Physics / Quantum Atomic & Molecular 100% Free Open Access
Chapter 4 β€’ Theory & Derivations

Rutherford-Bohr Atomic Model & Old Quantum Theory

Exhaustive treatment of atomic architecture: Thomson's plum-pudding model limitations; Rutherford alpha-particle scattering experiment, hyperbolic Coulomb orbit derivations, and nuclear dimension formulas; classical radiative collapse and the Larmor formula; Bohr's quantum model of hydrogenic atoms and spectral series; finite nuclear mass reduced mass corrections and the discovery of deuterium; Franck-Hertz inelastic electron impact excitation.

Β§4.1 Classical Atomic Models & The Rutherford Alpha-Scattering Experiment

1. J.J. Thomson's Plum-Pudding Model (1904)

Following his discovery of the electron (1897), J.J. Thomson proposed that an atom consists of a uniform sphere of positive electrostatic charge of atomic radius $R \sim 10^{-10}\,\text{m}$ (1 Γ…), within which tiny negative electrons are embedded like plums in a pudding.

Classical Predictions: Because positive charge is diffuse across the entire atomic volume, the maximum electric field inside the atom is weak ($E_{max} = \frac{Q}{4\pi\varepsilon_0 R^2} \sim 10^{11}\,\text{V/m}$). An energetic $\alpha$-particle ($q = +2e$, mass $m_\alpha \approx 7300 m_e$, $E_k \sim 5\text{ to }8\,\text{MeV}$) passing through a thin gold foil would experience only tiny deflections ($\theta < 1^\circ$). Multiple scattering could produce at most average deflections of a few degrees ($< 3^\circ$).

2. The Geiger-Marsden Experiments (1909–1911)

Under Ernest Rutherford's direction, Hans Geiger and Ernest Marsden directed a collimated beam of $5.5\,\text{MeV}$ $\alpha$-particles from a bismuth-214 radioactive source through an ultra-thin gold foil (thickness $\sim 400\,\text{nm}$, roughly $1000$ atoms thick) and recorded scintillations on a zinc sulfide ($\text{ZnS}$) phosphorescent screen.

The Shocking Discovery: While the vast majority of $\alpha$-particles passed straight through undeflected, roughly 1 in 8,000 $\alpha$-particles was deflected by angles greater than $90^\circ$, and some bounced directly backward ($\theta \approx 180^\circ$)! Rutherford famously recounted: "It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

Thomson's diffuse atom could never produce the colossal electrostatic field required to reverse a massive energetic $\alpha$-particle.

Β§4.2 Mathematical Derivation of Rutherford Scattering Cross-Section & Nuclear Dimensions

1. Coulomb Hyperbolic Trajectory Kinematics

Rutherford (1911) proposed that all the positive charge $+Z e$ and virtually all the mass of the atom are concentrated in a tiny central core termed the nucleus, surrounded by a cloud of orbiting electrons.

Consider an $\alpha$-particle (charge $q_1 = +2e$, mass $m_\alpha$) fired with initial speed $v_0$ and impact parameter $b$ (the perpendicular distance from the nucleus to the initial asymptotic velocity line) toward a stationary gold nucleus of charge $q_2 = +Z e$. The repulsive Coulomb force is:

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

Because the force is central, angular momentum $L = m_\alpha v_0 b = m_\alpha r^2 \dot{\phi}$ is strictly conserved. Integrating Newton's Second Law along the axis of symmetry yields the fundamental relation between impact parameter $b$ and scattering deflection angle $\theta$:

$$b = \frac{2 Z e^2}{4\pi\varepsilon_0 (m_\alpha v_0^2)} \cot\left( \frac{\theta}{2} \right) = \frac{z Z e^2}{4\pi\varepsilon_0 (2 K)} \cot\left( \frac{\theta}{2} \right)$$

where $K = \frac{1}{2} m_\alpha v_0^2$ is the kinetic energy of the incoming particle.

