Blackbody radiation from statistical mechanics — Planck distribution: Worked examples

2.4.19 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Blackbody radiation from statistical mechanics — Planck dist

Yeh page ek drill floor hai. Parent note ne machine banayi thi: $u (ν, T) d ν = \frac{8 π h ν ^{3}}{c ^{3}} \frac{1}{e ^{h ν / k_{B} T} - 1} d ν .$ Yahan hum isse har tarah ke input ke against test karte hain jo ek problem de sakti hai — small $ν$ , large $ν$ , exact peak, total integral, real starlight, aur kuch exam traps. Har worked example se pehle tumhe khud forecast karna hai ki answer kya hoga.

Scenario matrix

Har Planck-distribution problem in cells mein se kisi ek mein hoti hai. Neeche ke examples mein cell(s) ka label laga hai, aur milke yeh saari cells cover karte hain.

Cell	Input regime	Answer kya control karta hai	Example
A low- $ν$ limit	$x ≪ 1$	$⟨ ϵ ⟩ \to k_{B} T$ (Rayleigh–Jeans)	Ex 1
B high- $ν$ limit	$x ≫ 1$	exponential cutoff $e^{- x}$ (Wien)	Ex 2
C mid-range exact	$x \sim 1$	full formula, koi approximation nahi	Ex 3
D degenerate $ν \to 0$	$ν = 0$	mode mein energy nahi; $0/0$ check karo	Ex 3b (Ex 3 ke andar)
E peak (extremum)	$d u / d ν = 0$	transcendental $3 (1 - e^{- x}) = x$	Ex 4
F $ν$ -peak vs $λ$ -peak	Jacobian $c / λ^{2}$	alag transcendental, $x = 4.965$	Ex 5
G total (sab $ν$ integrate karo)	$\int_{0}^{\infty}$	Stefan–Boltzmann $T^{4}$	Ex 6
H real-world word problem	units ke saath numbers	sahi sub-law chuno	Ex 7 (the Sun)
I exam twist / ratio	constants cleverly cancel karo	dimensionless ratios	Ex 8

Hum puri tarah Density of states, Partition function, Rayleigh-Jeans law, Wien's law aur Stefan-Boltzmann law par rely karte hain.

Worked examples

Worked example Ex 1 — Cell A (low-frequency limit)

Ek FM radio wave hai jisme $ν = 1.0 \times 1 0^{8} Hz$ (100 MHz) hai. Cavity room temperature $T = 300 K$ par hai. Average energy per mode $⟨ ϵ ⟩$ kya hai, aur yeh classical $k_{B} T$ ke kitni kareeb hai?

Forecast: radio bahut kam frequency hai, isliye $x$ bahut chhota hoga aur hum lagbhag bilkul classical $k_{B} T$ par land karenge. Andaaza: essentially $k_{B} T$ kaafi digits tak.

$x = h ν / k_{B} T$ compute karo. Yeh step kyun? $x$ decide karta hai hum kis regime mein hain — hamesha pehle. $x = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 1.0 \times 1 0 ^{8} )}{( 1.381 \times 1 0 ^{- 23} ) ( 300 )} = 1.60 \times 1 0^{- 5} .$ Sach mein $x ≪ 1$ : Cell A confirmed.
Exact formula $⟨ ϵ ⟩ = \frac{h ν}{e ^{x} - 1}$ use karo (upar define ki gayi hai). Kyun? Hum dekhna chahte hain $k_{B} T$ ke kitna kareeb pohanchte hain. Numerically $h ν = 6.626 \times 1 0^{- 26} J$ aur $e^{x} - 1 \approx x = 1.60 \times 1 0^{- 5}$ , isliye $⟨ ϵ ⟩ \approx \frac{6.626 \times 1 0 ^{- 26}}{1.60 \times 1 0 ^{- 5}} = 4.14 \times 1 0^{- 21} J .$
$k_{B} T$ se compare karo. Kyun? Yahi classical prediction hai. $k_{B} T = (1.381 \times 1 0^{- 23}) (300) = 4.14 \times 1 0^{- 21} J .$ Yeh ~5 significant figures tak match karte hain.

