Bohr model of hydrogen — postulates, radius rₙ = 0.529 n² - Z Å, energy Eₙ = −13.6 Z² - n² eV

The three postulates (WHAT Bohr assumed)

WHY these? Postulate 3 rescues the atom from classical collapse (an accelerating charge should radiate — Bohr just forbids it in allowed orbits). Postulate 2 is the quantisation seed that makes energies discrete → discrete spectral lines.

Deriving the radius from scratch (HOW)

Step 1 — Force balance. Coulomb attraction = centripetal force.

$\frac{1}{4\pi\varepsilon_0}\frac{(Ze)(e)}{r^2} = \frac{mv^2}{r}$

Why this step? The electron doesn't fly off, so the inward electric pull must equal the mass×centripetal-acceleration needed to keep it circling. Nuclear charge is $+Ze$, electron $-e$.

Simplify (cancel one $r$): \frac{kZe^2}{r} = mv^2, \qquad k=\frac{1}{4\pi\varepsilon_0} \tag{1}

Step 2 — Quantisation. From postulate 2, $v = \dfrac{nh}{2\pi m r}$.

Why this step? We have two unknowns $v$ and $r$; the quantisation condition gives the second equation we need.

Step 3 — Substitute $v$ into (1): $\frac{kZe^2}{r} = m\left(\frac{nh}{2\pi m r}\right)^2 = \frac{n^2h^2}{4\pi^2 m r^2}$

Multiply both sides by $r^2$ and divide by $kZe^2$:

$\boxed{\,r_n = \frac{n^2 h^2}{4\pi^2 m k Z e^2} = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}\,}$

Step 4 — Plug numbers (for $n=1, Z=1$): everything except $n^2/Z$ is a bundle of constants that evaluates to $0.529\times10^{-10}$ m.

Deriving the energy (HOW)

Step 1 — Total energy = kinetic + potential. $E = \underbrace{\tfrac{1}{2}mv^2}_{KE} + \underbrace{\left(-\frac{kZe^2}{r}\right)}_{PE}$

Why this step? PE is negative because attraction lowers energy (bound = trapped in a well).

Step 2 — Use (1): $mv^2 = \dfrac{kZe^2}{r}$, so $KE = \dfrac{kZe^2}{2r}$.

E = \frac{kZe^2}{2r} - \frac{kZe^2}{r} = -\frac{kZe^2}{2r} \tag{2}

Why this step? Notice $E = -KE = \tfrac12 PE$ — a hallmark of the inverse-square force (virial theorem). Total energy is negative → electron is bound.

Step 3 — Insert $r_n$: $E_n = -\frac{kZe^2}{2}\cdot\frac{4\pi^2 mkZe^2}{n^2h^2} = -\frac{2\pi^2 m k^2 Z^2 e^4}{n^2 h^2}$

The constant bundle evaluates to $13.6$ eV.

Spectral lines (WHY hydrogen has a line spectrum)

A jump from $n_2 \to n_1$ (with $n_2>n_1$) emits a photon: $\Delta E = E_{n_2}-E_{n_1} = 13.6\,Z^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{eV}$ $\bar\nu = \frac{1}{\lambda}=R_H Z^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),\quad R_H=1.097\times10^7\,\text{m}^{-1}$

Because $n$ is discrete, only certain $\Delta E$ exist → sharp lines (Lyman $n_1=1$, Balmer $n_1=2$, etc.).

Worked examples

Common mistakes (Steel-man → fix)

Flashcards

Connections

• Rutherford model — Bohr fixes its stability problem.
• Hydrogen spectrum & Rydberg formula — direct application.
• Photoelectric effect & Planck's quantisation — same $E=h\nu$ quantum idea.
• de Broglie wavelength — standing-wave picture derives Bohr's $mvr=n\hbar$.
• Quantum mechanical model of atom — replaces orbits with orbitals; explains where Bohr fails.
• Ionisation energy — equals $|E_1|$ for hydrogen-like ions.

Concept Map

Hinglish (regional understanding)