Question 1 of 7: Electron Structure

Question I: Electron Structure (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

I.1(a) — Wave–Particle Duality

Every quantum entity — an electron, a photon — exhibits both wave-like behaviour (interference, diffraction, characterised by a wavelength $\lambda=h/mv$, de Broglie's relation) and particle-like behaviour (localized momentum and energy transfer on collision). Neither picture alone is complete: an electron diffracts around a crystal lattice exactly like a wave of wavelength $\lambda=h/mv$ (Davisson–Germer electron diffraction), yet strikes a detector as a single localized point, like a particle. Which behaviour dominates in an experiment depends on what is measured, not on the electron itself.

I.1(b) — Heisenberg Uncertainty Principle

The position and momentum of a quantum particle cannot both be known to arbitrary precision simultaneously: $\Delta x\cdot\Delta p\geq h/4\pi$ (given in the appendix). Narrowing the uncertainty in position $\Delta x$ (e.g. confining an electron to a smaller region) necessarily widens the uncertainty in its momentum $\Delta p$, and vice versa. This is the direct reason a Bohr-style "electron on a fixed circular orbit at a fixed radius and speed" cannot be physically exact — it would imply zero uncertainty in both $x$ and $p$ at once — and is why the quantum-mechanical model instead describes an electron's location only as a probability distribution (Question I.2).

I.1(c) — Hund's Rule

When filling a set of degenerate (equal-energy) orbitals, such as the three $2p$ or five $3d$ orbitals, electrons occupy separate orbitals singly, with parallel spins, before any orbital is doubly occupied. For example, carbon ($1s^2 2s^2 2p^2$) places its two $2p$ electrons in two different $p$ orbitals with parallel spin ($\uparrow\ \uparrow\ \_$), not paired in one orbital ($\uparrow\downarrow\ \_\ \_$); this minimizes electron–electron repulsion (electrons in different orbitals stay farther apart on average) and is the rule that fixes the number of unpaired electrons controlling paramagnetism in transition-metal ions.

I.2 — Bohr Model vs. Quantum-Mechanical Model

Bohr's model treats each electron as a point particle travelling on one of a discrete set of fixed, sharply-defined circular orbits about the nucleus, with quantized energy $E_n=-Z^2R_E/n^2$ (given in the appendix) and a definite radius and speed for each orbit $n$. The quantum-mechanical model instead solves the Schrödinger wave equation for the electron and interprets the squared wavefunction $|\psi|^2$ as a probability density: the electron's position is described only by an orbital, a three-dimensional probability cloud (e.g. the spherical $1s$, dumb-bell $2p$) with no single well-defined path or radius, consistent with $\Delta x\cdot\Delta p\geq h/4\pi$ from I.1(b). The quantum wave-mechanical model is the more accurate of the two: it correctly predicts the fine structure and relative intensities of atomic spectra, the shapes of chemical bonds, and multi-electron atoms, all of which the Bohr model (built only for one-electron systems and inconsistent with the uncertainty principle) gets wrong or cannot address at all.

I.3 — Potential-Energy and Force Curves; Young's Modulus

The potential energy $E(r)$ falls steeply at small $r$ (short-range electron-cloud/nuclear repulsion dominates), passes through a minimum $E_0$ at the equilibrium spacing $r_0$, and rises slowly back toward zero at large $r$ (the residual attractive term). The force curve is the negative slope of the energy curve, $F(r)=-dE/dr=-\partial E/\partial r$ (given in the appendix): it is attractive (net pulling the atoms together) for $r>r_0$, repulsive for $rinitial slope of the force curve at $r_0$: $E\propto\left(\dfrac{dF}{dr}\right)_{r_0}$, i.e. the stiffness of the interatomic "spring" right at equilibrium (equivalently, the curvature $d^2E/dr^2$ of the energy well at its minimum). A deep, sharply-curved energy well (steep force-curve slope at $r_0$) gives a stiff, high-modulus material (e.g. covalent ceramics, diamond); a shallow, broad well gives a compliant, low-modulus material (e.g. polymers, soft metals).

I.4 — Longest Ionising Wavelength for Potassium

Given. Ionisation energy $=419\ \text{kJ}\,\text{mol}^{-1}$; conversion factor (as supplied in the question) $1\ \text{kJ}\,\text{mol}^{-1}=1.67\times10^{-21}\ \text{J/atom}$; Planck's constant $h=6.63\times10^{-34}\ \text{J}\cdot\text{s}$; speed of light $c=3.0\times10^{8}\ \text{m/s}$ (standard value, appendix gives $\nu\lambda=c$).

Find. The longest wavelength of light $\lambda_{max}$ that still carries enough energy per photon to ionise a potassium atom.

Approach. Convert the molar ionisation energy to an energy per atom using the given factor, then use $E=h\nu=hc/\lambda$: the longest wavelength corresponds to the minimum photon energy that exactly equals the ionisation energy (any longer wavelength carries too little energy per photon to ionise the atom).

1. Ionisation energy per atom. $$E_{atom}=419\times1.67\times10^{-21}\ \text{J}=\boxed{6.997\times10^{-19}\ \text{J}}$$
2. Photon wavelength from $E=hc/\lambda$. Solving for $\lambda$ and substituting $h$, $c$ and $E_{atom}$: $$\lambda_{max}=\frac{hc}{E_{atom}}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{6.997\times10^{-19}}=\boxed{2.842\times10^{-7}\ \text{m} = 284.2\ \text{nm}}$$