Verify: fractional gap lagbhag $x /2 \approx 8 \times 1 0^{- 6}$ hai ( $e^{x} - 1 \approx x + x^{2} /2$ se). Units: J. Forecast confirmed — radio frequencies par equipartition essentially exact hai, aur yahi reason hai ki kisi ne long-wavelength measurements se catastrophe notice nahi ki.

Worked example Ex 2 — Cell B (high-frequency limit)

$T = 3000 K$ par ek cavity (ek glowing filament). Visible-blue frequency $ν = 7.0 \times 1 0^{14} Hz$ par energy density $u (ν)$ ko Rayleigh–Jeans ki (galat) prediction se compare karo. Sachchi Planck value Rayleigh–Jeans se kitne factor se kam hai?

Forecast: 3000 K par blue light "expensive" side par kaafi upar honi chahiye. Rayleigh–Jeans mein koi cutoff nahi hai, isliye woh overshoot karega. Andaaza: true value ek bade factor $\sim e^{- x} \cdot x$ se chhoti hogi.

Shuru karne se pehle, comparison object ka naam rakho. $u_{RJ}$ se denote karo Rayleigh-Jeans law dwara predict ki gayi energy density, yaani classical spectrum $u_{RJ} (ν) = \frac{8 π ν ^{2}}{c ^{3}} k_{B} T$ jo har mode ko poora $k_{B} T$ deta hai. Aur $u_{Planck}$ se sachchi Planck spectrum denote karo. Hum in dono ko compare karte hain.

$x$ compute karo. Kyun? Regime set karta hai. $x = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 7.0 \times 1 0 ^{14} )}{( 1.381 \times 1 0 ^{- 23} ) ( 3000 )} = 11.2.$ $x ≫ 1$ : Cell B.
Dono energies-per-mode likho. Kyun? Density of states $g (ν) = 8 π ν^{2} / c^{3}$ (parent note mein derive ki gayi — yeh unit volume aur frequency par modes ki count hai) Planck aur Rayleigh–Jeans dono spectra mein same hai, isliye ratio lete waqt yeh cancel ho jaata hai aur sirf $⟨ ϵ ⟩$ ka ratio bachta hai. $\frac{u _{Planck}}{u _{RJ}} = \frac{⟨ ϵ ⟩ _{Planck}}{k _{B} T} = \frac{h ν / ( e ^{x} - 1 )}{k _{B} T} = \frac{x}{e ^{x} - 1} .$ $h ν / k_{B} T$ kyon $x$ ban jaata hai? Kyunki definition se woh ratio hi $x$ hai.
$x = 11.2$ plug in karo. Kyun? Number nikalo. $\frac{x}{e ^{x} - 1} = \frac{11.2}{e ^{11.2} - 1} = \frac{11.2}{7.32 \times 1 0 ^{4}} = 1.53 \times 1 0^{- 4} .$

Verify: Rayleigh–Jeans $\sim 6500 \times$ over-predict kar raha hai. Exponential $e^{- x}$ ka $x$ prefactor ko crush karna ultraviolet catastrophe ka haarna hai — bilkul wahi mechanism jo parent note ne describe ki thi. Units: dimensionless ratio, achha.

Worked example Ex 3 — Cell C & D (mid-range exact, plus

$ν \to 0$ degenerate check) (C) $⟨ ϵ ⟩$ ko exactly evaluate karo jab $x = 1$ ho (yaani $h ν = k_{B} T$ ). (D) Phir dekho $⟨ ϵ ⟩$ ka kya hota hai jab $ν \to 0$ — formula naively $0/0$ hai; ise resolve karo.

Forecast (C): $x = 1$ crossover hai. Lump ki kimat lagbhag ek thermal kick ke barabar hai, isliye hum $k_{B} T$ se kuch kam expect karte hain — mode partly frozen hai. Andaaza: lagbhag $0.6 k_{B} T$ .

