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Chapter 2 • Theory & Derivations

Particle Properties of Waves, Photons & X-Ray Physics

Comprehensive quantum theory of electromagnetic radiation: Planck's blackbody quantum hypothesis; photoelectric effect and Einstein's work function formalism; relativistic Compton scattering kinematics and wavelength shift; pair production, pair annihilation, and threshold conservation laws; gravitational photon interactions and black holes; continuous and characteristic X-ray generation, Moseley's atomic law, and Bragg crystal diffraction.

§2.1 Electromagnetic Radiation, Blackbody Radiation & Planck's Quantum Hypothesis

1. Classical Failure & The Ultraviolet Catastrophe

Classical thermodynamics and electrodynamics proved completely unable to describe the spectral distribution of blackbody radiation. Rayleigh and Jeans treated radiation inside a cavity of volume $V$ as an ensemble of classical standing electromagnetic waves. The number of modes per unit volume in the frequency interval $[ u, u + d u]$ is:

$$g(\nu) d\nu = \frac{8\pi \nu^2}{c^3} d\nu$$

By the classical equipartition theorem, each harmonic oscillator mode possesses average thermal energy $\langle E \rangle = k_B T$. This gives the Rayleigh-Jeans Spectral Energy Density:

$$u(\nu) d\nu = \frac{8\pi \nu^2}{c^3} k_B T d\nu$$

As $\nu \to \infty$ in the ultraviolet and X-ray regimes, $u(\nu) \to \infty$. The total integrated energy density diverges to infinity ($\int_0^\infty u(\nu) d u = \infty$), predicting that every heated object should instantaneously incinerate the universe with an infinite burst of high-frequency radiation—the famous Ultraviolet Catastrophe.

2. Max Planck's Quantum Revolution (1900)

Max Planck resolved the catastrophe by proposing a revolutionary quantum postulate: the atomic oscillators in the cavity walls cannot absorb or emit radiation continuously, but only in discrete discrete packets (quanta) of energy proportional to frequency:

$$E_n = n h \nu \quad (n = 0, 1, 2, 3, \dots)$$

where $h = 6.626 \times 10^{-34}\,\text{J}\cdot\text{s}$ is Planck's constant.

Using Boltzmann statistics, the average energy $\langle E \rangle$ of an oscillator of frequency $\nu$ is derived via the discrete partition function:

$$\langle E \rangle = \frac{\sum_{n=0}^\infty (n h\nu) e^{-n h\nu / k_B T}}{\sum_{n=0}^\infty e^{-n h\nu / k_B T}} = \frac{h\nu}{e^{h\nu / k_B T} - 1}$$

Multiplying by the mode density $g(\nu)$ delivers Planck's Blackbody Radiation Law:

$$u(\nu) d\nu = \frac{8\pi h \nu^3}{c^3} \frac{1}{e^{h\nu / k_B T} - 1} d\nu$$

At high frequencies ($h\nu \gg k_B T$), the exponential denominator explodes, suppressing mode occupancy to zero and cleanly eliminating the ultraviolet catastrophe.

§2.2 Photoelectric Effect: Stopping Potentials & Einstein's Photoelectric Equation

1. Experimental Observations of the Photoelectric Effect

When ultraviolet light shines on clean metallic surfaces (e.g., zinc, sodium, cesium), electrons are ejected (photoelectrons). Philipp Lenard (1902) discovered experimental features that completely contradicted classical wave theory:

  1. Absence of Time Lag: Photoelectrons are emitted instantaneously ($t < 10^{-9}\,\text{s}$) upon illumination, even under ultra-weak light intensities where classical wave theory requires hours of continuous wave-front soaking to accumulate the necessary escape energy.
  2. Threshold Frequency ($\nu_0$): For each metal, there exists a distinct threshold frequency $\nu_0$. If $\nu < \nu_0$, zero electrons are emitted regardless of how intense the light beam is.
  3. Independence of Kinetic Energy on Intensity: The maximum kinetic energy $K_{max}$ of the ejected electrons depends strictly on the frequency $ u$ of the incident light, and is entirely independent of light intensity.
  4. Linear Dependence on Intensity: Increasing the light intensity increases only the rate of photoelectron emission (photocurrent), not their individual kinetic energies.

2. Einstein's Photon Hypothesis (1905)

Albert Einstein explained these observations by postulating that light itself propagates and interacts as localized particle-like energy packets called photons, each carrying discrete energy $E = h\nu$.

