Bohr's Model for Hydrogen Atom

Bohr's Model (1913)

Niels Bohr combined Planck's quantum theory with Rutherford's nuclear model to propose a new model for the hydrogen atom. He addressed the key question: Why doesn't the electron spiral into the nucleus?

Bohr's Postulates

Postulate 1 — Quantised Orbits: The electron in a hydrogen atom revolves around the nucleus in certain fixed circular orbits called stationary states or allowed energy levels. While in these orbits, the electron does not radiate energy.

Postulate 2 — Angular Momentum Quantisation: The angular momentum of the electron is quantised:

$$\boxed{mvr = \frac{nh}{2\pi}} \quad (n = 1, 2, 3, \ldots)$$

where $n$ is the principal quantum number, $m$ is mass of the electron, $v$ is velocity, $r$ is radius of the orbit, and $h$ is Planck's constant.

This means only those orbits are allowed where the angular momentum is an integral multiple of $h/2\pi$.

Postulate 3 — Energy Transitions: The electron can jump from one orbit to another. When it jumps:

• From a higher to a lower orbit → energy is emitted as a photon
• From a lower to a higher orbit → energy is absorbed

The energy of the photon emitted or absorbed: $$\Delta E = E_2 - E_1 = h\nu$$

Key Formulas from Bohr's Model

Radius of the $n^{\text{th}}$ Orbit

For hydrogen atom ($Z = 1$): $$\boxed{r_n = n^2 \times a_0}$$

where $a_0 = 0.529$ Å $= 52.9$ pm is the Bohr radius (radius of the first orbit).

For hydrogen-like species (one electron, nuclear charge $Z$): $$r_n = \frac{n^2 a_0}{Z}$$

Key observations:

• Radius increases as $n^2$ — orbits get farther apart as $n$ increases
• For the same $n$, higher $Z$ gives smaller radius (stronger nuclear pull)

Velocity of Electron in $n^{\text{th}}$ Orbit

$$v_n = \frac{2.18 \times 10^6 \times Z}{n} \text{ m/s}$$

• Velocity decreases as $n$ increases (electron moves slower in outer orbits)
• For the same $n$, higher $Z$ gives higher velocity

Energy of Electron in $n^{\text{th}}$ Orbit

For hydrogen ($Z = 1$): $$\boxed{E_n = -\frac{R_H}{n^2} = -\frac{13.6}{n^2} \text{ eV per atom} = -\frac{2.18 \times 10^{-18}}{n^2} \text{ J per atom}}$$

For hydrogen-like species: $$E_n = -\frac{13.6 \times Z^2}{n^2} \text{ eV}$$

The negative sign means the electron is bound to the nucleus. The energy is zero when $n = \infty$ (the electron is free).

Level ($n$)	Energy (eV)	Name
1	$-13.6$	Ground state (K shell)
2	$-3.4$	First excited state (L shell)
3	$-1.51$	Second excited state (M shell)
4	$-0.85$	Third excited state (N shell)
$\infty$	$0$	Ionised (free electron)

Energy Level Diagram and Transitions

Understanding the Energy Levels

• Energy levels get closer together as $n$ increases
• The energy gap between successive levels decreases: $E_2 - E_1 > E_3 - E_2 > E_4 - E_3 > \ldots$
• The ground state ($n = 1$) has the most negative (lowest) energy — the electron is most tightly bound
• Ionisation energy = energy to remove the electron from $n = 1$ to $n = \infty$ = 13.6 eV for hydrogen

Linking Bohr's Model to the Rydberg Formula

The energy of the photon emitted in a transition from $n_2$ to $n_1$: $$\Delta E = E_{n_2} - E_{n_1} = 13.6\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ eV}$$

Since $\Delta E = h\nu = hc/\lambda$: $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

This is exactly the Rydberg formula! Bohr's model successfully derived the Rydberg formula from first principles.

Hydrogen-like Species

Bohr's model also applies to ions with only one electron:

• $\text{He}^+$ ($Z = 2$), $\text{Li}^{2+}$ ($Z = 3$), $\text{Be}^{3+}$ ($Z = 4$), etc.

For these species: $$E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}, \quad r_n = \frac{n^2 a_0}{Z}, \quad \frac{1}{\lambda} = R_H Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

[JEE Tip] $\text{He}^+$ has the same energy levels as hydrogen scaled by $Z^2 = 4$. So the ground state energy of $\text{He}^+$ is $-13.6 \times 4 = -54.4$ eV.

Limitations of Bohr's Model

Despite its success with hydrogen, Bohr's model has significant limitations:

1. Multi-electron Atoms

The model fails for atoms with more than one electron. It cannot explain the spectra of helium or any higher element. This is because it doesn't account for electron-electron repulsion.

2. Fine Structure of Spectral Lines

High-resolution spectroscopy shows that each spectral line is actually composed of several closely spaced lines (fine structure). Bohr's model predicts only single lines.

3. Zeeman and Stark Effects

• Zeeman effect: Splitting of spectral lines in a magnetic field
• Stark effect: Splitting of spectral lines in an electric field

Bohr's model cannot explain these splittings.

4. Three-dimensional Nature

Bohr assumed electrons move in flat circular orbits (2D). In reality, electrons occupy three-dimensional regions of space (orbitals).