(C) Exact value. Exact formula kyun? $x = 1$ par koi bhi limit apply nahi hoti. $⟨ ϵ ⟩ = \frac{h ν}{e ^{1} - 1} = \frac{k _{B} T \cdot 1}{e - 1} = \frac{k _{B} T}{1.71828} = 0.5820 k_{B} T .$ Toh crossover par ek mode classical share ka sirf ~58% rakhta hai.
(D) $ν \to 0$ limit. Iska khayal kyun rakhein? Radio, microwave, DC — sab $ν = 0$ ke paas rehte hain; formula ko theek behave karna chahiye. $⟨ ϵ ⟩ = \frac{h ν}{e ^{h ν / k_{B} T} - 1}$ likho. Jab $ν \to 0$ toh numerator aur denominator dono $\to 0$ : ek sachcha $0/0$ .
Taylor series $e^{x} - 1 = x + \frac{x ^{2}}{2} + \dots$ se resolve karo. Taylor kyun, L'Hôpital nahi? Yahan same baat hai, lekin series agla correction bhi dikhati hai: $⟨ ϵ ⟩ = \frac{k _{B} T x}{x + \frac{x ^{2}}{2} + \dots} = k_{B} T (1 - \frac{x}{2} + \dots) x \to 0 k_{B} T .$

Verify: degenerate input $ν = 0$ ek finite, sensible $⟨ ϵ ⟩ \to k_{B} T$ deta hai — koi blow-up nahi, koi divide-by-zero disaster nahi. Number $1/ (e - 1) = 0.5820$ directly check kiya ja sakta hai (VERIFY dekho). Yeh crossover behaviour neeche graph kiya gaya hai.

Figure (s01) — Average energy per mode $⟨ ϵ ⟩ / k_{B} T$ versus $x = h ν / k_{B} T$ . Chalk-blue curve exact factor

x / (e^{x} - 1)

hai; dashed pale-yellow line classical ceiling

k_{B} T

hai. Left par (

x \to 0

, Cell A) curve

k_{B} T

se chipki rehti hai — har low-frequency mode poori tarah "on" hai. Pink dot par (

x = 1

, is example ka Cell C) yeh

0.582 k_{B} T

tak aa chuki hai: mode partly frozen hai. Right par (

x ≫ 1

, Cell B) yeh zero ki taraf collapse karti hai — high-frequency modes frozen out hain. Yeh ek picture hi poori kahani hai ki ultraviolet catastrophe kyun kabhi nahi hoti: ise left-to-right padho jaise "modes switch off ho rahe hain jaise woh mehange hote jaate hain".

Us chalk-blue curve ko dekho: yeh $k_{B} T$ par flat shuru hoti hai (Cell A), marked $x = 1$ par $0.582 k_{B} T$ se guzarti hai (Cell C), aur daayein zero ki taraf dive karti hai (Cell B). Ek picture, teen cells.

Worked example Ex 4 — Cell E (frequency peak, ek extremum problem)

$T = 5800 K$ par (roughly Sun ki surface) $u (ν, T)$ jo frequency $ν_{m a x}$ par sabse zyada hai, woh dhundho.

Forecast: $ν^{3} / (e^{x} - 1)$ ka derivative zero karne se ek fixed dimensionless $x$ milega; parent note ne $x \approx 2.82$ quote kiya tha. Toh $ν_{m a x} \propto T$ . Andaaza: kuch $\times 1 0^{14}$ Hz (visible).