When a photon strikes an electron in the metal, it transfers its entire energy $h\nu$ instantaneously in a 1-to-1 collision. A portion of this energy, the work function $W_0 = \Phi$, is expended to overcome the electrostatic binding potential holding the electron in the metal lattice. The remaining energy emerges as the electron's maximum kinetic energy $K_{max}$:

$$h\nu = W_0 + K_{max} \implies K_{max} = h\nu - W_0 = h(\nu - \nu_0)$$

where the threshold frequency $\nu_0$ and threshold wavelength $\lambda_0$ are:

$$\nu_0 = \frac{W_0}{h}, \quad \lambda_0 = \frac{c}{\nu_0} = \frac{hc}{W_0}$$

3. Stopping Potential Measurement

By applying a retarding negative potential to the collector plate, electrons are decelerated. The exact potential $V_0$ that reduces the photocurrent to zero is the stopping potential:

$$e V_0 = K_{max} = h\nu - W_0 \implies V_0 = \left( \frac{h}{e} \right) \nu - \frac{W_0}{e}$$

Robert Millikan (1916) plotted $V_0$ versus frequency $\nu$ across numerous alkali metals. The resulting graphs were straight lines with universal slope $h/e$, verifying Einstein's equation with high precision and earning Einstein the 1921 Nobel Prize in Physics.

§2.3 Compton Scattering: Relativistic Kinematics & Compton Shift Derivation

1. Arthur Compton's Discovery (1923)

When monochromatic X-rays ($\lambda \sim 0.07\text{ nm}$) were scattered from a graphite target, Arthur Compton discovered that the scattered radiation contained not only the incident wavelength $\lambda$, but also an additional component shifted toward longer wavelengths $\lambda' > \lambda$, with the shift depending only on the scattering angle $\theta$. Classical Thomson scattering predicted $\lambda' = \lambda$.

2. Rigorous Relativistic Kinematic Derivation

Treat the process as an elastic collision between an incident photon and a stationary free electron of rest mass $m_0$.

  • Before Collision:
    • Incident photon four-momentum: $P_i = (E/c, \vec{p}) = (h u/c, \frac{h u}{c} \hat{x})$
    • Target electron at rest: $P_e = (m_0 c, \vec{0})$
  • After Collision:
    • Scattered photon: Energy $E' = h u'$, momentum $\vec{p}'$ at angle $\theta$ relative to $\hat{x}$
    • Recoil electron: Energy $E_e = \gamma m_0 c^2$, momentum $\vec{p}_e$ at angle $\phi$

By conservation of four-momentum: $P_i + P_e = P_f + P_e'$. Isolating the recoil electron four-momentum:

$$P_e' = P_i + P_e - P_f$$

Taking the invariant scalar product of both sides with itself:

$$(P_e')^2 = (P_i + P_e - P_f)^2$$
$$m_0^2 c^2 = P_i^2 + P_e^2 + P_f^2 + 2 P_i \cdot P_e - 2 P_i \cdot P_f - 2 P_e \cdot P_f$$

Since photons are massless ($P_i^2 = P_f^2 = 0$) and the initial electron is at rest ($P_e^2 = m_0^2 c^2$):

$$m_0^2 c^2 = m_0^2 c^2 + 2 P_i \cdot P_e - 2 P_i \cdot P_f - 2 P_e \cdot P_f$$
$$P_i \cdot P_f = P_e \cdot (P_i - P_f)$$

Evaluating the four-vector products:

  • $P_i \cdot P_f = \left(\frac{h\nu}{c}\right)\left(\frac{h\nu'}{c}\right) - \vec{p} \cdot \vec{p}' = \frac{h^2 \nu \nu'}{c^2} (1 - \cos\theta)$
  • $P_e \cdot (P_i - P_f) = (m_0 c) \left( \frac{h\nu}{c} - \frac{h\nu'}{c} \right) = m_0 h (\nu - \nu')$

Equating both expressions:

$$\frac{h^2 \nu \nu'}{c^2} (1 - \cos\theta) = m_0 h (\nu - \nu')$$

Dividing both sides by $m_0 h u u'$:

$$\frac{h}{m_0 c^2} (1 - \cos\theta) = \frac{\nu - \nu'}{\nu \nu'} = \frac{1}{\nu'} - \frac{1}{\nu}$$

