Fermi-Dirac statistics — fermions, Fermi energy: Worked examples

2.4.17 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Fermi-Dirac statistics — fermions, Fermi energy

Yeh page ek drill hai. Hum parent topic lete hain aur har tarah ke question ko tackle karte hain jo yeh topic throw kar sakta hai: energy ka chemical potential ke relative har regime, do temperature limits, degenerate inputs, real-world numbers, aur ek exam-style twist. Pehle hum cases ka poora grid banate hain, phir har cell ke liye ek example karte hain.

Kuch bhi shuru karne se pehle, do cheezein jo hum baar baar use karte rahenge:

The scenario matrix

Neeche har cell ek alag cheez hai jo topic test kar sakta hai. Har numbered example ek (ya do) cell ko target karta hai.

Cell	Case class	Kya cheez alag banati hai	Example
A	$ε = μ$ exactly	exponent $0$ hai; answer $T$ -independent hai	Ex 1
B	$ε > μ$ (sea ke upar)	positive exponent; $f < \frac{1}{2}$ , decay karta hai	Ex 2
C	$ε < μ$ (sea ke neeche)	negative exponent; $f > \frac{1}{2}$	Ex 2
D	$T \to 0$ limit	step function; degenerate answer	Ex 3
A∩D	$ε = μ$ aur $T \to 0$ (the corner)	conventional $f = \frac{1}{2}$ by limit	Ex 1 (note)
E	High-energy tail $ε - μ ≫ k_{B} T$	Maxwell-Boltzmann mein reduce ho jaata hai	Ex 4
F	$f$ ki $μ$ ke baare mein symmetry (particle↔hole)	$f (μ + δ) + f (μ - δ) = 1$	Ex 5
G	Real-world density → $E_{F}$ , $T_{F}$ , $v_{F}$	real $n$ plug karo (sodium)	Ex 6
H	Degenerate/zero input: $n \to 0$ ya $n$ double	scaling law $E_{F} \propto n^{2/3}$	Ex 7
I	$f$ ko invert karke $T$ nikalna (exam twist)	unknown temperature ke liye solve karo	Ex 8
J	Thermal smear ki width $\approx 4.4 k_{B} T$	curve ka geometric read	Ex 9 (figure)

Example 1 — Cell A: Fermi level par bilkul sahi

Worked example Exactly

$ε = μ$ par occupation Ek quantum state energy par hai jo exactly chemical potential ke barabar hai, $T = 300$ K par. Woh kitne fraction time ke liye occupied hai? Kya answer badlega agar $T = 3000$ K ho?

Forecast: Padhne se pehle guess karo — kya zyada garam hone ka matlab zyada full, zyada khali, ya koi change nahi?

Step 1. Exponent $x = (ε - μ) / k_{B} T$ likho. Kyunki $ε = μ$ hai, numerator $0$ hai, isliye $x = 0$ kisi bhi $T > 0$ ke liye. Yeh step kyun? Saari temperature dependence usi exponent ke andar rehti hai. Agar numerator zero hai, to $T$ matter nahi kar sakta.

Step 2. Plug in karo: $f = \frac{1}{e ^{0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$ . Yeh step kyun? $e^{0} = 1$ exactly; yahi $μ$ ka definitional anchor hai.

Verify: $\frac{1}{2}$ dimensionless hai aur $[0, 1]$ mein hai ✓. Yeh $T$ ke saath badla nahi — isliye sahi forecast tha "koi change nahi". Exactly isi tarah experimenters $μ$ locate karte hain: woh energy dhundho jo half-filled rehti hai.

Example 2 — Cells B & C: sea ke upar aur neeche

Worked example

$μ$ ke aas paas do states $T = 300$ K par do states lo: ek $μ$ se $0.05$ eV upar, ek $μ$ se $0.05$ eV neeche. Dono ke liye $f$ nikalo.

Forecast: Ek $\frac{1}{2}$ se zyada hoga, ek kam. Kaun sa kaun sa hai, aur kya woh related hain?

Step 0 (thermal scale). Upar wale box se, $T = 300$ K par, $k_{B} T = (8.617 \times 1 0^{- 5}) (300) = 0.02585$ eV. Yeh step kyun? Hum eV offset ko is eV energy se divide karenge, isliye dono eV mein hone chahiye — exactly yahi $k_{B}$ ka eV/K roop hamare liye karta hai.