Extremum set up karo. Differentiate kyun? Maximum wahan hota hai jahan slope zero ho. $u \propto \frac{ν ^{3}}{e ^{x} - 1}$ likho jahan $x = h ν / k_{B} T$ hai, toh fixed $T$ par $ν \propto x$ aur hum $f (x) = \frac{x ^{3}}{e ^{x} - 1}$ maximize karte hain.
$f$ differentiate karo aur zero karo. $x$ mein kyun? Isse saare constants remove ho jaate hain — peak ki $x$ mein location universal hai. $f^{'} (x) = \frac{3 x ^{2} ( e ^{x} - 1 ) - x ^{3} e ^{x}}{( e ^{x} - 1 ) ^{2}} = 0 \Rightarrow 3 (e^{x} - 1) = x e^{x} \Rightarrow 3 (1 - e^{- x}) = x .$
Transcendental equation numerically solve karo. Numerically kyun? Yeh ek polynomial aur ek exponential ko mix karta hai — koi closed form nahi. Iterate karne par $x = 2.8214$ milta hai.
$x$ ko $ν_{m a x}$ mein convert karo. Kyun? Hum ek physical frequency chahte hain. $ν_{m a x} = \frac{x k _{B} T}{h} = \frac{2.8214 ( 1.381 \times 1 0 ^{- 23} ) ( 5800 )}{6.626 \times 1 0 ^{- 34}} = 3.41 \times 1 0^{14} Hz .$

Verify: $ν_{m a x} \propto T$ (Wien) ✓. Woh frequency $λ = c / ν \approx 880 nm$ (near-infrared) se correspond karti hai. Note karo yeh Sun ki aam quoted 500 nm peak nahi hai — kyunki woh quote $λ$ -peak hai. Yahi mismatch bilkul Cell F hai, agle mein.

Worked example Ex 5 — Cell F (

$λ$ -peak vs $ν$ -peak trap) Usi Sun ke liye ( $T = 5800 K$ ), woh wavelength $λ_{m a x}$ dhundho jahan $u (λ)$ peak karta hai, aur confirm karo ki yeh $c / ν_{m a x}$ se alag hai.

Forecast: parent note ne warn kiya tha ki Jacobian ki wajah se dono peaks alag hote hain. Ek alag transcendental root expect karo, $x \approx 4.97$ , jisse $λ_{m a x} \approx 500 nm$ mile (green — woh famous number).

Jacobian ke saath variables change karo. Kyun? Ek density transform hoti hai taaki total energy unchanged rahe: $u (ν) d ν = u (λ) d λ$ , isliye $u (λ) = u (ν) \frac{d ν}{d λ} = u (ν) \frac{c}{λ ^{2}}$ . Woh extra $λ^{- 2}$ (yaani $ν^{2}$ ) peak ko shift karta hai — yahi poora trap hai.
$u (λ) \propto \frac{λ ^{- 5}}{e ^{y} - 1}$ likho, jahan $y = \frac{h c}{λ k _{B} T}$ hai. $λ^{- 5}$ kyun? Kyunki $ν^{3} \cdot c / λ^{2} = λ^{- 3} λ^{- 2}$ constants tak. (Yahan $y$ sirf $x$ hai wavelength ke liye rewritten: kyunki $ν = c / λ$ hai, $y = h ν / k_{B} T = h c / (λ k_{B} T)$ milta hai — note karo poora product $λ k_{B} T$ denominator mein hai.)
Maximize karo: Ex 4 jaisi hi algebra lekin power 3 ki jagah 5: $5 (1 - e^{- y}) = y \Rightarrow y = 4.9651.$ 5 kyun, 3 nahi? Jacobian ne $λ^{- 1}$ ki do powers add ki.
$λ_{m a x}$ solve karo. Kyun? Hum woh physical wavelength chahte hain jo ek spectrometer report karega, $c / ν_{m a x}$ se compare karne ke liye. $λ_{m a x} = \frac{h c}{y k _{B} T} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{4.9651 ( 1.381 \times 1 0 ^{- 23} ) ( 5800 )} = 499 nm .$

Verify: $c / ν_{m a x} = (3.0 \times 1 0^{8}) / (3.41 \times 1 0^{14}) = 880 nm \neq = 499 nm$ . Dono peaks genuinely alag hain — same spectrum, alag variable. Yeh wavelength form mein Wien's displacement law hai: $λ_{m a x} T = (h c) / (4.9651 k_{B}) = 2.90 \times 1 0^{- 3} m\cdotK$ ✓ (textbook Wien constant).

Worked example Ex 6 — Cell G (saare

$ν$ integrate karo: Stefan–Boltzmann) Total energy density $U (T)$ derive karo aur $T = 5800 K$ ke liye Stefan–Boltzmann-style prefactor numerically evaluate karo.