2. Derivation of the Rutherford Differential Cross-Section

Particles incident with impact parameters between $b$ and $b + db$ through an annular ring of area $d\sigma = 2\pi b |db|$ are scattered into solid angle $d\Omega = 2\pi \sin\theta d\theta$:

$$\frac{d\sigma}{d\Omega} = \frac{2\pi b |db|}{2\pi \sin\theta d\theta} = \frac{b}{\sin\theta} \left| \frac{db}{d\theta} \right|$$

Differentiating $b(\theta)$:

$$\frac{db}{d\theta} = - \frac{z Z e^2}{8\pi\varepsilon_0 K} \csc^2\left( \frac{\theta}{2} \right) \frac{1}{2}$$

Substituting into the cross-section and using $\sin\theta = 2 \sin(\theta/2) \cos(\theta/2)$:

$$\frac{d\sigma}{d\Omega} = \left( \frac{z Z e^2}{4\pi\varepsilon_0 \cdot 4 K} \right)^2 \frac{1}{\sin^4\left( \frac{\theta}{2} \right)}$$

This is the celebrated Rutherford Scattering Formula. Key testable predictions verified experimentally by Geiger and Marsden:

  1. Scattering intensity scales inversely as the fourth power of the sine of half the angle: $N(\theta) \propto \csc^4(\theta/2)$.
  2. Scattering scales inversely with the square of incident kinetic energy: $N \propto K^{-2}$.
  3. Scattering scales with the square of target nuclear charge: $N \propto Z^2$.
  4. Scattering scales linearly with foil thickness $t$.

3. Nuclear Dimension: Distance of Closest Approach ($d_{min}$)

In a head-on collision ($b = 0, \theta = 180^\circ$), the $\alpha$-particle decelerates until its entire initial kinetic energy is converted into electrostatic potential energy at the turning point $d_{min}$:

$$K = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{d_{min}} \implies d_{min} = \frac{2 Z e^2}{4\pi\varepsilon_0 K}$$

For a $7.7\,\text{MeV}$ $\alpha$-particle fired at Gold ($Z = 79$):

$$d_{min} = \frac{2 \times 79 \times (1.602 \times 10^{-19})^2}{4\pi (8.854 \times 10^{-12}) \times (7.7 \times 1.602 \times 10^{-13}\,\text{J})} \approx 2.95 \times 10^{-14}\,\text{m} = 29.5\,\text{fm}$$

Since Coulomb's law held precisely down to $d_{min}$, the atomic nucleus must have radius $R_{nuc} < 30\,\text{fm} = 3 \times 10^{-14}\,\text{m}$β€”more than 10,000 times smaller than the overall atomic radius ($10^{-10}\,\text{m}$)! The atom is almost entirely empty space.

Β§4.3 Classical Radiative Collapse & The Need for Quantum Atomic Postulates

1. The Classical Instability of Rutherford's Planetary Atom

Although Rutherford's nuclear model triumphed in explaining scattering, it suffered from a fatal theoretical catastrophe under classical electrodynamics.

In Rutherford's model, electrons orbit the central nucleus like planets around the Sun, held by Coulomb attraction:

$$\frac{m_e v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2} \implies a = \frac{v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 m_e r^2}$$

An orbiting electron undergoing centripetal acceleration is an accelerated electric charge. According to classical electrodynamics, any accelerated charge radiates electromagnetic power governed by the Larmor Radiation Formula:

$$P = \frac{e^2 a^2}{6\pi\varepsilon_0 c^3} = \frac{e^2}{6\pi\varepsilon_0 c^3} \left( \frac{e^2}{4\pi\varepsilon_0 m_e r^2} \right)^2 = \frac{e^6}{96 \pi^3 \varepsilon_0^3 m_e^2 c^3 r^4}$$

2. Derivation of the Collapse Time ($\tau$)

The total mechanical energy of an electron in a circular orbit of radius $r$ is:

$$E = K + U = \frac{1}{2} m_e v^2 - \frac{e^2}{4\pi\varepsilon_0 r} = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = - \frac{e^2}{8\pi\varepsilon_0 r}$$

Differentiating with respect to time:

$$\frac{dE}{dt} = \frac{e^2}{8\pi\varepsilon_0 r^2} \frac{dr}{dt} = - P = - \frac{e^6}{96 \pi^3 \varepsilon_0^3 m_e^2 c^3 r^4}$$

Solving for the orbital decay rate $dr/dt$:

$$\frac{dr}{dt} = - \frac{e^4}{12 \pi^2 \varepsilon_0^2 m_e^2 c^3} \frac{1}{r^2}$$