Step 1 (Cell B, upar). $x_{+} = \frac{+ 0.05}{0.02585} = 1.934$ . Phir $f_{+} = \frac{1}{e ^{1.934} + 1} = \frac{1}{6.92 + 1} = 0.1263$ . Yeh step kyun? $μ$ ke upar numerator positive hai, isliye $x > 0$ , isliye $e^{x} > 1$ , isliye $f < \frac{1}{2}$ : sea ke upar ke states mostly empty hain.

Step 2 (Cell C, neeche). $x_{-} = \frac{- 0.05}{0.02585} = - 1.934$ . Phir $f_{-} = \frac{1}{e ^{- 1.934} + 1} = \frac{1}{0.1445 + 1} = 0.8737$ . Yeh step kyun? $μ$ ke neeche numerator negative hai, isliye $e^{x} < 1$ , isliye $f > \frac{1}{2}$ : sea ke neeche ke states mostly full hain.

Verify: $f_{+} + f_{-} = 0.1263 + 0.8737 = 1.000$ exactly. Yeh particle–hole symmetry (Cell F) chhup ke aa gayi: $μ$ se $δ$ distance upar ek khali slot utni hi likely hai jitni $δ$ neeche ek filled slot. Dono values $[0, 1]$ mein hain ✓.

Example 3 — Cell D: $T = 0$ step

Worked example Thanda metal, sharp edge

$T = 0$ par, $E_{F}$ se $0.5$ eV neeche ek state ke liye aur $E_{F}$ se $0.5$ eV upar ek state ke liye $f$ evaluate karo.

Forecast: Absolute zero par fuzzy edge gayab ho jaani chahiye — do numbers guess karo.

Step 1 (neeche). Jab $T \to 0$ , $ε < E_{F}$ ke liye exponent $(ε - E_{F}) / k_{B} T \to \frac{negative}{0 ^{+}} \to - \infty$ , isliye $e^{- \infty} = 0$ aur $f = \frac{1}{0 + 1} = 1$ . Yeh step kyun? Fermi energy ke neeche har state filled hoti hai — exclusion principle electrons ko ground state mein pile hone ke bajaye stack karne par majboor karta hai.

Step 2 (upar). $ε > E_{F}$ ke liye exponent $\to + \infty$ , isliye $e^{+ \infty} = \infty$ aur $f = \frac{1}{\infty + 1} = 0$ . Yeh step kyun? Sea ke upar zero temperature par un slots ko fill karne ke liye kuch nahi hai.

Verify: Humne step function recover kar liya: $f = 1$ neeche, $f = 0$ upar, $E_{F}$ par jump ke saath. Ex 1 ke half-filling limit ke saath consistent hai jo dono sides se approach ki gayi ✓ (aur exactly $E_{F}$ par value $\frac{1}{2}$ hai, A∩D corner).

Example 4 — Cell E: classical (Maxwell–Boltzmann) tail

Worked example Jab Fermi-Dirac classical ban jaata hai

Ek state $μ$ se $0.30$ eV upar hai $T = 300$ K par. Dikhao ki $f$ ko Maxwell-Boltzmann form $e^{- (ε - μ) / k_{B} T}$ se achhi tarah approximate kiya ja sakta hai, aur relative error nikalo.

Forecast: Tail par bahut door, kya " $+ 1$ " abhi bhi matter karta hai?

Step 1 — algebraic reduction. $f$ ke numerator aur denominator dono ko $e^{- x}$ se multiply karo (jahan $x = (ε - μ) / k_{B} T$ ): $f = \frac{1}{e ^{x} + 1} = \frac{e ^{- x}}{1 + e ^{- x}} .$ Ab jab $x ≫ 1$ , to denominator mein $e^{- x}$ term tiny hai, isliye $1 + e^{- x} \to 1$ aur isliye $f ≃ \frac{e ^{- x}}{1} = e^{- x} = e^{- (ε - μ) / k_{B} T} .$ Yeh step kyun? Yeh Fermi-Dirac se Maxwell-Boltzmann tak ka exact algebraic bridge hai — approximation hai "denominator mein $e^{- x}$ ko $1$ ke against drop karo", jo precisely isliye valid hai kyunki high-energy tail mein $e^{- x}$ chhota hota hai.

