⚛️ Bohr’s Atomic Model
Postulates · Step‑by‑Step Derivation of Radius and Energy · Spectral Lines · Limitations
In 1913, Niels Bohr proposed a revolutionary model of the hydrogen atom, combining the quantum theory of radiation with classical mechanics. Bohr’s model successfully explained the stability of atoms and the origin of discrete spectral lines. It introduced the concepts of stationary orbits and quantized angular momentum.
📐 DIAGRAM 1: Bohr’s atom – central nucleus with electrons revolving in stationary orbits (K, L, M)
[Concentric circles: n=1 (K), n=2 (L), n=3 (M); electron jumps between orbits with absorption/emission of photons]
📜 Postulates of Bohr’s Atomic Model
Postulate 1 Nuclear atom: The atom has a small, massive, positively charged nucleus at its centre, containing all protons and neutrons. The size of the nucleus is extremely small compared to the size of the atom.
Postulate 2 Stationary orbits: Electrons revolve around the nucleus only in certain discrete, non‑radiating orbits called stationary orbits (energy levels). In these orbits, electrons do not lose energy.
Postulate 3 Force balance: The electrostatic attraction between the nucleus (charge \(Ze\)) and the electron (charge \(-e\)) provides the centripetal force:
\[
\frac{1}{4\pi\epsilon_0} \cdot \frac{Ze^2}{r^2} = \frac{mv^2}{r}
\]
Postulate 4 Quantization of angular momentum: The angular momentum of an electron in a stationary orbit is an integer multiple of \(\frac{h}{2\pi}\):
\[
m v r = n \frac{h}{2\pi} \quad (n = 1,2,3,\dots)
\]
where \(h\) is Planck’s constant, \(m\) electron mass, \(v\) velocity, \(r\) orbit radius. The integer \(n\) is the principal quantum number (n=1: K, n=2: L, n=3: M, …).
Postulate 5 Energy transitions: An electron can jump from one stationary orbit to another by absorbing or emitting radiation. Emission occurs when an electron falls from a higher energy orbit to a lower one.
Postulate 6 Photon energy: The radiation is emitted/absorbed as a single photon whose energy equals the difference in energies between the two orbits:
\[
h\nu = \Delta E = E_{\text{higher}} – E_{\text{lower}}
\]
Postulate 7 Ground state: The lowest energy state (\(n=1\)) is the ground state. Excited states (\(n>1\)) are unstable; electrons spontaneously return to lower levels, emitting photons.
🔢 Step‑by‑Step Derivation of Radius of nth Orbit
We start with two fundamental equations from postulates 3 and 4.
Step 1: Coulomb force = centripetal force
\[ \frac{1}{4\pi\epsilon_0} \cdot \frac{Ze^2}{r^2} = \frac{mv^2}{r} \] Multiply both sides by \(r\): \[ \frac{Ze^2}{4\pi\epsilon_0 r} = mv^2 \quad \Rightarrow \quad v^2 = \frac{Ze^2}{4\pi\epsilon_0 m r} \quad \text{(1)} \]
\[ \frac{1}{4\pi\epsilon_0} \cdot \frac{Ze^2}{r^2} = \frac{mv^2}{r} \] Multiply both sides by \(r\): \[ \frac{Ze^2}{4\pi\epsilon_0 r} = mv^2 \quad \Rightarrow \quad v^2 = \frac{Ze^2}{4\pi\epsilon_0 m r} \quad \text{(1)} \]
Step 2: Quantization of angular momentum: \( m v r = n \frac{h}{2\pi} \).
Solve for \(v\): \[ v = \frac{n h}{2\pi m r} \quad \text{(2)} \]
Solve for \(v\): \[ v = \frac{n h}{2\pi m r} \quad \text{(2)} \]
Step 3: Square equation (2):
\[
v^2 = \frac{n^2 h^2}{4\pi^2 m^2 r^2}
\]
Step 4: Equate \(v^2\) from (1) and the squared expression:
\[
\frac{Ze^2}{4\pi\epsilon_0 m r} = \frac{n^2 h^2}{4\pi^2 m^2 r^2}
\]
Step 5: Simplify by cancelling common factors. Multiply both sides by \(4\pi^2 m^2 r^2\):
\[
\frac{Ze^2}{4\pi\epsilon_0 m r} \cdot 4\pi^2 m^2 r^2 = n^2 h^2
\]
\[
\frac{Ze^2 \cdot 4\pi^2 m^2 r^2}{4\pi\epsilon_0 m r} = n^2 h^2
\]
\[
\frac{Ze^2 \cdot \pi m r}{\epsilon_0} = n^2 h^2
\]
Step 6: Solve for \(r\):
\[
r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}
\]
This is the radius of the nth orbit. For hydrogen (\(Z=1\)), the first orbit radius (Bohr radius) is:
\[
a_0 = \frac{h^2 \epsilon_0}{\pi m e^2} \approx 0.529 \, \text{Å}
\]
💡 Step‑by‑Step Derivation of Energy of nth Orbit
The total energy \(E_n\) = kinetic energy + potential energy.