Integrating from initial radius $r_0 \approx 0.53 \times 10^{-10}\,\text{m}$ down to $r = 0$:

$$\int_{r_0}^0 r^2\, dr = - \frac{e^4}{12 \pi^2 \varepsilon_0^2 m_e^2 c^3} \int_0^\tau dt \implies \frac{r_0^3}{3} = \frac{e^4 \tau}{12 \pi^2 \varepsilon_0^2 m_e^2 c^3}$$
$$\tau = \frac{4 \pi^2 \varepsilon_0^2 m_e^2 c^3 r_0^3}{e^4} \approx 1.56 \times 10^{-11}\,\text{s}$$

The Classical Paradox: Classical electrodynamics predicts that every atom in the universe must collapse into its nucleus within a hundredth of a nanosecond, emitting a continuous burst of radiation! In reality, atoms are stable for billions of years and emit sharp, discrete line spectra.

Β§4.4 Bohr Theory of Hydrogenic Atoms: Radii, Energy Levels & Spectral Series

1. Niels Bohr's Quantum Postulates (1913)

To overcome classical collapse, Niels Bohr introduced three bold quantum postulates for hydrogenic atoms (one electron orbiting a nucleus of charge $+Ze$):

  1. Stationary States Postulate: Electrons move in discrete, non-radiating circular orbits called stationary states. While in these orbits, the electron accelerates but does NOT radiate electromagnetic energy.
  2. Angular Momentum Quantization: The orbital angular momentum $L$ of the electron is quantized in integer multiples of $\hbar = h / (2\pi)$:
    $$L = m_e v r = n \hbar = n \frac{h}{2\pi} \quad (n = 1, 2, 3, \dots)$$
  3. Frequency Postulate (Bohr Transition Rule): Radiation is emitted or absorbed only when an electron transitions discontinuously between two stationary states ($n_i \to n_f$). The photon frequency is:
    $$h\nu = E_i - E_f$$

2. Derivation of Orbit Radii and Speeds

Equating the Coulomb electrostatic force to the centripetal force:

$$\frac{m_e v^2}{r} = \frac{Z e^2}{4\pi\varepsilon_0 r^2} \implies v = \frac{Z e^2}{4\pi\varepsilon_0 m_e v r} = \frac{Z e^2}{4\pi\varepsilon_0 n \hbar}$$

Substituting $v$ into the quantization condition $m_e v r = n\hbar$ yields the quantized radius $r_n$:

$$r_n = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2} \frac{n^2}{Z} = n^2 \frac{a_0}{Z}$$

where the Bohr radius of hydrogen ($Z = 1, n = 1$) is:

$$a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2} = \frac{4\pi (8.854 \times 10^{-12})(1.0546 \times 10^{-34})^2}{(9.109 \times 10^{-31})(1.602 \times 10^{-19})^2} \approx 0.529177\,\text{\AA} = 0.05292\,\text{nm}$$

3. Quantized Energy Levels & Rydberg Formula

The total energy $E_n = K + U = \frac{1}{2} m_e v^2 - \frac{Z e^2}{4\pi\varepsilon_0 r_n} = - \frac{Z e^2}{8\pi\varepsilon_0 r_n}$:

$$E_n = - \frac{m_e Z^2 e^4}{32 \pi^2 \varepsilon_0^2 \hbar^2} \frac{1}{n^2} = - \frac{13.606\,\text{eV} \times Z^2}{n^2}$$

For a transition from initial level $n_i$ to final level $n_f$ ($n_i > n_f$), the wavenumber $\bar{\nu} = 1/\lambda$ of the emitted photon is:

$$\frac{1}{\lambda} = \frac{E_i - E_f}{h c} = R_\infty Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$

where the Rydberg constant for infinite mass is:

$$R_\infty = \frac{m_e e^4}{8 \varepsilon_0^2 h^3 c} \approx 1.097373 \times 10^7\,\text{m}^{-1} = 109,737.3\,\text{cm}^{-1}$$