Multiplying by $c$ ($c/ u' = \lambda'$, $c/ u = \lambda$) yields the celebrated Compton Wavelength Shift Formula:

$$\Delta \lambda = \lambda' - \lambda = \frac{h}{m_0 c} (1 - \cos\theta) = \lambda_c (1 - \cos\theta)$$

where the Compton wavelength of the electron is:

$$\lambda_c = \frac{h}{m_0 c} = \frac{6.626 \times 10^{-34}\,\text{J}\cdot\text{s}}{(9.109 \times 10^{-31}\,\text{kg})(2.998 \times 10^8\,\text{m/s})} = 2.426 \times 10^{-12}\,\text{m} = 0.02426\,\text{\AA}$$

The maximum shift occurs in backscattering ($\theta = 180^\circ, \cos\theta = -1$):

$$\Delta \lambda_{max} = 2 \lambda_c \approx 0.0485\,\text{\AA}$$

§2.4 Pair Production, Annihilation & Photons in Gravitational Fields

1. Pair Production ($h u \to e^- + e^+$)

In pair production, a high-energy gamma-ray photon materializes into an electron-positron pair: $\gamma \to e^- + e^+$.

Conservation Laws and Threshold Energy:

$$E_{th} = 2 m_0 c^2 = 2 \times 0.511\,\text{MeV} = 1.022\,\text{MeV}$$

Any excess photon energy beyond $1.022\,\text{MeV}$ appears as kinetic energy of the produced pair: $K_{e^-} + K_{e^+} = h u - 2 m_0 c^2$.

Impossibility in Vacuum: Pair production cannot occur in empty space. In the center-of-momentum frame of the electron-positron pair, net momentum is zero. A single photon, however, always carries momentum $p = E/c \neq 0$. Energy and momentum cannot be simultaneously conserved in vacuum; a heavy atomic nucleus must be present to absorb the recoil momentum.

2. Pair Annihilation ($e^- + e^+ \to 2\gamma$)

The inverse process occurs when a positron encounters an electron, forming a short-lived hydrogen-like atom called positronium before annihilating into pure radiation. To conserve momentum in the rest frame, the pair must annihilate into at least two collinear, opposite-traveling gamma photons:

$$e^- + e^+ \to \gamma_1 + \gamma_2$$
$$E_{\gamma 1} = E_{\gamma 2} = m_0 c^2 = 511\,\text{keV}$$

This 511 keV coincidence radiation is the foundational operating principle of medical Positron Emission Tomography (PET) scanning.

3. Photons and Gravity: Gravitational Redshift & Black Holes

Although photons have zero rest mass, they carry effective gravitational mass $m_g = E/c^2 = h u / c^2$ by the Equivalence Principle.

When a photon climbs upward against gravity through height $H$ in a gravitational field $g$, it loses gravitational potential energy:

$$\Delta E = m_g g H = \left( \frac{h\nu}{c^2} \right) g H = h \Delta \nu$$
$$\frac{\Delta \nu}{\nu} = - \frac{g H}{c^2}$$

This fractional frequency drop is the Gravitational Redshift, verified experimentally by Pound and Rebka (1959) using Mössbauer gamma-ray spectroscopy over a $22.5\,\text{m}$ tower at Harvard.

Black Holes: If a mass $M$ is compressed until the escape velocity equals the speed of light ($v_{esc} = \sqrt{2GM/R} = c$), no light or matter can escape. This defines the Schwarzschild Radius:

$$R_s = \frac{2 G M}{c^2}$$

For the Sun, $R_s \approx 2.95\,\text{km}$; for Earth, $R_s \approx 8.87\,\text{mm}$.

§2.5 X-Ray Production: Bremsstrahlung, Duane-Hunt Cutoff & Moseley's Law

1. X-Ray Generation Mechanism (Coolidge Tube)

In a Coolidge X-ray tube, electrons emitted from a thermionic tungsten filament are accelerated through a colossal high-voltage potential difference $V$ ($20\,\text{kV} - 150\,\text{kV}$) and crash into a heavy metal anode target (tungsten, molybdenum, copper).

The resulting emission spectrum consists of two distinct components:

  1. Continuous X-Ray Spectrum (Bremsstrahlung / Braking Radiation): High-speed electrons decelerate rapidly in the Coulomb fields of target nuclei. The lost kinetic energy is emitted as continuous radiation.
  2. Characteristic X-Ray Spectrum: Bombarding electrons knock out inner-shell electrons ($K, L, M$ shells) from target atoms. When outer-shell electrons fall into the vacancies, sharp spectral lines ($K_\alpha, K_\beta, L_\alpha$) characteristic of the anode element are emitted.