5. Wave Nature of Electron

Bohr treated the electron as a particle moving in a definite orbit. But de Broglie later showed that electrons also have wave nature. You can't have a particle in a precise orbit if it's also a wave!

6. Heisenberg Uncertainty Principle

Bohr's model assumes we can know the exact position and velocity of the electron simultaneously. This violates the Heisenberg uncertainty principle (covered in Section 8).

Key Point: Bohr's model was a crucial stepping stone — it got the energy levels right for hydrogen and introduced the concept of quantised orbits. But a fundamentally different approach (quantum mechanics) was needed for a complete theory.

Solved Examples

Example 1: Radius of an Orbit

Calculate the radius of the third orbit ($n = 3$) in a hydrogen atom.

Solution: $$r_n = n^2 \times a_0 = 3^2 \times 0.529 \text{ Å} = 9 \times 0.529 = 4.761 \text{ Å}$$

In pm: $r_3 = 9 \times 52.9 = 476.1$ pm.

Answer: $r_3 = 4.761$ Å = 476.1 pm.

Example 2: Energy of an Orbit

Calculate the energy of the electron in the second orbit of hydrogen.

Solution: $$E_n = -\frac{13.6}{n^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$$

Answer: $E_2 = -3.4$ eV.

Example 3: Ionisation Energy of Hydrogen

Calculate the ionisation energy of hydrogen from the ground state.

Solution: Ionisation energy = energy to move electron from $n = 1$ to $n = \infty$: $$IE = E_\infty - E_1 = 0 - (-13.6) = 13.6 \text{ eV}$$

In kJ/mol: $IE = 13.6 \times 96.49 = 1312 \text{ kJ/mol}$

Answer: IE = 13.6 eV per atom = 1312 kJ/mol.

Example 4: Energy of Transition

Calculate the energy emitted when an electron in hydrogen transitions from $n = 3$ to $n = 1$.

Solution: $$\Delta E = 13.6\left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 13.6\left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} = 12.09 \text{ eV}$$

Answer: Energy emitted = 12.09 eV.

Example 5: Wavelength of Emitted Photon

Calculate the wavelength of the photon emitted in the transition $n = 5 \to n = 2$ in hydrogen.

Solution: $$\frac{1}{\lambda} = R_H\left(\frac{1}{4} - \frac{1}{25}\right) = 1.097 \times 10^7 \times \frac{21}{100}$$ $$= 2.304 \times 10^6 \text{ m}^{-1}$$ $$\lambda = 434 \text{ nm}$$

Answer: $\lambda = 434$ nm (this is the H$_\gamma$ line — blue-violet, Balmer series).

Example 6: Hydrogen-like Species — He$^+$

Calculate the radius of the first orbit and energy of the ground state of $\text{He}^+$ ($Z = 2$).

Solution: Radius: $r_1 = \frac{1^2 \times 0.529}{2} = 0.2645$ Å

Energy: $E_1 = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4$ eV

Answer: $r_1 = 0.2645$ Å, $E_1 = -54.4$ eV.

Example 7: Which Orbit has the Same Radius?

Find the orbit number in $\text{Li}^{2+}$ that has the same radius as the first Bohr orbit of hydrogen.

Solution: For hydrogen: $r_1 = a_0$

For $\text{Li}^{2+}$ ($Z = 3$): $r_n = \frac{n^2 a_0}{3}$

Setting equal: $\frac{n^2 a_0}{3} = a_0 \implies n^2 = 3 \implies n = \sqrt{3}$

Since $n$ must be an integer, no orbit of $\text{Li}^{2+}$ has exactly the same radius as the first Bohr orbit of hydrogen.

Answer: No allowed orbit matches exactly.

Example 8: Velocity of Electron

Calculate the velocity of the electron in the first Bohr orbit of hydrogen.

Solution: $$v_1 = \frac{2.18 \times 10^6 \times Z}{n} = \frac{2.18 \times 10^6 \times 1}{1} = 2.18 \times 10^6 \text{ m/s}$$

This is about 1/137 of the speed of light.

Answer: $v_1 = 2.18 \times 10^6$ m/s.

Example 9: Comparing Energy Levels

In which hydrogen-like species is the energy of the electron in the $n = 2$ orbit equal to the energy of the electron in the $n = 4$ orbit of hydrogen?

Solution: Energy of hydrogen in $n = 4$: $$E = -\frac{13.6}{16} = -0.85 \text{ eV}$$

For a hydrogen-like species with atomic number $Z$ at $n = 2$: $$E = -\frac{13.6 Z^2}{4}$$

Equating: $$-\frac{13.6 Z^2}{4} = -0.85$$ $$\frac{13.6 Z^2}{4} = 0.85$$ $$Z^2 = \frac{0.85 \times 4}{13.6} = 0.25$$ $$Z = 0.5$$

But $Z$ must be a positive integer for a real hydrogen-like species.

Answer: No hydrogen-like species satisfies this condition.

Example 10: Excitation Energy

Calculate the energy required to excite a hydrogen atom from the ground state to the first excited state.

Solution: $$\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \text{ eV}$$

Answer: Excitation energy = 10.2 eV.