Exercises — Understand conductors, insulators, and semiconductors

WHAT we compare against (from the parent table):

• Conductor $\rho \approx 10^{-8}\Omega \cdot \text{m}$
• Semiconductor $\rho \approx 10^{-4}\text{–}10^3\Omega \cdot \text{m}$
• Insulator $\rho \approx 10^{10}\text{–}10^{16}\Omega \cdot \text{m}$

WHY these ranges separate the classes: resistivity is set by the free-carrier density $n$ through $\rho =1/(ne\mu )$, and $n$ itself is decided by the band gap $E_g$. A conductor has $E_g\approx 0$, so $n$ is enormous → tiny $\rho$ near $10^{-8}$. An insulator has $E_g>5$ eV, so almost no electron climbs the gap → $n\approx 0$ → gigantic $\rho$ near $10^{14}$. A semiconductor's $E_g\approx 1$ eV lets a small, tunable $n$ exist, landing it in the wide middle band. The ranges are separated by roughly 8 to 12 orders of magnitude precisely because $n$ swings that hugely across the classes — so a single $\rho$ reading almost always drops cleanly into one bucket.

(a) $1.7\times 10^{-8}$ → conductor (this is copper). (b) $0.46$ → sits in the wide semiconductor band → semiconductor. (c) $10^{14}$ → insulator. (d) $2.8\times 10^{-8}$ → conductor (this is aluminium).

• Copper → bands overlap, $E_g\approx 0$ → no ladder to climb → conductor.
• Silicon → $E_g\approx 1.1$ eV → short ladder heat can climb → semiconductor.
• Glass → $E_g>5$ eV → huge ladder → insulator.

What the figure below shows: three vertical energy diagrams side by side. In each, a solid black block at the bottom is the valence band (full of electrons) and an open black block at the top is the conduction band (where free carriers live). The ==red bracket marks the band gap $E_g$== — the vertical distance an electron must jump. Reading left to right, watch that red gap grow from nothing (copper: the blocks touch, labelled "bands overlap"), to a short red bracket (silicon, $1.1$ eV), to a tall red bracket (glass, $6.5$ eV). The single visual takeaway: the taller the red gap, the fewer electrons reach the top, the worse the material conducts.

False. An insulator is full of electrons — but they are tightly bound in the valence band and cannot cross the large band gap $E_g$. It has almost no free carriers ($n\approx 0$), which is different from having no electrons at all.

WHY $R=\rho L/A$: we want a shaped object's resistance, so geometry (length, area) matters. $R=\frac{(1.7\times 10^{-8})(5)}{2\times 10^{-6}}=\frac{8.5\times 10^{-8}}{2\times 10^{-6}}=4.25\times 10^{-2}\Omega =0.0425\Omega .$ Sanity check: a few centimetres of an ohm — near-zero, exactly what a conductor should give.

WHAT changed: $L$ doubled ($\times 2$) and $A$ halved ($\times \frac{1}{2}$). Since $R\propto L/A$, the factor is $2÷\frac{1}{2}=4$. $R_{\text{new}}=4\times 0.0425=0.17\Omega .$ Direct check: $R=\frac{(1.7\times 10^{-8})(10)}{1\times 10^{-6}}=0.17\Omega .$ ✓

WHY $\sigma =ne\mu$: conductivity is carrier density × charge × how mobile each carrier is. $\sigma =(8.5\times 10^{28})(1.6\times 10^{-19})(4.4\times 10^{-3})\approx 5.99\times 10^7\text{S/m}.$ $\rho =\frac{1}{\sigma }\approx 1.67\times 10^{-8}\Omega \cdot \text{m}.$ That $\rho$ lands right on copper — the numbers were copper's all along.

Chain of reasoning:

1. Silicon is a semiconductor; carrier density obeys $n\propto e^{-E_g/(2k_{B}T)}$, where $E_g$ is the band gap (defined at the top of the page) and $k_B=8.617\times 10^{-5}$ eV/K is the Boltzmann constant.
2. Raising $T$ makes the exponent less negative → $n$ grows (fast).
3. $\sigma =ne\mu$ rises → $\rho =1/\sigma$ falls → resistance falls.

Contrast: a copper wire does the opposite — heating it scatters electrons more, shrinking collision time $\tau$, so $\rho$ rises.

When one electron jumps the gap $E_g$ it leaves behind a hole in the valence band. Both are created together as a pair, so the gap energy $E_g$ is effectively split between the two carriers. Dividing $E_g$ by 2 in the exponent encodes "the cost is shared by the electron–hole pair." This is the signature of intrinsic (undoped) generation. (Here $k_B$ is again the Boltzmann constant from the top of the page.)

Resistance increasing with temperature ⇒ metal (conductor). Using the definition above, $\alpha =\frac{1}{R}\frac{\mathrm{d}R}{\mathrm{d}T}$ is positive: a $0.4\%$ rise per °C gives $\alpha =+0.004/°\text{C}$. A semiconductor would show a negative $\alpha$ (resistance drops as it warms).

