E1-4. The Bohr Model of Hydrogen and Angular Momentum QuantizationHL Only
1. Postulates of the Bohr Model
Classical electrodynamics predicted that an electron orbiting a positive nucleus would constantly emit electromagnetic radiation due to its centripetal acceleration. This continuous loss of energy would cause the electron to rapidly spiral into the nucleus within a fraction of a second, making stable atoms impossible. To resolve this paradox, Niels Bohr proposed three revolutionary postulates:
- Electrons move around the positive nucleus in specific circular orbits called stationary states. While moving within these allowed orbits, they do not radiate any energy.
- Radiation is emitted or absorbed only when an electron jumps from one allowed stationary orbit to another. The energy of the photon matches the energy difference between the two states: $\Delta E = |E_i - E_f| = hf$.
- The orbital angular momentum ($L$) of the electron is strictly quantized. It can only exist in integer multiples of Dirac's reduced Planck constant ($\hbar = \dfrac{h}{2\pi}$): $$L = mvr = \frac{nh}{2\pi} \quad (n = 1, 2, 3...)$$
2. Louis de Broglie's Wave Explanation of Bohr's Quantization
Bohr's third postulate was originally an unproven assumption designed to match experimental line spectra. A decade later, Louis de Broglie provided a physical explanation by proposing that all matter possesses a wave-particle duality. An electron traveling with momentum $p = mv$ behaves like a matter wave with a wavelength given by:
For an electron to orbit a nucleus indefinitely without destroying itself through destructive interference, its matter wave must form a stable, continuous standing wave. This means the total circumference of the circular orbit ($2\pi r$) must equal an exact integer number ($n$) of wavelengths ($\lambda$):
Let's derive Bohr's quantization condition by substituting de Broglie's matter wavelength into this standing wave condition:
Rearranging the variables to isolate the angular momentum ($L = mvr$) yields:
This proof provides a clear physical meaning for the principal quantum number $n$: it represents the exact count of standing matter wavelengths that fit inside that specific electronic orbit!
3. Comprehensive Theoretical Derivation of Hydrogen Energy Levels
By combining Bohr's angular momentum quantization with classical mechanics, we can derive the precise radius and total energy equation for any orbital level in a hydrogen-like atom:
Step 1: Classical Force Balance
The electrostatic Coulomb force between a nucleus with charge $+Ze$ and an electron with charge $-e$ provides the necessary centripetal force holding the electron in its circular path of radius $r$:
Step 2: Substitute Velocity from Quantization
From Bohr's condition, isolate the orbital velocity: $v = \frac{nh}{2\pi m r}$. Substitute this into the force balance equation from Step 1:
$$m \left(\frac{nh}{2\pi m r}\right)^2 = \frac{k Z e^2}{r} \implies \frac{n^2 h^2}{4\pi^2 m r^2} = \frac{k Z e^2}{r}$$
Isolating the radius $r$ yields the allowed orbital radii (the Bohr Radii):
$$r_n = \left(\frac{h^2}{4\pi^2 k m e^2}\right) \frac{n^2}{Z}$$
For the ground state of hydrogen ($n=1, Z=1$), this simplifies to the Bohr radius constant $a_0 \approx 0.529 \times 10^{-10}\text{ m}$ ($0.529\text{ \AA}$).
Step 3: Derive Total Energy
The total mechanical energy $E$ is the sum of the electron's kinetic energy and its electric potential energy:
$$E = E_k + E_p = \frac{1}{2}mv^2 - \frac{k Z e^2}{r}$$
From our Step 1 force balance, we know $\frac{1}{2}mv^2 = \frac{kZe^2}{2r}$. Substituting this gives:
$$E = \frac{k Z e^2}{2r} - \frac{k Z e^2}{r} = -\frac{k Z e^2}{2r}$$
Finally, substituting our $r_n$ radius expression from Step 2 into this total energy equation gives the energy value for level $n$:
$$E_n = -\left(\frac{2\pi^2 k^2 m e^4}{h^2}\right) \frac{Z^2}{n^2}$$
Plugging in the standard values for an electron orbiting a hydrogen nucleus ($Z=1$), the constants group together perfectly to yield the classic energy level formula:
The negative sign signifies that the electron is stably bound to the nucleus. An energy of $0\text{ eV}$ represents a completely free electron ($n \rightarrow \infty$), meaning the ionization energy required to tear an electron away from the hydrogen ground state is exactly $+13.6\text{ eV}$.
4. Examples
Example 1: State Transition & Photon Frequency in the Balmer Series
Problem: An electron in an excited hydrogen atom drops from the $n = 4$ energy level down to the $n = 2$ energy level, emitting a single photon that forms a distinct line in the visible Balmer series. Calculate the exact energy change ($\Delta E$) in electronvolts, and determine the precise frequency ($f$) of the emitted light.
