Thermal effects on conductivity

1.3.8 · Hardware › Materials & Atomic Structure

KYA ho raha hai (definitions)

KYUN metals aur semiconductors opposite behave karte hain

Master equation $\sigma =nq\mu$ mein do temperature-dependent parts hain: $n$ aur $\mu$.

• Metals: $n$ basically fixed hai — free electrons ka "sea" pehle se hi maujood hai aur temperature use barely change karti hai. Sirf mobility $\mu$ girta hai kyunki hotter lattice = zyada vibrations = zyada scattering. Result: $\sigma \downarrow$, $\rho \uparrow$.
• Semiconductors: $\mu$ bhi thoda girta hai, LEKIN $n$ exponentially badhta hai jab electrons band gap cross karte hain. $n$ ki bahut badi growth dominate karti hai. Result: $\sigma \uparrow$, $\rho \downarrow$.

KAISE metal law derive karte hain (scratch se)

Step 1 — Newton se Drift. Collisions ke beech field $E$ acceleration deta hai $a=qE/m$. Average time $\tau$ mein drift velocity gain hoti hai: $v_d=a\tau =\frac{qE\tau }{m}$ Ye step kyun? $\tau$ woh "memory" hai jo electron rakhta hai collision se pehle jo use randomize kar deta hai.

Step 2 — Mobility define karo. Mobility hai drift per field: $\mu =\frac{v_d}{E}=\frac{q\tau }{m}$ Ye step kyun? Ye material/temperature property ko applied field se alag kar deta hai.

Step 3 — $\sigma$ mein plug karo. $\boxed{\sigma =nq\mu =\frac{n{q}^2\tau }{m}}$ Ye step kyun? Ab temperature ki poori kahani sirf $\tau$ (aur $n$) mein hai.

Step 4 — $\tau$ $T$ pe kaise depend karta hai? Collision rate $\propto$ vibration amplitude squared, aur lattice thermal energy $\propto T$, toh scattering rate $1/\tau \propto T$. Isliye $\tau \propto 1/T$ aur ek metal ke liye (constant $n$): $\rho =\frac{m}{n{q}^2\tau }\propto T$ Isliye metal resistivity room temperature ke paas roughly $T$ mein linear hoti hai: $\rho (T)={\rho }_0[1+\alpha (T-T_0)]$

KAISE semiconductor law derive karte hain

Carrier creation ko energy $E_g$ (band gap) chahiye. Electrons ka wo fraction jinke paas itni thermal energy hai woh Boltzmann statistics follow karta hai, jo deta hai: $n\propto e^{-E_g/(2k_{B}T)}$ Factor 2 kyun? Har excited electron ek hole chhod jaata hai; pair-creation probability band gap energy ko do carriers mein split kar deti hai, isliye exponent mein $E_g/2$ aata hai.

Isliye: $\boxed{\sigma \propto e^{-E_g/(2k_{B}T)}}$ Jab $T\uparrow$, exponent ka magnitude shrink hota hai, $\sigma$ explosively badhta hai. Ye exponential mild $\mu$ decrease ko beat karta hai.

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Connections

• Free Electron Model of Metals
• Band Theory & Energy Gaps
• Intrinsic vs Extrinsic Semiconductors
• Mobility and Drift Velocity
• Superconductivity (τ → ∞ limit at low T)
• Doping and Carrier Concentration

Concept Map