Forecast: $ν^{3} / (e^{x} - 1)$ ko saari frequencies par integrate karne se $\propto T^{4}$ milna chahiye (dimensional argument: sirf $k_{B} T / h$ scale set kar sakta hai). Andaaza: $U \sim$ kuch $J/m^{3}$ .

Integral set up karo. Kyun? "Total" = har mode par sum. $U = \int_{0}^{\infty} \frac{8 π h ν ^{3}}{c ^{3}} \frac{d ν}{e ^{h ν / k_{B} T} - 1} .$
$x = h ν / k_{B} T$ substitute karo toh $ν = (k_{B} T / h) x$ , $d ν = (k_{B} T / h) d x$ . Kyun? Isse saari $T$ -dependence aage aa jaati hai aur bacha hua integral ek pure number ban jaata hai. $U = \frac{8 π h}{c ^{3}} (\frac{k _{B} T}{h})^{4} \int_{0}^{\infty} \frac{x ^{3} d x}{e ^{x} - 1} .$
Standard integral use karo $\int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x = \frac{π ^{4}}{15}$ . Yeh value kyun? Yeh $Γ (4) ζ (4) = 6 \cdot \frac{π ^{4}}{90} = \frac{π ^{4}}{15}$ hai. $U = \frac{8 π ^{5} k _{B}^{4}}{15 c ^{3} h ^{3}} T^{4} .$
$T = 5800$ K par numbers plug karo. Kyun? Physical density nikalo. Coefficient $a = \frac{8 π ^{5} k _{B}^{4}}{15 c ^{3} h ^{3}} = 7.566 \times 1 0^{- 16} J\cdotm^{- 3} K^{- 4}$ , isliye $U = 7.566 \times 1 0^{- 16} (5800)^{4} = 8.57 \times 1 0^{- 1} J/m^{3} .$

Verify: $\int_{0}^{\infty} x^{3} / (e^{x} - 1) d x = π^{4} /15 = 6.4939$ ✓. Units: $[J\cdotm^{- 3} K^{- 4}] \cdot K^{4} = J/m^{3}$ ✓. Related surface flux $σ T^{4}$ hai jahan $σ = a c /4 = 5.67 \times 1 0^{- 8}$ — Stefan constant, Cell G confirm karta hai.

Worked example Ex 7 — Cell H (real-world word problem: Sun ka color)

Ek telescope ek door tare ka spectrum measure karta hai jo (wavelength mein) $λ_{m a x} = 290 nm$ (ultraviolet) par peak karta hai. Uski surface temperature kya hai, aur Sun (5800 K) se kaise compare karta hai?

Forecast: chhoti peak wavelength ⇒ zyada garam (Wien). Sun ki peak wavelength ke aadhe par, roughly double temperature expect karo, ~10000 K, ek blue-white star.

Sahi law chuno. Wien-displacement kyun? Humein ek peak diya gaya hai, total flux nahi, isliye Ex 5 se $λ_{m a x} T = b$ use karo jahan $b = 2.898 \times 1 0^{- 3} m\cdotK$ hai.
$T$ solve karo. Rearrange kyun? $T$ unknown hai. $T = \frac{b}{λ _{m a x}} = \frac{2.898 \times 1 0 ^{- 3}}{290 \times 1 0 ^{- 9}} = 9993 K \approx 1.0 \times 1 0^{4} K .$
Sun se ratio. Kyun? Question comparison maang raha hai. $\frac{T _{star}}{T _{⊙}} = \frac{9993}{5800} = 1.72.$
Brightness interpret karo. Kyun? Total output $T^{4}$ se scale karta hai (Stefan–Boltzmann), isliye ek garam surface per unit area disproportionately zyada radiate karta hai. $(\frac{T _{star}}{T _{⊙}})^{4} = 1.7 2^{4} = 8.7,$ toh per unit area yeh Sun ki power ka ~8.7× radiate karta hai.