2. The Duane-Hunt Cutoff Limit

An electron gives up its entire kinetic energy $e V$ in a single head-on collision, producing a photon of maximum frequency $ u_{max}$ and minimum wavelength $\lambda_{min}$:

$$e V = h \nu_{max} = \frac{h c}{\lambda_{min}}$$
$$\lambda_{min} = \frac{h c}{e V} = \frac{1.2398 \times 10^{-6}\,\text{V}\cdot\text{m}}{V} = \frac{12,398}{V\,(\text{in Volts})}\,\text{\AA}$$

The short-wavelength cutoff $\lambda_{min}$ depends solely on the accelerating voltage $V$, and is completely independent of the target material.

3. Moseley's Law & The Concept of Atomic Number (1913)

Henry Moseley systematically measured the characteristic $K_\alpha$ X-ray lines across the periodic table from aluminum to gold. He discovered that the square root of the frequency $ u$ was strictly linear with atomic number $Z$:

$$\sqrt{\nu} = a (Z - \sigma)$$

where $a$ is a proportionality constant and $\sigma$ is the screening constant ($\sigma = 1$ for the $K$-series, due to the single remaining electron screening the nucleus). Using Bohr's formula for a transition from $n = 2$ to $n = 1$:

$$\nu = c R_\infty (Z - 1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4} c R_\infty (Z - 1)^2$$
$$\sqrt{\nu} = \sqrt{\frac{3}{4} c R_\infty} \, (Z - 1)$$

Historic Importance: Moseley proved that the periodic table is ordered by nuclear charge / atomic number ($Z$) rather than atomic weight, resolving historical anomalies (such as Argon-Potassium and Cobalt-Nickel) and predicting the existence of undiscovered elements ($Z = 43, 61, 72, 75$).

§2.6 X-Ray Diffraction: Bragg's Law & Crystal Structure Determination

1. Crystals as Natural Three-Dimensional Diffraction Gratings

Because X-ray wavelengths ($\lambda \sim 0.1\text{ nm}$) match the interatomic lattice spacings $d$ of crystalline solids, Max von Laue (1912) recognized that crystals act as natural three-dimensional diffraction gratings.

2. Bragg's Law Derivation

William Henry Bragg and William Lawrence Bragg (1913) modeled diffraction as specular reflection from parallel atomic planes separated by spacing $d_{hkl}$.

Consider a monochromatic parallel X-ray beam striking parallel planes at glancing angle $\theta$ (measured relative to the crystal surface, not the normal). The path difference between waves reflected from adjacent planes is:

$$\Delta = AB + BC = d \sin\theta + d \sin\theta = 2 d \sin\theta$$

For constructive interference, this path difference must equal an integer number of wavelengths $n\lambda$:

$$2 d \sin\theta = n \lambda \quad (n = 1, 2, 3, \dots)$$

This is Bragg's Law. Since $\sin\theta \le 1$, diffraction can only occur if $\lambda \le 2d$. Visible light ($\lambda \approx 500\text{ nm}$) cannot produce crystal diffraction because its wavelength is thousands of times larger than lattice spacings.

3. Experimental Methods of X-Ray Crystallography

  • Laue Method: Uses a single stationary crystal illuminated by a continuous "white" polychromatic X-ray beam. Each plane selects the specific wavelength satisfying Bragg's law, producing a pattern of diffraction spots revealing crystal symmetry.
  • Bragg Spectrometer: Uses monochromatic X-rays and rotates a single crystal on a precision goniometer to measure scattering angles $\theta$ and determine lattice planes.
  • Debye-Scherrer Powder Method: Uses a fine polycrystalline powder containing millions of tiny randomly oriented micro-crystallites. The diffraction beams form concentric cones of constructive interference, recorded as concentric rings on photographic film.
Solved Problem Example 2.1: Photoelectric Effect Stopping Potential & Work Function Analysis

When a metallic cesium surface is illuminated with ultraviolet light of wavelength $\lambda_1 = 300\text{ nm}$, the stopping potential is measured to be $V_{01} = 2.24\text{ V}$. When illuminated with light of wavelength $\lambda_2 = 400\text{ nm}$, the stopping potential is $V_{02} = 1.20\text{ V}$. (a) Using Einstein's photoelectric equation, determine Planck's constant $h$ from these data. (b) Calculate the work function $\Phi$ of cesium in electron-volts. (c) Determine the threshold cutoff frequency $\nu_0$ and threshold wavelength $\lambda_0$.