Solution:
Step 1: Compute the energy of the individual levels: $$E_4 = -\frac{13.6\text{ eV}}{4^2} = -\frac{13.6}{16} = -0.85\text{ eV}$$ $$E_2 = -\frac{13.6\text{ eV}}{2^2} = -\frac{13.6}{4} = -3.40\text{ eV}$$
Step 2: Find the absolute energy change: $$\Delta E = |E_4 - E_2| = |-0.85\text{ eV} - (-3.40\text{ eV})| = \mathbf{2.55\text{ eV}}$$
Step 3: Convert energy to Joules and calculate the frequency using $\Delta E = hf$: $$\Delta E = 2.55 \times (1.60 \times 10^{-19}\text{ J}) = 4.08 \times 10^{-19}\text{ J}$$ $$f = \frac{\Delta E}{h} = \frac{4.08 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = \mathbf{6.15 \times 10^{14}\text{ Hz}}$$ Verification: This frequency matches a wavelength of $\lambda = \dfrac{c}{f} \approx 487\text{ nm}$, which corresponds to the blue-green line in the visible Balmer series.
Example 2: Velocity of the Electron in the First Bohr Orbit
Problem: Determine the orbital speed ($v$) of an electron in its ground state orbit ($n = 1$) around a hydrogen atom ($Z = 1$). Use the values: electron mass $m_e = 9.11 \times 10^{-31}\text{ kg}$, ground state radius $r_1 = 0.529 \times 10^{-10}\text{ m}$, and Planck's constant $h = 6.63 \times 10^{-34}\text{ J s}$. Then, calculate what fraction of the speed of light this value represents to verify if relativistic effects can be safely ignored.
Solution:
Step 1: Isolate velocity from Bohr's quantization condition: $$m_e v r = \frac{nh}{2\pi}$$ For the ground state, set $n = 1$: $$v = \frac{h}{2\pi m_e r_1}$$
Step 2: Substitute the constants: $$v = \frac{6.63 \times 10^{-34}\text{ J s}}{2 \times \pi \times (9.11 \times 10^{-31}\text{ kg}) \times (0.529 \times 10^{-10}\text{ m})}$$ Calculate the denominator: $$\text{Denominator} = 2 \times \pi \times 9.11 \times 10^{-31} \times 0.529 \times 10^{-10} = 3.0278 \times 10^{-40}$$ Divide the numerator by this result: $$v = \frac{6.63 \times 10^{-34}}{3.0278 \times 10^{-40}} = \mathbf{2.19 \times 10^6\text{ m s}^{-1}}$$
Step 3: Calculate the fraction of the speed of light ($c = 3.00 \times 10^8\text{ m s}^{-1}$): $$\text{Fraction} = \frac{v}{c} = \frac{2.19 \times 10^6}{3.00 \times 10^8} \approx \mathbf{0.0073} \quad (\mathbf{0.73\%})$$ Because the electron moves at less than $1\%$ of the speed of light, its relativistic momentum correction is negligible. This confirms that classical momentum equations work perfectly for modeling the ground state of a hydrogen atom.
Example 3: Singly-Ionized Helium (${\text{He}^{+}}$ Scaling)
Problem: A helium atom stripped of one electron ($\text{He}^{+}$, $Z = 2$) behaves exactly like a hydrogen atom but features a more powerful nuclear charge.
(a) Find the total energy of an electron in the ground state ($n = 1$) of a $\text{He}^{+}$ ion.
(b) Calculate the minimum frequency photon required to completely ionize this electron from its ground state.
Solution:
Part (a): Apply the full structural scaling law derived in our step-by-step proof, where $E_n = -13.6\text{ eV} \cdot \dfrac{Z^2}{n^2}$. For $\text{He}^{+}$, substitute $Z = 2$ and $n = 1$: $$E_1 = -13.6\text{ eV} \cdot \frac{2^2}{1^2} = -13.6 \times 4 = \mathbf{-54.4\text{ eV}}$$ Analysis: Because the nucleus contains two protons instead of one, it pulls the electron into a tighter orbit with four times the binding energy of standard hydrogen.
Part (b): To ionize the electron, it must be lifted from its ground state ($E_1 = -54.4\text{ eV}$) up to a completely free state ($E_{\infty} = 0\text{ eV}$). This requires an input energy of exactly $\Delta E = +54.4\text{ eV}$.
Convert this ionization energy to standard Joules:
$$\Delta E = 54.4 \times (1.60 \times 10^{-19}\text{ J}) = 8.704 \times 10^{-18}\text{ J}$$
Now, use the relation $\Delta E = hf$ to find the required photon frequency:
$$f = \frac{\Delta E}{h} = \frac{8.704 \times 10^{-18}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = \mathbf{1.31 \times 10^{16}\text{ Hz}}$$
This university-level scaling example demonstrates that ionizing a single electron from helium requires a high-frequency photon deep within the extreme ultraviolet range.