Question 1 of 7: Atomic Structure

Question 1: Atomic 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.

1.1 — Four Concepts of Atomic Structure

a. Four quantum numbers. An electron's state in an atom is fixed by four quantum numbers. The principal number $n=1,2,3,\ldots$ sets the principal energy level and dominates the energy. The azimuthal (angular-momentum) number $l=0,1,\ldots,n-1$ sets the sub-level shape ($l=0,1,2,3 \to s,p,d,f$). The magnetic number $m_l=0,\pm1,\ldots,\pm l$ sets the orbital's spatial orientation ($2l+1$ orbitals per sub-level). The spin number $m_s=\pm1/2$ sets the electron's intrinsic angular momentum. Example: an electron in a $3p$ orbital has $n=3$, $l=1$, $m_l\in\{-1,0,1\}$, $m_s=\pm1/2$.

b. Isotopes. Isotopes are atoms of the same element (same atomic number $Z$, i.e. same proton count, hence identical chemistry) that differ in neutron number and therefore in mass number $A$. Example: $^{12}\text{C}$ (6 neutrons) and $^{14}\text{C}$ (8 neutrons) are both carbon but differ in mass and in nuclear stability ($^{14}$C is radioactive).

c. Wave-mechanical atomic model. Rather than Bohr's fixed circular orbits, the wave-mechanical model describes each electron by a wavefunction $\psi$ (a solution of Schrödinger's equation) whose squared magnitude $|\psi|^2$ gives the probability of finding the electron at a location — an "electron cloud" rather than a trajectory. This is consistent with the Heisenberg uncertainty principle (appendix: $\Delta x\cdot\Delta p\ge h/4\pi$), which forbids a simultaneously exact position and momentum, so no fixed orbit can exist.

d. Radioactive decay. An unstable (energetically unfavourable) nucleus spontaneously transforms toward a more stable configuration, emitting $\alpha$, $\beta$ or $\gamma$ radiation. Decay is a first-order random process, $N(t)=N_0 e^{-\lambda t}$, characterized by a half-life $t_{1/2}=\ln 2/\lambda$ independent of the sample's age or history.

1.2 — Electronic Structures of Ions

Each ion's configuration is obtained from the parent atom's ground-state Aufbau filling, then removing (cation) or adding (anion) electrons from the outermost, highest-energy sub-level first.

a. Fe (Z = 26): neutral configuration $1s^22s^22p^63s^23p^63d^64s^2$ = [Ar]$3d^64s^2$. Fe³⁺ removes the two $4s$ electrons (higher energy, more diffuse) and one $3d$ electron: $$\text{Fe}^{3+}: 1s^22s^22p^63s^23p^63d^5 = [\text{Ar}]3d^5$$

b. Nb (Z = 41): niobium is one of the Aufbau exceptions — its ground state is [Kr]$4d^45s^1$ (not [Kr]$4d^35s^2$), because a half-filled-adjacent $4d$ configuration is lower in energy. Nb³⁺ removes the single $5s$ electron and two $4d$ electrons: $$\text{Nb}^{3+}: [\text{Kr}]4d^2$$

c. S (Z = 16): neutral configuration $1s^22s^22p^63s^23p^4$ = [Ne]$3s^23p^4$. S²⁻ gains two electrons to fill the $3p$ sub-level: $$\text{S}^{2-}: 1s^22s^22p^63s^23p^6 = [\text{Ne}]3s^23p^6 \;(\text{isoelectronic with Ar})$$

1.3 — Wave-Particle Duality and Heisenberg Uncertainty

Given.

Find. The de Broglie wavelength $\lambda=h/mv$ of each, and the position uncertainty $\Delta x$ implied by $\Delta x\cdot\Delta p\ge h/4\pi$ for a 1% speed uncertainty.

Approach. Apply $\lambda=h/(mv)$ directly to each object; then compute $\Delta p=m\,\Delta v$ with $\Delta v=0.01v$ and invert the uncertainty relation for $\Delta x$.

1. Electron wavelength. $$\lambda_e=\frac{h}{m_ev}=\frac{6.63\times10^{-34}}{(9.11\times10^{-31})(6\times10^{7})}=\boxed{1.213\times10^{-11}\ \text{m} = 12.1\ \text{pm}}$$ — comparable to an atomic/X-ray length scale, so wave behaviour is significant.
2. Baseball wavelength. $$\lambda_{\text{ball}}=\frac{h}{mv}=\frac{6.63\times10^{-34}}{(0.142)(42.91)}=\boxed{1.088\times10^{-34}\ \text{m}}$$ — roughly $10^{20}$ times smaller than a nucleus; no observable wave behaviour.
3. Position uncertainty, electron. With $\Delta v_e=0.01(6\times10^7)=6\times10^5$ m/s, $\Delta p_e=m_e\Delta v_e=5.47\times10^{-25}$ kg·m/s. Solving $\Delta x\cdot\Delta p\ge h/4\pi$ for equality: $$\Delta x_e=\frac{h}{4\pi\,\Delta p_e}=\boxed{9.65\times10^{-11}\ \text{m}\approx0.097\ \text{nm}}$$ — a few atomic diameters: at 1% speed precision the electron's position is fundamentally, not just practically, uncertain.
4. Position uncertainty, baseball. $\Delta v_{\text{ball}}=0.4291$ m/s, $\Delta p_{\text{ball}}=0.0609$ kg·m/s. $$\Delta x_{\text{ball}}=\frac{h}{4\pi\,\Delta p_{\text{ball}}}=\boxed{8.66\times10^{-34}\ \text{m}}$$ — utterly negligible; classical mechanics applies without any measurable limitation.