WHAT we compute: the ratio kills the proportionality constant, leaving pure exponentials. $\frac{n(350)}{n(300)}=\exp \left[-\frac{E_g}{2k_B}\left(\frac{1}{350}-\frac{1}{300}\right)\right].$ Inside: $\frac{1}{350}-\frac{1}{300}=0.0028571-0.0033333=-4.762\times 10^{-4}{\text{K}}^{-1}$. Prefactor: $\frac{E_g}{2k_B}=\frac{1.1}{2(8.617\times 10^{-5})}=6383\text{K}$. Exponent: $-(6383)(-4.762\times 10^{-4})=+3.039$. $\frac{n(350)}{n(300)}=e^{3.039}\approx \mathrm{20.9.}$ A modest 50 K rise multiplies free carriers by ~21× — that steep exponential is why semiconductors are so temperature-sensitive.

WHY the ratio is just $n$-ratio: $\sigma =ne\mu$, and $e,\mu$ are fixed, so $\sigma \propto n$. $\frac{{\sigma }_{\text{doped}}}{{\sigma }_i}=\frac{n}{n_i}=\frac{1.5\times 10^{22}}{1.5\times 10^{16}}=10^6.$ A million-fold rise from a pinch of impurity. Adding donor atoms (extra electrons) makes this n-type silicon — see Semiconductor Diodes and Transistors.

WHY set the resistances equal: the problem demands both wires have identical $R$, so we write each wire's $R=\rho L/A$ and force them to match — that single equation ties the two geometries together. WHY $L$ drops out: the lengths are the same, so $L$ appears on both sides and cancels, leaving a clean ratio of the two remaining quantities $\rho$ and $A$. ${\rho }_{\text{Cu}}\frac{L}{A_{\text{Cu}}}={\rho }_{\text{Al}}\frac{L}{A_{\text{Al}}}\Rightarrow \frac{{\rho }_{\text{Cu}}}{A_{\text{Cu}}}=\frac{{\rho }_{\text{Al}}}{A_{\text{Al}}}.$ WHY the algebra flips: cross-multiplying to solve for the area ratio puts the resistivities on top: $\frac{A_{\text{Al}}}{A_{\text{Cu}}}=\frac{{\rho }_{\text{Al}}}{{\rho }_{\text{Cu}}}=\frac{2.8\times 10^{-8}}{1.7\times 10^{-8}}\approx \mathrm{1.65.}$ Aluminium must be ~1.65× fatter to match copper's resistance — the more resistive material needs more lanes, which is the physical meaning of the ratio.

WHY rearrange for $L$: we already know the target $R$, the material $\rho$, and the wire's area $A$ — the only unknown is the length, so we isolate it. WHAT the algebra does: start from $R=\rho L/A$; multiply both sides by $A$ to clear the fraction, then divide by $\rho$ to leave $L$ alone: $L=\frac{RA}{\rho }=\frac{(100)(5\times 10^{-8})}{1.1\times 10^{-6}}=\frac{5\times 10^{-6}}{1.1\times 10^{-6}}\approx 4.55\text{m}.$ WHY it makes sense: a big target $R$ with a thin, high-$\rho$ wire needs plenty of length — that long, thin coil is exactly why heater elements look like tightly wound springs.

Sample 1: $R$ rises with $T$ (200 → 260 Ω). Fixed carrier count, more scattering → this is a metal / conductor (positive temperature coefficient $\alpha$). Sample 2: $R$ falls sharply with $T$ (5000 → 900 Ω). Only a semiconductor does this: $n\propto e^{-E_g/(2k_{B}T)}$ grows fast, $\sigma$ up, $\rho$ down. This is a semiconductor (negative temperature coefficient $\alpha$ — a thermistor).

WHY this order ($\sigma \to \rho \to R$): conductivity is the material quantity we can build straight from carriers; resistivity is just its inverse; and resistance only appears once we bring in the sample's shape. So we walk from microscopic to macroscopic.

Step 1 — conductivity ($\sigma =ne\mu$). Why this formula first: it turns the three microscopic numbers into one material property. $\sigma =(1\times 10^{20})(1.6\times 10^{-19})(0.14)=2.24\text{S/m}.$ Step 2 — resistivity ($\rho =1/\sigma$). Why invert: the geometry formula $R=\rho L/A$ needs $\rho$, not $\sigma$. $\rho =\frac{1}{\sigma }=\frac{1}{2.24}\approx 0.446\Omega \cdot \text{m}.$ Step 3 — resistance ($R=\rho L/A$). Why last: only now do we feed in the shape ($L$ and $A$). $R=\frac{(0.446)(0.01)}{1\times 10^{-6}}=4.46\times 10^3\Omega \approx 4.46\text{k}\Omega .$ Classify: $\rho \approx 0.45\Omega \cdot \text{m}$ sits in the semiconductor band → semiconductor (moderately doped).

Copper's conductivity is fixed — you cannot make part of a copper block behave differently from another part. Silicon's carrier density $n$ is tunable by doping (n-type vs p-type), and joining the two regions builds a p–n junction that lets current pass one way and blocks it the other. Millions of such switchable junctions form transistors and logic gates (Semiconductor Diodes and Transistors) — computation is controlled switching, and only a controllable material can switch.

$R=\rho L/A$ ::: shaped-object resistance (geometry matters) $\sigma =ne\mu$ ::: microscopic conductivity ($n$ separates the classes) $\rho =1/\sigma$ ::: resistivity is the inverse of conductivity $E_g$ ::: band-gap energy (eV) — the ladder height that sets $n$ $\alpha =\frac{1}{R}\frac{dR}{dT}$ ::: temperature coefficient of resistance (sign flags metal vs semiconductor) $n\propto e^{-E_g/(2k_{B}T)}$ ::: why semiconductor resistance falls as it heats