4. The Hydrogen Spectral Series

Series Name Lower State $n_f$ Upper State $n_i$ Spectral Region First Line $\lambda$
Lyman Series $n_f = 1$ $2, 3, 4, \dots$ Ultraviolet $121.6\,\text{nm}$ ($L_\alpha$)
Balmer Series $n_f = 2$ $3, 4, 5, \dots$ Visible / Near-UV $656.3\,\text{nm}$ ($H_\alpha$, Red)
Paschen Series $n_f = 3$ $4, 5, 6, \dots$ Infrared $1875.1\,\text{nm}$
Brackett Series $n_f = 4$ $5, 6, 7, \dots$ Infrared $4051.2\,\text{nm}$
Pfund Series $n_f = 5$ $6, 7, 8, \dots$ Far Infrared $7457.8\,\text{nm}$

Β§4.5 Finite Nuclear Mass, Reduced Mass & The Discovery of Deuterium

1. Finite Nuclear Mass & Reduced Mass Formulation

In Bohr's elementary derivation, the nucleus was assumed to have infinite mass ($M = \infty$) and remain strictly fixed at the center of the orbit. In reality, both the electron (mass $m_e$) and nucleus (mass $M$) orbit around their common center of mass.

To account for finite nuclear motion, the electron mass $m_e$ in all dynamical formulas is replaced by the two-body Reduced Mass $\mu$:

$$\mu = \frac{m_e M}{m_e + M} = \frac{m_e}{1 + \frac{m_e}{M}}$$

Because $\mu < m_e$, the actual Rydberg constant $R_M$ for an atom with nuclear mass $M$ is slightly smaller than the theoretical infinite-mass constant $R_\infty$:

$$R_M = R_\infty \left( \frac{\mu}{m_e} \right) = \frac{R_\infty}{1 + \frac{m_e}{M}}$$

For ordinary Hydrogen (${}^1\text{H}$, $M = m_p \approx 1836.15 m_e$):

$$R_H = \frac{R_\infty}{1 + \frac{1}{1836.15}} = \frac{R_\infty}{1.0005446} \approx 1.096776 \times 10^7\,\text{m}^{-1}$$

2. Harold Urey's Discovery of Deuterium (1932)

Heavy hydrogen, or Deuterium (${}^2\text{H}$ or $\text{D}$), possesses a nucleus (deuteron) containing one proton and one neutron, giving it a nuclear mass approximately double that of protium: $M_D \approx 2 m_p \approx 3670.48 m_e$.

The Rydberg constant for deuterium is:

$$R_D = \frac{R_\infty}{1 + \frac{1}{3670.48}} = \frac{R_\infty}{1.0002724} \approx 1.097074 \times 10^7\,\text{m}^{-1}$$

Because $R_D > R_H$, every spectral line of deuterium is shifted slightly toward shorter wavelengths (higher frequencies) relative to hydrogen. For the red $H_\alpha$ Balmer line ($n = 3 \to n = 2$):

$$\Delta \lambda = \lambda_H - \lambda_D = \lambda_H \left( 1 - \frac{R_H}{R_D} \right) \approx (656.3\,\text{nm}) \times (2.72 \times 10^{-4}) \approx 0.179\,\text{nm} = 1.79\,\text{\AA}$$

By evaporating liquid hydrogen down to its triple point to concentrate heavy isotopes, Harold Urey and collaborators photographed a faint satellite line shifted by precisely $1.79\,\text{\AA}$ from the $H_\alpha$ line, confirming the existence of deuterium and earning Urey the 1934 Nobel Prize in Chemistry.

Β§4.6 The Franck-Hertz Experiment: Direct Proof of Quantized Energy Levels

1. Objective of the Franck-Hertz Experiment (1914)

While atomic emission spectroscopy revealed discrete light wavelengths, James Franck and Gustav Hertz provided the first direct non-optical proof that internal atomic energy states are quantized, demonstrating discrete energy absorption through inelastic electron collisions with mercury vapor atoms.

2. Experimental Apparatus & Operational Mechanics

The apparatus consists of a heated glass tube filled with mercury ($\text{Hg}$) vapor at low pressure ($P \approx 1\text{ to }10\,\text{Torr}$):

  • Cathode (K): Heated filament emitting electrons with negligible initial thermal energy.
  • Grid Anode (G): Fine mesh accelerated by variable forward voltage $V_a$ ($0\text{ to }30\,\text{V}$).
  • Collector Plate (P): Positioned beyond the grid, biased with a small retarding reverse voltage $V_r \approx 1.0\text{ to }1.5\,\text{V}$ relative to the grid.

