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.

Part 1. Three organizing principles for how electrons fill the atomic energy levels of a many-electron atom:

a. Aufbau principle. Electrons fill the available orbitals starting from the lowest energy orbital upward, so the ground-state configuration is built one electron at a time into the lowest still-vacant state. Example: carbon ($Z=6$) fills $1s^2\,2s^2\,2p^2$—$1s$ and $2s$ are completely filled before any electron enters $2p$.

b. Pauli's exclusion principle. No two electrons in the same atom may share an identical set of all four quantum numbers ($n,\ell,m_\ell,m_s$); an orbital (fixed $n,\ell,m_\ell$) therefore holds at most two electrons, and only if their spins are opposite. Example: the $1s$ orbital of helium holds exactly two electrons, $1s^2$, with $m_s=+1/2$ and $m_s=-1/2$—a third electron cannot enter $1s$.

c. Electronegativity. A relative measure of how strongly an atom attracts the shared electron pair of a covalent bond toward itself. Example: in H–F, fluorine ($\chi\approx 4.0$) is far more electronegative than hydrogen ($\chi\approx 2.1$), so the bonding electron density is pulled toward F, giving the molecule a permanent dipole (polar covalent bond).

Part 2 — Given. Hydrogen atom, transition $n_i=4\to n_f=3$, Bohr energy levels $E_n=-13.6\,\text{eV}/n^2$.

Find. $\Delta E$ for the transition, and whether the photon is absorbed or emitted.

1. Level energies. $E_4=-13.6/4^2=-0.850\ \text{eV}$, $E_3=-13.6/3^2=-1.511\ \text{eV}$.
2. Energy change. $$\Delta E=E_3-E_4=-1.511-(-0.850)=\boxed{-0.661\ \text{eV}}.$$ The negative sign means the atom loses energy: the electron drops from the higher energy level ($n=4$) to the lower energy level ($n=3$), so a photon of energy $0.661$ eV is emitted, not absorbed.

Part 3(a) — Given. Radiation frequency $f=1\ \text{Hz}$ (as printed on the exam), $h=6.63\times10^{-34}\ \text{J}\cdot\text{s}$, $c=3.00\times10^{8}\ \text{m/s}$.

Find. The energy of a single quantum, and its wavelength.

1. Quantum energy. $$E=hf=6.63\times10^{-34}\times1=\boxed{6.63\times10^{-34}\ \text{J}}.$$
2. Wavelength. $$\lambda=\frac{c}{f}=\frac{3.00\times10^{8}}{1} =\boxed{3.00\times10^{8}\ \text{m}}.$$ (For scale, this "wavelength" is comparable to the Earth–Moon distance—a direct consequence of using an everyday, non-thermal frequency in the formula.)

Part 3(b) — Given. Electron speed $v=c/6$; position uncertainty $\Delta x=1\%$ of the electron's own de Broglie wavelength $\lambda_{dB}$ (the only length scale the sub-part establishes, via the de Broglie question asked immediately before it).

Find. What de Broglie's hypothesis states, and the minimum uncertainty in the electron's speed, $\Delta v_{\min}$.

de Broglie's hypothesis: every moving particle has an associated wave of wavelength $\lambda=h/p$, where $p=mv$ is its momentum—matter, not just light, exhibits wave–particle duality.

1. de Broglie wavelength. $v=c/6=5.00\times10^{7}\ \text{m/s}$, $p=m_ev=9.11\times10^{-31}\times5.00\times10^{7}=4.555\times10^{-23}\ \text{kg}\cdot\text{m/s}$. $$\lambda_{dB}=\frac{h}{p}=\frac{6.63\times10^{-34}}{4.555\times10^{-23}} =\boxed{1.456\times10^{-11}\ \text{m}=14.56\ \text{pm}}.$$
2. Position uncertainty. $\Delta x=0.01\times1.456\times10^{-11} =1.456\times10^{-13}\ \text{m}$.
3. Minimum speed uncertainty. Using $\Delta x\,\Delta p\ge h/(4\pi)$, $$\Delta p_{\min}=\frac{h}{4\pi\,\Delta x}=\frac{6.63\times10^{-34}}{4\pi\times1.456\times10^{-13}} =3.62\times10^{-22}\ \text{kg}\cdot\text{m/s},$$ $$\Delta v_{\min}=\frac{\Delta p_{\min}}{m_e}=\boxed{3.98\times10^{8}\ \text{m/s}}.$$