Q.Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level $n$ to level $(n-1)$. For large $n$, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Why the Bohr Model Gives Those Energy Levels

The Bohr model is a beautiful piece of physics because it takes a simple, almost desperate idea — "electrons only exist in certain orbits" — and derives the entire hydrogen spectrum from it. The key is that Bohr didn't just assume the energy levels; he forced them to be consistent with classical physics in one specific way, then broke with it in another.

The Two Non-Negotiable Pieces

First, the electron moves in a circle around the proton. That's pure classical mechanics: the Coulomb attraction provides the centripetal force.

This gives you a relation between speed $v$ and radius $r$:

Second, the total energy of the electron is the sum of its kinetic and potential energies. Potential energy for a Coulomb force is $-\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}$ (negative because the force is attractive, and we set zero at infinity).

Substitute $v^2$ from above:

So far, nothing is quantised. Any radius $r$ gives a valid classical orbit, and the energy just follows from that radius. The problem is that a classical electron in a curved path radiates energy and spirals into the nucleus — atoms should collapse. Bohr needed a rule to pick out stable orbits.

The Quantisation Condition

Bohr's revolutionary step was to postulate that the angular momentum of the electron is quantised in units of $\hbar = h/2\pi$:

Why this particular rule? Bohr later said it was the simplest way to get the right answer. But there's a deeper physical motivation: if you think of the electron as a wave (de Broglie's idea, which came a decade later), the condition that a standing wave fits exactly around the circumference $2\pi r = n \lambda$ gives $m v r = n \hbar$ directly. So the quantisation condition is really a wave condition imposed on a particle picture.

Deriving the Allowed Radii and Energies

From $m v r = n \hbar$, we get $v = n\hbar / (mr)$. Substitute this into the centripetal force equation:

Solve for $r$:

The constant in front is the Bohr radius $a_0 \approx 0.529 \, \text{Å}$. So the radii are $r_n = a_0 n^2$.

Now plug $r_n$ back into the energy expression $E = -\frac12 \frac{e^2}{4\pi\varepsilon_0 r}$:

Substitute $a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2}$:

That's the famous result. The $1/n^2$ dependence comes directly from the $n^2$ dependence of the radius, which came from the angular momentum quantisation.

Why the Negative Sign Matters

The energy is negative because the electron is bound. To remove the electron from the atom (ionise it), you need to add $+13.6 \text{ eV}$ to get it to $E=0$ (free electron at rest). The ground state ($n=1$) is the most tightly bound; higher $n$ states are less negative, meaning they're closer to being free.

The Physical Picture

The Bohr model gives you a ladder of energies because the electron can only exist in orbits whose angular momentum is an integer multiple of $\hbar$. Each orbit has a specific radius, and therefore a specific energy. When the electron jumps from a higher orbit to a lower one, the energy difference is emitted as a photon of frequency $f = (E_i - E_f)/h$ — which exactly matches the hydrogen spectral lines.

The model fails for multi-electron atoms and doesn't explain why angular momentum is quantised in the first place. But for hydrogen, it's remarkably accurate — and the derivation shows that the $1/n^2$ energy law is a direct consequence of combining classical circular motion with a single quantisation postulate.