Step 2 — numbers. $k_{B} T = 0.02585$ eV ke saath (same $T = 300$ K pehle jaisa, $k_{B}$ ka eV/K roop): $x = \frac{0.30}{0.02585} = 11.60$ . Phir $f_{MB} = e^{- x} = e^{- 11.60} = 9.14 \times 1 0^{- 6}$ , jabki exact $f = \frac{1}{e ^{11.60} + 1} = 9.14 \times 1 0^{- 6}$ (paanch figures tak agree karte hain). Yeh step kyun? Isliye classical statistics dilute/high-energy states ke liye kaam karta hai — yeh Fermi-Dirac ka $ε - μ ≫ k_{B} T$ limit hai.

Verify: Relative error $= \frac{f _{MB} - f}{f} = e^{- x} \approx 9 \times 1 0^{- 6}$ , yaani lagbhag $0.0009%$ — bilkul negligible ✓.

Example 5 — Cell F: particle–hole symmetry, general taur par

Worked example Mirror prove karo

Dikhao ki kisi bhi $δ$ aur kisi bhi $T$ ke liye, $f (μ + δ) + f (μ - δ) = 1$ .

Forecast: Ex 2 ne numerically hint diya tha — kya yeh sabhi $δ$ ke liye exact hai?

Step 1. $y = e^{δ / k_{B} T}$ lo. Phir $f (μ + δ) = \frac{1}{y + 1}$ aur $f (μ - δ) = \frac{1}{y ^{- 1} + 1}$ . Yeh step kyun? Dono ko same base $y$ par rakhne se unhe cleanly add karna aasaan ho jaata hai.

Step 2. $f (μ - δ) = \frac{1}{\frac{1}{y} + 1} = \frac{y}{1 + y}$ . Add karo: $\frac{1}{y + 1} + \frac{y}{y + 1} = \frac{1 + y}{1 + y} = 1$ . Yeh step kyun? Dono fractions simplify hone ke baad ek common denominator share karte hain, aur unke numerators usse sum karte hain.

Verify: Symbolically identically $1$ hai (neeche check kiya), aur yeh Ex 2 ka $0.1263 + 0.8737 = 1$ reproduce karta hai ✓. Physically: probability ki ek state $μ + δ$ par khali hai, probability ke barabar hai ki woh $μ - δ$ par filled hai — electrons upar aur holes neeche ek doosre ko mirror karte hain.

Example 6 — Cell G: real metal (sodium)

Worked example Sodium ki Fermi energy, temperature, velocity

Sodium mein $n = 2.65 \times 1 0^{28} m^{- 3}$ free electrons hain. $E_{F}$ (eV mein), $T_{F}$ , aur $v_{F}$ nikalo.

Forecast: Copper (parent note) ne $n = 8.5 \times 1 0^{28}$ par $\sim 7$ eV diya. Sodium roughly $3 \times$ kam dense hai — guess karo ki $E_{F}$ bada hoga ya chhota.

Step 1 — $k_{F}$ banao. $k_{F} = (3 π^{2} n)^{1/3} = (3 \cdot 9.8696 \cdot 2.65 \times 1 0^{28})^{1/3} = (7.846 \times 1 0^{29})^{1/3} = 9.225 \times 1 0^{9} m^{- 1}$ . Yeh step kyun? $E_{F}$ Fermi wavevector $k_{F}$ ke through sabse clean hai; cube root sphere volume $\frac{4}{3} π k_{F}^{3}$ ko undo karta hai. $k_{F}$ mein full precision rakho — baad mein square karne par rounding magnify ho jaati hai.

Step 2 — joules mein energy. $E_{F} = \frac{ℏ ^{2} k _{F}^{2}}{2 m} = \frac{( 1.055 \times 1 0 ^{- 34} ) ^{2} ( 9.225 \times 1 0 ^{9} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} = 5.20 \times 1 0^{- 19} J$ . Yeh step kyun? $ε = ℏ^{2} k^{2} /2 m$ free-particle energy hai; $k = k_{F}$ par evaluate karne se sea ki top milti hai. ( $k_{F}$ ko sirf 3 figures mein round karne se yeh kuch $%$ shift ho jaata, isliye humne fuller value carry ki.)