Electrons reaching the grid must possess kinetic energy $K > e V_r$ to overcome the retarding barrier and register as collector current $I_c$.

3. Elastic vs. Inelastic Collisions & Current Drops

  • Low Accelerating Voltages ($V_a < 4.9\,\text{V}$): The kinetic energy of the electrons is insufficient to excite a mercury atom from its ground state ($6^1S_0$) to its first excited state ($6^3P_1$, which lies $4.9\,\text{eV}$ higher). The electrons undergo purely elastic collisions. Because a mercury atom is $\sim 360,000$ times heavier than an electron, the electron bounces off with essentially zero kinetic energy loss. As $V_a$ increases, collector current $I_c$ rises steadily.
  • First Inelastic Threshold ($V_a = 4.9\,\text{V}$): As soon as $V_a$ reaches $4.9\,\text{V}$, electrons near the grid attain exactly $4.9\,\text{eV}$ of kinetic energy. They undergo inelastic collisions with mercury atoms, transferring their entire $4.9\,\text{eV}$ to excite mercury electrons. Left with near-zero kinetic energy ($K \approx 0$), these electrons cannot overcome the $1.5\,\text{V}$ retarding potential. Collector current $I_c$ plunges sharply!
  • Periodic Successive Inelastic Drops ($V_a = 9.8\,\text{V}, 14.7\,\text{V}, 19.6\,\text{V}$): As $V_a$ increases further, electrons gain enough energy to undergo a second inelastic collision at $9.8\,\text{V}$ ($2 \times 4.9\,\text{V}$), a third at $14.7\,\text{V}$ ($3 \times 4.9\,\text{V}$), and so on.

The resulting $I_c$ versus $V_a$ curve exhibits a striking series of periodic peaks and valleys separated by exactly $\Delta V = 4.9\,\text{V}$.

Optical Confirmation: When mercury atoms de-excite back to the ground state, they emit ultraviolet photons of wavelength:

$$\lambda = \frac{h c}{\Delta E} = \frac{1240\,\text{eV}\cdot\text{nm}}{4.9\,\text{eV}} \approx 253\,\text{nm}$$

Franck and Hertz observed the simultaneous appearance of $253.7\,\text{nm}$ UV luminescence as soon as $V_a$ exceeded $4.9\,\text{V}$, clinching the 1925 Nobel Prize in Physics.

Solved Problem Example 4.1: Rutherford Scattering Closest Approach & Cross-Section Calculation

An alpha particle beam with kinetic energy $K = 7.7\text{ MeV}$ is directed at a thin gold foil ($Z = 79$, density $\rho = 19.3\text{ g/cm}^3$, thickness $t = 1.0\ \mu\text{m}$, molar mass $M = 197\text{ g/mol}$). (a) Calculate the distance of closest approach $d_{\text{min}}$ for a head-on collision (scattering angle $\theta = 180^\circ$). (b) Compute the impact parameter $b$ corresponding to a scattering angle $\theta = 60.0^\circ$. (c) Determine the fraction $f$ of incident alpha particles scattered through angles greater than $60.0^\circ$.

Step 1: Calculate Distance of Closest Approach $d_{min}$

$$K = 7.7\,\text{MeV} = 7.7 \times 10^6 \times 1.6022 \times 10^{-19}\,\text{J} = 1.2337 \times 10^{-12}\,\text{J}$$
$$d_{min} = \frac{2 Z e^2}{4\pi\varepsilon_0 K} = \frac{2 \times 79 \times (1.6022 \times 10^{-19})^2}{4\pi (8.854 \times 10^{-12}) \times (1.2337 \times 10^{-12}\,\text{J})} = \frac{4.056 \times 10^{-36}}{1.373 \times 10^{-22}} \approx 2.954 \times 10^{-14}\,\text{m} = 29.54\,\text{fm}$$