Step 1: Kinetic energy \(K = \frac{1}{2} m v^2\). From the force balance equation:
\[
\frac{Ze^2}{4\pi\epsilon_0 r} = m v^2
\]
Therefore,
\[
K = \frac{1}{2} m v^2 = \frac{1}{2} \cdot \frac{Ze^2}{4\pi\epsilon_0 r} = \frac{Ze^2}{8\pi\epsilon_0 r}
\]
Step 2: Potential energy \(U\) for an electron at distance \(r\) from nucleus of charge \(+Ze\):
\[
U = -\frac{1}{4\pi\epsilon_0} \cdot \frac{Ze^2}{r}
\]
Step 3: Total energy:
\[
E_n = K + U = \frac{Ze^2}{8\pi\epsilon_0 r} – \frac{Ze^2}{4\pi\epsilon_0 r} = -\frac{Ze^2}{8\pi\epsilon_0 r}
\]
Step 4: Substitute the expression for \(r\) from the radius derivation:
\[
r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}
\]
Then:
\[
\frac{1}{r} = \frac{\pi m Z e^2}{n^2 h^2 \epsilon_0}
\]
Step 5: Plug into \(E_n\):
\[
E_n = -\frac{Ze^2}{8\pi\epsilon_0} \cdot \frac{\pi m Z e^2}{n^2 h^2 \epsilon_0}
= -\frac{Z^2 e^4 m}{8 \epsilon_0^2 h^2 n^2}
\]
\[
E_n = -\frac{m Z^2 e^4}{8 \epsilon_0^2 h^2} \cdot \frac{1}{n^2}
\]
Step 6: For hydrogen (\(Z=1\)), evaluate constant:
\[
\frac{m e^4}{8 \epsilon_0^2 h^2} = 13.6 \, \text{eV}
\]
Hence,
\[
E_n = -\frac{13.6}{n^2} \, \text{eV}
\]
The negative sign indicates a bound state; \(n=1\) gives \(-13.6\) eV (ground state).
📊 DIAGRAM 2: Energy level diagram for hydrogen (not to scale). Transitions between levels give Lyman, Balmer, Paschen series.
[Vertical axis: energy increasing upward; horizontal lines at n=1,2,3,…; arrows showing photon emission]
🌈 Explanation of Hydrogen Spectral Series
When an electron jumps from an outer orbit \(n_2\) to an inner orbit \(n_1\) (\(n_2 > n_1\)), the emitted photon wavelength is given by:
\[
\frac{1}{\lambda} = \frac{E_{n_2} – E_{n_1}}{hc} = \frac{13.6 \, \text{eV}}{hc} \left( \frac{1}{n_1^2} – \frac{1}{n_2^2} \right)
\]
Using constants, \( \frac{13.6 \, \text{eV}}{hc} = R_H = 1.097 \times 10^7 \, \text{m}^{-1} \). Thus:
\[
\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} – \frac{1}{n_2^2} \right)
\]
| Series | \(n_1\) | \(n_2\) | Spectral region |
|---|---|---|---|
| Lyman | 1 | 2,3,4,… | Ultraviolet |
| Balmer | 2 | 3,4,5,… | Visible |
| Paschen | 3 | 4,5,6,… | Infrared |
| Brackett | 4 | 5,6,7,… | Far infrared |
| Pfund | 5 | 6,7,8,… | Far infrared |
✅ Successes of Bohr’s Model
- Explained stability of atoms (no radiation in stationary orbits).
- Predicted hydrogen line spectrum with high accuracy.
- Introduced quantum numbers and energy quantization.
- Provided a theoretical value of the Rydberg constant.
⚠️ Limitations of Bohr’s Model
- Only works for single‑electron atoms (H, He⁺, Li²⁺). Fails for multi‑electron atoms.
- Cannot explain fine structure or Zeeman effect.
- Violates Heisenberg’s uncertainty principle (definite orbits).
- Does not incorporate the wave nature of electrons.
- Unable to explain relative intensities of spectral lines.
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