Step 3 — eV mein convert karo. Conversion factor $1 eV = 1.602 \times 1 0^{- 19} J$ se divide karo: $E_{F} = \frac{5.20 \times 1 0 ^{- 19} J}{1.602 \times 1 0 ^{- 19} J/eV} = 3.24 eV .$ Yeh step kyun? Joules SI energy hai; eV electrons ke liye natural scale hai — conversion factor hi number ko human-readable banata hai.

Step 4 — temperature & velocity. $T_{F} = E_{F} / k_{B} = \frac{5.20 \times 1 0 ^{- 19} J}{1.381 \times 1 0 ^{- 23} J/K} = 3.77 \times 1 0^{4} K$ . $v_{F} = ℏ k_{F} / m = \frac{( 1.055 \times 1 0 ^{- 34} ) ( 9.225 \times 1 0 ^{9} )}{9.11 \times 1 0 ^{- 31}} = 1.068 \times 1 0^{6} m/s$ . Yeh step kyun? $T_{F}$ batata hai gas kitni degenerate hai ( $T_{F} ≫ 300$ K ⇒ deeply degenerate); $v_{F}$ sea surface par electrons ki speed hai. Note karo ki hum yahan $E_{F}$ joules mein use karte hain taaki $k_{B}$ ke joules cancel ho jayein.

Verify: Units: $J / (J/K) = K$ ✓; $v_{F}$ light speed ka $\sim 0.4%$ hai (non-relativistic, isliye free-electron formula valid hai) ✓. Copper se kam density ⇒ kam $E_{F}$ (3.24 < 7 eV), $E_{F} \propto n^{2/3}$ scaling se match karta hai — sahi forecast. (Standard tables sodium $E_{F} \approx 3.24$ eV quote karte hain.)

Example 7 — Cell H: degenerate inputs (density double karo, box khali karo)

Worked example Scaling aur edge cases

(a) Agar electron density $n \to 2 n$ kar do fixed volume par, to $E_{F}$ kitne factor se change hoga? (b) $n \to 0$ limit mein $E_{F}$ kya hai?

Forecast: $E_{F} \propto n^{2/3}$ . $n$ double karna — factor $2$ se bada hoga ya chhota?

Step 1 (a). $\frac{E _{F} ( 2 n )}{E _{F} ( n )} = (\frac{2 n}{n})^{2/3} = 2^{2/3} = 1.587$ . Yeh step kyun? Sirf density $E_{F}$ mein enter karti hai, aur woh $\frac{2}{3}$ power se karta hai; constants ratio mein cancel ho jaate hain.

Step 2 (b). Jab $n \to 0$ , $E_{F} = const \cdot n^{2/3} \to 0$ . Yeh step kyun? Ek khale box mein koi Fermi sea nahi hoti — ek non-existent stack ki top zero energy par hoti hai. Yahi degenerate/zero-input check hai jo matrix ne demand kiya tha.

Verify: $2^{2/3} = 1.587 < 2$ — doubling se chhota, kyunki exponent $1$ se kam hai (sahi forecast). Aur $n \to 0 \Rightarrow E_{F} \to 0$ finite aur sensible hai ✓.

Example 8 — Cell I: exam twist, temperature ke liye solve karo

Worked example Occupation diya hua hai,

$T$ nikalo $μ$ se $0.10$ eV upar ek state mein occupation $f = 0.20$ observe kiya gaya. System kis temperature par hai?

Forecast: Zyada thermal smearing (higher $T$ ) ek fixed offset par $f$ ko $\frac{1}{2}$ ke paas push karta hai. Kyunki $0.20 < \frac{1}{2}$ , guess karo: kya $T$ chhota hai ya bada?

Step 1 — $f$ invert karo. $f = \frac{1}{e ^{x} + 1}$ se: $e^{x} = \frac{1}{f} - 1 = \frac{1}{0.20} - 1 = 4$ , isliye $x = ln 4 = 1.386$ . Yeh step kyun? Hum output $f$ jaante hain aur input $x$ chahiye; algebraically $e^{x}$ isolate karna phir $ln$ lena distribution ka inverse hai.

Step 2 — $T$ ke liye solve karo. $x = \frac{ε - μ}{k _{B} T} \Rightarrow T = \frac{ε - μ}{k _{B} x} = \frac{0.10 eV}{( 8.617 \times 1 0 ^{- 5} eV/K ) ( 1.386 )} = 837 K$ . Yeh step kyun? Hum $k_{B}$ eV/K mein use karte hain (woh roop jo humne top par banaya, $8.617 \times 1 0^{- 5}$ ) taaki numerator ka eV cancel ho jaaye, kelvin bachta hai.