Step 1: Set Up Einstein's Equations

For two frequencies $ u_1 = c/\lambda_1$ and $ u_2 = c/\lambda_2$:

$$e V_{01} = h \nu_1 - W_0 = \frac{h c}{\lambda_1} - W_0$$
$$e V_{02} = h \nu_2 - W_0 = \frac{h c}{\lambda_2} - W_0$$

Subtracting the second equation from the first eliminates the work function $W_0$:

$$e (V_{01} - V_{02}) = h c \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$$
$$V_{01} - V_{02} = 2.24\,\text{V} - 0.86\,\text{V} = 1.38\,\text{V}$$
$$\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{300 \times 10^{-9}} - \frac{1}{450 \times 10^{-9}} = 3.333 \times 10^6 - 2.222 \times 10^6 = 1.111 \times 10^6\,\text{m}^{-1}$$

Step 2: Calculate Planck's Constant $h$

$$h = \frac{e (V_{01} - V_{02})}{c \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)} = \frac{(1.602 \times 10^{-19}\,\text{C})(1.38\,\text{V})}{(3.0 \times 10^8\,\text{m/s})(1.111 \times 10^6\,\text{m}^{-1})} = \frac{2.2108 \times 10^{-19}}{3.333 \times 10^{14}} \approx 6.63 \times 10^{-34}\,\text{J}\cdot\text{s}$$

Step 3: Calculate Work Function $W_0$

$$\frac{h c}{\lambda_1} = \frac{1240\,\text{eV}\cdot\text{nm}}{300\,\text{nm}} \approx 4.133\,\text{eV}$$
$$W_0 = \frac{h c}{\lambda_1} - e V_{01} = 4.133\,\text{eV} - 2.24\,\text{eV} = 1.893\,\text{eV} \approx 1.89\,\text{eV}$$

Step 4: Threshold Frequency and Wavelength

$$\lambda_0 = \frac{h c}{W_0} = \frac{1240\,\text{eV}\cdot\text{nm}}{1.893\,\text{eV}} \approx 655\,\text{nm}$$
$$\nu_0 = \frac{c}{\lambda_0} = \frac{3.0 \times 10^8\,\text{m/s}}{655 \times 10^{-9}\,\text{m}} \approx 4.58 \times 10^{14}\,\text{Hz}$$

Step 5: Maximum Speed of Photoelectrons

$$K_{max} = e V_{01} = 2.24\,\text{eV} = 2.24 \times 1.602 \times 10^{-19}\,\text{J} = 3.588 \times 10^{-19}\,\text{J}$$
$$v_{max} = \sqrt{\frac{2 K_{max}}{m_e}} = \sqrt{\frac{2(3.588 \times 10^{-19}\,\text{J})}{9.109 \times 10^{-31}\,\text{kg}}} = \sqrt{7.878 \times 10^{11}} \approx 8.88 \times 10^5\,\text{m/s}$$
Solved Problem Example 2.2: Compton Scattering Energy & Recoil Electron Dynamics

A gamma-ray photon with energy $E = 0.511\text{ MeV}$ (equal to the rest mass energy of an electron) collides with a stationary free electron and is scattered through an angle $\theta = 60.0^\circ$. (a) Determine the Compton shift $\Delta\lambda$ and the scattered photon wavelength $\lambda'$. (b) Calculate the energy of the scattered photon $E'$ in keV. (c) Compute the kinetic energy $K_e$ and scattering recoil angle $\phi$ of the recoil electron.