Step 2: Calculate Impact Parameter $b$ for $\theta = 60^\circ$

Using the impact parameter relation $b = \frac{d_{min}}{2} \cot(\theta/2)$:

$$b = \frac{29.54\,\text{fm}}{2} \cot(30^\circ) = (14.77\,\text{fm}) \times \sqrt{3} = 14.77 \times 1.732 \approx 25.58\,\text{fm}$$

Step 3: Calculate Fraction Scattered Beyond $90^\circ$

Scattering at angles $\theta > 90^\circ$ corresponds to impact parameters $b < b_{90}$:

$$b_{90} = \frac{d_{min}}{2} \cot(45^\circ) = \frac{29.54\,\text{fm}}{2} \times 1.0 = 14.77\,\text{fm} = 1.477 \times 10^{-14}\,\text{m}$$

The cross-section for scattering beyond $90^\circ$ is $\sigma(> 90^\circ) = \pi b_{90}^2$:

$$\sigma(> 90^\circ) = \pi (1.477 \times 10^{-14}\,\text{m})^2 \approx 6.853 \times 10^{-28}\,\text{m}^2 = 0.6853\,\text{barns}$$

Number density of target gold atoms $n$:

$$n = \frac{\rho N_A}{A} = \frac{(19.3 \times 10^3\,\text{kg/m}^3)(6.022 \times 10^{23}\,\text{mol}^{-1})}{0.197\,\text{kg/mol}} \approx 5.90 \times 10^{28}\,\text{atoms/m}^3$$

The fraction $f$ scattered by foil thickness $t = 1.0\,\mu\text{m} = 1.0 \times 10^{-6}\,\text{m}$ is:

$$f = n t \sigma(> 90^\circ) = (5.90 \times 10^{28}\,\text{m}^{-3})(1.0 \times 10^{-6}\,\text{m})(6.853 \times 10^{-28}\,\text{m}^2) \approx 4.04 \times 10^{-5}$$

Approximately 1 in every 24,700 incident alpha particles is scattered by more than $90^\circ$.

Solved Problem Example 4.2: Bohr Model Hydrogenic Transitions & Ionized Helium Spectra

For singly-ionized helium ($\text{He}^+, Z = 2$): (a) Calculate the radius of the ground state orbit $r_1$ and the ground state energy $E_1$ using the Bohr model. (b) Find the wavelength of the photon emitted in the transition from $n = 4$ to $n = 2$ in $\text{He}^+$ and determine in what spectral region this line lies. (c) Compare this transition to the $n = 2 \to n = 1$ Lyman-alpha transition of atomic hydrogen.

Step 1: Ground State Radius and Energy of $\text{He}^+$

$$r_1(\text{He}^+) = \frac{a_0}{Z} = \frac{0.5292\,\text{\AA}}{2} = 0.2646\,\text{\AA} = 0.02646\,\text{nm}$$
$$E_1(\text{He}^+) = - (13.606\,\text{eV}) \frac{Z^2}{1^2} = - 13.606 \times 4 = - 54.424\,\text{eV}$$

Step 2: Transition from $n = 4$ to $n = 2$ in $\text{He}^+$

$$\frac{1}{\lambda} = R_\infty Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_\infty (4) \left( \frac{1}{4} - \frac{1}{16} \right) = R_\infty (4) \left( \frac{3}{16} \right) = \frac{3}{4} R_\infty$$
$$\lambda = \frac{4}{3 R_\infty} = \frac{4}{3 \times 1.09737 \times 10^7\,\text{m}^{-1}} \approx 1.215 \times 10^{-7}\,\text{m} = 121.5\,\text{nm}$$

This emission lies in the ultraviolet spectrum.

Step 3: Comparison with Hydrogen ($Z = 1$)

For Hydrogen ($n = 4 \to n = 2$, the $H_\beta$ Balmer line):

$$\frac{1}{\lambda_H} = R_\infty (1)^2 \left( \frac{3}{16} \right) = \frac{3}{16} R_\infty \implies \lambda_H = \frac{16}{3 R_\infty} = 4 \times \lambda(\text{He}^+) = 4 \times 121.5\,\text{nm} = 486.1\,\text{nm} \text{ (Cyan Visible)}$$

Because wavelengths scale as $1/Z^2$, $\text{He}^+$ lines are shifted by a factor of 4 toward shorter wavelengths relative to corresponding hydrogen lines.