Verify: Plug back karo: $x = (0.10) / (8.617 \times 1 0^{- 5} \cdot 837) = 1.386$ , aur $f = 1/ (e^{1.386} + 1) = 1/ (4 + 1) = 0.20$ ✓. Yeh $T$ , $T_{F}$ se kaafi kam hai, $μ$ ke upar mostly-empty state ke saath consistent hai.

Example 9 — Cell J: thermal smear ki geometry (figure)

Worked example Fuzzy edge kitni wide hai?

$T = 1000$ K par, kis energy width mein $f$ $0.90$ se $0.10$ tak girta hai? Curve se read karo.

Forecast: Parent note claim karta hai $\approx 4.4 k_{B} T$ . Pehle eV mein width guess karo.

Figure kaise padhein: moti blue S-shaped curve $f$ ko scaled energy $x = (ε - μ) / k_{B} T$ ke against plot karti hai. Vertical dashed gray line $ε = μ$ ( $x = 0$ ) mark karta hai, jahan gray dot half-filled value $f = 0.5$ par baitha hai. Curve ke shoulders par do round orange markers $f = 0.90$ (upper-left) aur $f = 0.10$ (lower-right) par hain; yeh woh do occupation levels hain jo question poochh raha hai. Green horizontal double-headed arrow apne shaded band ke saath un do markers ke beech horizontal distance span karta hai — woh span hi woh width hai jo hum chahte hain, $k_{B} T$ ki units mein maapi gayi.

Step 1 — do edge exponents. $f = 0.90 \Rightarrow e^{x} = \frac{1}{0.9} - 1 = 0.111 \Rightarrow x = ln (0.111) = - 2.197$ . $f = 0.10 \Rightarrow e^{x} = 9 \Rightarrow x = + 2.197$ . Yeh step kyun? Yeh $x$ values exactly figure mein do orange shoulder markers ki horizontal positions hain. $x$ mein unka span hai $2.197 - (- 2.197) = 4.394$ — green band ki length.

Step 2 — energy mein convert karo. Scaled units mein width hai $Δ x = 4.394$ , isliye $Δ ε = 4.394 k_{B} T$ . $k_{B} T = (8.617 \times 1 0^{- 5} eV/K) (1000 K) = 0.0862$ eV ke saath: $Δ ε = 4.394 \times 0.0862 = 0.379$ eV. Yeh step kyun? $x$ ko $k_{B} T$ ki units mein measure kiya jaata hai, isliye $x$ -span (band ki length) ko $k_{B} T$ se multiply karne par "shoulder width" dimensionless se eV mein convert ho jaati hai. Yahi woh quantitative read-off hai jo figure deliver karne ke liye set up tha.

Verify: $4.394 \approx 4.4$ , parent note ke $4.4 k_{B} T$ claim se match karta hai ✓. Numerically $Δ ε = 0.379$ eV $1000$ K par. $E_{F} \sim 3$ – $7$ eV wale metal mein yeh $0.38$ eV smear top ke paas ek patli sliver hai — sirf us band ke electrons thermally/electrically active hain. Units: (dimensionless)×eV = eV ✓, cell J close.

Recall Quick self-test

$ε - μ = - k_{B} T$ par ek state ki occupation kya hai? ::: $f = 1/ (e^{- 1} + 1) = 1/1.368 = 0.731$ Agar $f = 0.5$ aur $ε > 0$ , to $μ$ kahan hai? ::: exactly $ε$ par (Cell A) $n$ double karne par $E_{F}$ kitne se multiply hota hai? ::: $2^{2/3} \approx 1.587$ (Cell H) $f$ ka high-energy tail kaun sa classical law ban jaata hai? ::: Maxwell-Boltzmann $e^{- (ε - μ) / k_{B} T}$ (Cell E)

Yeh bhi dekho: Density of States ( $f$ ko particle numbers mein convert karta hai), Sommerfeld Expansion ( $μ$ finite $T$ par $E_{F}$ se kaise drift karta hai), aur Electron Degeneracy Pressure & White Dwarfs (filled sea kya thaame hue hai).