Verify: UV peak ⇒ ~10000 K, Sun se zyada garam ✓ ("blue = hot" intuition se match karta hai). Units: $m\cdotK / m = K$ ✓. Wien's law link ne tool diya; $T^{4}$ jump Stefan-Boltzmann law hai. Cell H poori tarah worked.

Worked example Ex 8 — Cell I (exam twist: ek clean ratio jo constants ko khatam kar deta hai)

Temperature $T$ par, $ν_{2}$ par energy density ka $ν_{1}$ ke energy density se ratio kya hai, jahan $x_{2} = 6$ aur $x_{1} = 2$ hain? (Dono same $T$ par.) Sirf $x$ 's ke terms mein express karo — $h, k_{B}, c$ mein se koi survive nahi karna chahiye.

Forecast: zyada frequency ( $x = 6$ ) expensive region mein deep hai; ratio 1 se bahut kam hona chahiye chahe $ν^{3}$ uske favor mein ho. Andaaza: order $1 0^{- 2}$ .

Poora ratio likho. Kyun? Fixed $T$ par constants $h, k_{B}, c$ dono spectra ko equally multiply karte hain, isliye ratio unhe wipe out kar deta hai — yahi problem ka poora trick hai. $\frac{u ( ν _{2} )}{u ( ν _{1} )} = \frac{ν _{2}^{3} / ( e ^{x_{2}} - 1 )}{ν _{1}^{3} / ( e ^{x_{1}} - 1 )} .$
$ν$ ko $x$ mein convert karo. Kyun? Fixed $T$ par, $x = h ν / k_{B} T$ ka matlab hai $ν \propto x$ , isliye $ν_{2} / ν_{1} = x_{2} / x_{1}$ aur bacha hua pure- $x$ hai, exactly jaisa demand tha. $\frac{u ( ν _{2} )}{u ( ν _{1} )} = (\frac{x _{2}}{x _{1}})^{3} \frac{e ^{x_{1}} - 1}{e ^{x_{2}} - 1} .$
$x_{1} = 2, x_{2} = 6$ plug karo. Kyun? Khatam karo aur ek number par land karo. $= (\frac{6}{2})^{3} \cdot \frac{e ^{2} - 1}{e ^{6} - 1} = 27 \cdot \frac{6.389}{402.4} = 27 \times 0.01588 = 0.4287.$

Verify: $ν^{3}$ factor ne ek healthy $27 \times$ boost diya, lekin exponential $e^{x}$ ne ise $0.43$ tak knock down kar diya — $ν^{3}$ -up/ $e^{x}$ -down tug-of-war explicitly dikh raha hai. Har physical constant cancel ho gaya ✓, exactly jo problem ne demand kiya tha. Final answer $0.4287$ dimensionless hai, jaisa ek ratio ke liye hona chahiye. Cell I done.

Recall Self-test: pehle cell ka naam batao, phir solve karo

$x = 0.01$ diya gaya, kya $⟨ ϵ ⟩$ zyada $k_{B} T$ ke kareeb hai ya $0$ ke? ::: $k_{B} T$ ke kareeb — Cell A, $⟨ ϵ ⟩ \approx k_{B} T (1 - x /2) = 0.995 k_{B} T$ . Kaun sa peak $x = 4.965$ use karta hai, $ν$ -peak ya $λ$ -peak? ::: $λ$ -peak (Cell F); $ν$ -peak $x = 2.821$ use karta hai. Woh ek dimensionless number konsa hai jo tumhe hamesha pehle compute karna chahiye? ::: $x = h ν / k_{B} T$ . Ek hi star do alag peak wavelengths kyun deta hai? ::: $u (ν) d ν = u (λ) d λ$ se Jacobian $c / λ^{2}$ maximum ko shift karta hai (Cell F).

Related building blocks: Bose-Einstein statistics ( $1/ (e^{x} - 1)$ kyun hai), Equipartition theorem (classical $k_{B} T$ ), Density of states ( $ν^{2}$ counting), aur failed Rayleigh-Jeans law jo is poore page mein repair hota hai.