Step 4: Lyman Series Limit for $\text{He}^+$ ($n_i = \infty, n_f = 1$)

$$\frac{1}{\lambda_{limit}} = R_\infty (2^2) \left( \frac{1}{1^2} - \frac{1}{\infty} \right) = 4 R_\infty$$
$$\lambda_{limit} = \frac{1}{4 R_\infty} = \frac{1}{4 \times 1.09737 \times 10^7\,\text{m}^{-1}} \approx 2.278 \times 10^{-8}\,\text{m} = 22.78\,\text{nm}$$
Solved Problem Example 4.3: Hydrogen-Deuterium Isotope Shift in the Balmer Alpha Line

Given the proton mass $m_p = 1.67262 \times 10^{-27}\text{ kg}$, deuteron mass $m_d = 3.34358 \times 10^{-27}\text{ kg}$, and electron mass $m_e = 9.10938 \times 10^{-31}\text{ kg}$: (a) Calculate the reduced mass of ordinary hydrogen $\mu_H$ and deuterium $\mu_D$. (b) Using the Rydberg constant $R = \mu e^4 / (8 \varepsilon_0^2 h^3 c)$, compute the wavelength of the Balmer alpha line ($n = 3 \to 2$) for hydrogen ($\lambda_H$) and deuterium ($\lambda_D$). (c) Compute the isotope wavelength shift $\Delta\lambda = \lambda_H - \lambda_D$ in angstroms and explain how this shift led to the experimental discovery of deuterium.

Step 1: Calculate Reduced Masses $\mu_H$ and $\mu_D$

$$\frac{m_e}{m_p} = \frac{9.10938 \times 10^{-31}}{1.67262 \times 10^{-27}} \approx 5.44617 \times 10^{-4}$$
$$\mu_H = \frac{m_e}{1 + \frac{m_e}{m_p}} = \frac{m_e}{1 + 0.000544617} = \frac{m_e}{1.000544617} \approx 0.9994557 m_e$$
$$\frac{m_e}{m_d} = \frac{9.10938 \times 10^{-31}}{3.34358 \times 10^{-27}} \approx 2.72444 \times 10^{-4}$$
$$\mu_D = \frac{m_e}{1 + \frac{m_e}{m_d}} = \frac{m_e}{1.000272444} \approx 0.9997276 m_e$$

Step 2: Calculate Rydberg Constants $R_H$ and $R_D$

$$R_H = R_\infty \left( \frac{\mu_H}{m_e} \right) = (1.097373 \times 10^7\,\text{m}^{-1}) \times 0.9994557 \approx 1.096776 \times 10^7\,\text{m}^{-1}$$
$$R_D = R_\infty \left( \frac{\mu_D}{m_e} \right) = (1.097373 \times 10^7\,\text{m}^{-1}) \times 0.9997276 \approx 1.097074 \times 10^7\,\text{m}^{-1}$$

Step 3: Calculate $H_\alpha$ Wavelengths ($n = 3 \to n = 2$)

For $n = 3 \to n = 2$: $\frac{1}{\lambda} = R \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36} R \implies \lambda = \frac{36}{5 R}$:

$$\lambda_H = \frac{36}{5 \times 1.096776 \times 10^7\,\text{m}^{-1}} = \frac{7.2}{1.096776 \times 10^7} \approx 6.56469 \times 10^{-7}\,\text{m} = 6564.69\,\text{\AA}$$
$$\lambda_D = \frac{36}{5 \times 1.097074 \times 10^7\,\text{m}^{-1}} = \frac{7.2}{1.097074 \times 10^7} \approx 6.56291 \times 10^{-7}\,\text{m} = 6562.91\,\text{\AA}$$

Step 4: Calculate Isotope Shift $\Delta \lambda$

$$\Delta \lambda = \lambda_H - \lambda_D = 6564.69\,\text{\AA} - 6562.91\,\text{\AA} = 1.78\,\text{\AA} = 0.178\,\text{nm}$$

This $1.78\,\text{\AA}$ spectral separation is easily resolved with standard diffraction grating spectrometers, confirming Urey's discovery of deuterium.

EXAM SUCCESS WORKSHOP

Solved University Examination Problems

Step-by-step mathematical solutions to classic university honors examination questions.