Bohr model — derive expression for radius of nth orbit and energy levels of hydrogen

Question

Using Bohr’s postulates, derive expressions for the radius of the $n$ th orbit and the energy of the electron in the $n$ th orbit of a hydrogen atom.

(NCERT Class 12, Chapter 12)

Solution — Step by Step

Centripetal force = Coulomb force:

\frac{mv^2}{r} = \frac{ke^2}{r^2} \quad \cdots (i)

Bohr’s quantization condition:

mvr = \frac{nh}{2\pi} \quad \cdots (ii)

where $k = \frac{1}{4\pi\varepsilon_0}$ .

From (ii): $v = \frac{nh}{2\pi mr}$

Substitute into (i):

\frac{m}{r}\left(\frac{nh}{2\pi mr}\right)^2 = \frac{ke^2}{r^2}

\frac{n^2h^2}{4\pi^2 mr} = ke^2

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

For $n = 1$ (Bohr radius): $r_1 = a_0 = 0.529\,\text{\AA}$

General relation: $r_n = n^2 a_0$

From (ii): $v_n = \frac{nh}{2\pi m r_n} = \frac{ke^2}{nh/2\pi} = \frac{2\pi ke^2}{nh}$

So $v_n \propto 1/n$ — electrons in higher orbits move slower.

Kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{ke^2}{2r}$ (from equation (i))

Potential energy: $PE = -\frac{ke^2}{r}$ (Coulomb potential)

Total energy: $E = KE + PE = \frac{ke^2}{2r} - \frac{ke^2}{r} = -\frac{ke^2}{2r}$

Substituting $r_n$ :

\boxed{E_n = -\frac{me^4}{8\varepsilon_0^2 h^2} \cdot \frac{1}{n^2} = -\frac{13.6}{n^2}\,\text{eV}}

Why This Works

Bohr’s model combines classical mechanics (circular orbit) with a quantum condition (angular momentum quantization). The quantization restricts the electron to specific orbits — it can’t spiral inward continuously.

The negative energy means the electron is bound to the nucleus. The ground state ( $n = 1$ ) has $E = -13.6\,\text{eV}$ , meaning you need to supply $13.6\,\text{eV}$ to ionise the atom. As $n \to \infty$ , $E \to 0$ (free electron).

The $1/n^2$ dependence explains the hydrogen spectrum. Transitions between levels give photons of energy $\Delta E = 13.6\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ eV — the Rydberg formula.

Alternative Method — Using dimensional analysis for scaling

Once you know $r_1 = 0.529\,\text{\AA}$ and $E_1 = -13.6\,\text{eV}$ :

r_n = n^2 \times 0.529\,\text{\AA}, \quad E_n = \frac{-13.6}{n^2}\,\text{eV}

For hydrogen-like ions (charge $Ze$ ): $r_n = \frac{n^2 a_0}{Z}$ and $E_n = \frac{-13.6 Z^2}{n^2}\,\text{eV}$ .

For JEE, memorise: $r \propto n^2/Z$ , $v \propto Z/n$ , $E \propto Z^2/n^2$ . These scaling laws let you handle He $^+$ , Li $^{2+}$ , and other hydrogen-like ions without re-deriving anything.

Common Mistake

Students often write $E_n = -ke^2/(2r_n)$ and stop — forgetting to substitute the expression for $r_n$ to get the $1/n^2$ dependence. The final answer must be in terms of $n$ and fundamental constants (or numerically as $-13.6/n^2$ eV). Also, don’t drop the negative sign — the energy is negative because the electron is bound. Writing $E_n = 13.6/n^2$ (positive) implies the electron is unbound.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using dimensional analysis for scaling

Common Mistake

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