Step 1: Calculate Incident Photon Wavelength $\lambda$

Since $E = h u = m_0 c^2$, the incident wavelength equals precisely the Compton wavelength $\lambda_c$:

$$\lambda = \frac{h c}{E} = \frac{h c}{m_0 c^2} = \frac{h}{m_0 c} = \lambda_c = 0.02426\,\text{\AA} = 2.426\,\text{pm}$$

Step 2: Calculate Compton Shift $\Delta \lambda$ and Scattered Wavelength $\lambda'$

$$\Delta \lambda = \lambda_c (1 - \cos 60^\circ) = \lambda_c (1 - 0.5) = 0.5 \lambda_c = 0.5 \times 2.426\,\text{pm} = 1.213\,\text{pm}$$
$$\lambda' = \lambda + \Delta \lambda = 2.426\,\text{pm} + 1.213\,\text{pm} = 3.639\,\text{pm} = 1.5 \lambda_c$$

Step 3: Calculate Scattered Photon Energy $E'$

$$E' = \frac{h c}{\lambda'} = \frac{h c}{1.5 \lambda_c} = \frac{E}{1.5} = \frac{0.511\,\text{MeV}}{1.5} \approx 0.3407\,\text{MeV} = 340.7\,\text{keV}$$

Step 4: Calculate Kinetic Energy of Recoil Electron $K_e$

$$K_e = E - E' = 511.0\,\text{keV} - 340.7\,\text{keV} = 170.3\,\text{keV}$$

Step 5: Calculate Recoil Angle $\phi$

From momentum conservation perpendicular and parallel to the incident beam:

$$\tan\phi = \frac{p' \sin\theta}{p - p' \cos\theta} = \frac{\frac{E'}{c} \sin\theta}{\frac{E}{c} - \frac{E'}{c} \cos\theta} = \frac{\sin 60^\circ}{\frac{E}{E'} - \cos 60^\circ} = \frac{0.8660}{1.5 - 0.5} = \frac{0.8660}{1.0} = 0.8660$$
$$\phi = \arctan(0.8660) = 40.89^\circ$$
Solved Problem Example 2.3: Duane-Hunt Cutoff & Moseley's Law X-Ray Tube Analysis

An X-ray tube operates with an accelerating voltage of $V = 50\text{ kV}$. (a) Calculate the Duane-Hunt minimum wavelength cutoff $\lambda_{\text{min}}$ in angstroms. (b) The $K_\alpha$ characteristic line for a target with atomic number $Z = 42$ (Molybdenum) occurs at $\lambda = 0.071\text{ nm}$. Using Moseley's empirical law $\sqrt{\nu} = a(Z - \sigma)$ with screening constant $\sigma = 1$, predict the $K_\alpha$ wavelength for a target with $Z = 74$ (Tungsten).

Step 1: Calculate Duane-Hunt Minimum Wavelength $\lambda_{min}$

$$\lambda_{min} = \frac{h c}{e V} = \frac{12,398\,\text{V}\cdot\text{\AA}}{50,000\,\text{V}} = 0.2480\,\text{\AA} = 0.0248\,\text{nm} = 24.8\,\text{pm}$$

Step 2: Calculate Maximum Photon Frequency $\nu_{max}$

$$\nu_{max} = \frac{c}{\lambda_{min}} = \frac{3.0 \times 10^8\,\text{m/s}}{2.480 \times 10^{-11}\,\text{m}} \approx 1.21 \times 10^{19}\,\text{Hz}$$

Step 3: Determine Atomic Number $Z$ via Moseley's Law

The frequency of the $K_\alpha$ line is:

$$\nu = \frac{c}{\lambda_{K\alpha}} = \frac{3.0 \times 10^8\,\text{m/s}}{0.1542 \times 10^{-9}\,\text{m}} \approx 1.9455 \times 10^{18}\,\text{Hz}$$

From Bohr's formula for $K_\alpha$ ($n = 2 \to n = 1$ with $\sigma = 1$):

$$\nu = \frac{3}{4} c R_\infty (Z - 1)^2$$

Using Rydberg constant $R_\infty = 1.09737 \times 10^7\,\text{m}^{-1}$:

$$\frac{3}{4} c R_\infty = \frac{3}{4} \times (2.9979 \times 10^8\,\text{m/s}) \times (1.09737 \times 10^7\,\text{m}^{-1}) = 2.4674 \times 10^{15}\,\text{Hz}$$
$$(Z - 1)^2 = \frac{\nu}{\frac{3}{4} c R_\infty} = \frac{1.9455 \times 10^{18}}{2.4674 \times 10^{15}} = 788.48$$
$$Z - 1 = \sqrt{788.48} \approx 28.08 \implies Z \approx 29$$

Atomic number $Z = 29$ corresponds precisely to Copper ($\text{Cu}$), whose celebrated $K_\alpha$ emission line at $1.5418\,\text{\AA}$ is the universal laboratory standard for X-ray diffraction worldwide.

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