $n=1$.

Step 4: Calculate the energy difference

In joules:

$\Delta E = 10.2 \times 1.6 \times 10^{-19} = 1.632 \times 10^{-18} \ \text{J}$ …

Common Mistakes on This Question (Photon Energy & Classical Hydrogen)

This question is a classic trap — it asks you to use classical electromagnetic theory to calculate something that classical theory cannot correctly describe. The very act of doing the calculation reveals why classical physics fails for the atom. Here are the mistakes students make most often.

Mistake 1: Forgetting that classical theory predicts a continuous spectrum, not a single frequency

Students often try to calculate one "initial frequency" and stop there. But classical electrodynamics says an accelerating charge radiates at the instantaneous orbital frequency of the electron. Since the electron spirals inward as it loses energy, this frequency changes continuously — there is no single answer.

How to avoid: Recognise that the question is asking for the frequency at the start, when the electron is in its ground-state orbit (Bohr radius $a_0$). You must calculate the orbital frequency from the centripetal force condition, then state that this is the initial frequency of the emitted radiation — and that it will increase as the electron spirals in.

Mistake 2: Using the wrong radius or velocity

Some students plug in $r = 0.529 \, \text{Å}$ without deriving it, or they use the Bohr radius but forget it comes from quantisation. Others use $r = 10^{-10} \, \text{m}$ as a guess.

How to avoid: Derive the orbital radius from the Coulomb force and circular motion:

This gives $v = \sqrt{\frac{ke^2}{mr}}$. But you still need $r$. In classical theory, there is no fixed $r$ — so you must choose the ground-state Bohr radius $a_0 = 0.529 \times 10^{-10} \, \text{m}$ as the starting point. State this assumption clearly.

Mistake 3: Confusing orbital frequency with photon frequency

Students sometimes calculate the orbital period $T$ and then say the photon frequency is $f = 1/T$. That is correct for classical radiation — the emitted wave has the same frequency as the orbital motion. But then they stop, not realising this is the initial frequency only.

How to avoid: After finding $f = \frac{v}{2\pi r}$, explicitly note: "According to classical theory, the radiation frequency equals the orbital frequency at that instant."

Mistake 4: Arithmetic errors in the final calculation

The numbers are messy: $k = 9 \times 10^9$, $e = 1.6 \times 10^{-19}$, $m = 9.1 \times 10^{-31}$, $a_0 = 5.29 \times 10^{-11}$. Students often misplace exponents or forget to square $e$.

How to avoid: Work step by step with symbols first, then substitute once. The cleanest path:

1. From $\frac{mv^2}{r} = \frac{ke^2}{r^2}$, get $v = \sqrt{\frac{ke^2}{mr}}$.
2. Orbital frequency: $f = \frac{v}{2\pi r} = \frac{1}{2\pi} \sqrt{\frac{ke^2}{mr^3}}$.
3. Substitute $r = a_0$ and compute.

Plugging in:

Compute the denominator inside the square root first: $m a_0^3 = 9.1 \times 10^{-31} \times 1.48 \times 10^{-31} \approx 1.35 \times 10^{-61}$. Then numerator: $ke^2 = 9 \times 10^9 \times 2.56 \times 10^{-38} = 2.30 \times 10^{-28}$. The ratio is about $1.70 \times 10^{33}$, square root gives $4.12 \times 10^{16}$, divided by $2\pi$ gives $f \approx 6.6 \times 10^{15} \, \text{Hz}$. …