Blackbody radiation — Planck's quantum hypothesis: Exercises

2.3.1 · D4 · Physics › Modern Physics › Blackbody radiation — Planck's quantum hypothesis

Shuru karne se pehle, wo constants jo hum baar baar use karenge (exams ke liye yaad kar lene layak hain):

Level 1 — Recognition

L1.1 — Kaun sa curve kaun sa temperature hai?

Do blackbodies glow kar rahe hain. Body A ka spectrum $λ_{A} = 1000 nm$ (infrared) par peak karta hai, Body B ka $λ_{B} = 500 nm$ (green) par. Kaun sa body zyada hot hai, aur kitne factor se?

Figure — Blackbody radiation — Planck's quantum hypothesis

Recall Solution

KYA use karein: Wien's displacement law, $λ_{peak} T = b$ . Ye law kehta hai: jis wavelength par ek blackbody sabse zyada bright hota hai, wo uski temperature ke inversely proportional hoti hai. Zyada hot → shorter (bluer) peak.

Inverse kyun: agar $λ_{peak} T$ ek fixed constant $b$ hai, toh $T$ ko upar push karne par $λ_{peak}$ ko neeche aana padega taaki product constant rahe.

Kyunki $b$ dono bodies ke liye same hai: $λ_{A} T_{A} = λ_{B} T_{B} \Rightarrow \frac{T _{B}}{T _{A}} = \frac{λ _{A}}{λ _{B}} = \frac{1000}{500} = 2.$ Body B zyada hot hai, exactly 2 ke factor se. Peak wavelength jitni chhoti (figure mein red curve dekho), body utni hi zyada hot.

L1.2 — Classical law ko pehchano

Tumhe spectral energy density ke do formulas dikhaye gaye hain: $(P) \frac{8 π ν ^{2}}{c ^{3}} \cdot \frac{h ν}{e ^{h ν / k_{B} T} - 1}, (Q) \frac{8 π ν ^{2}}{c ^{3}} k_{B} T .$ Kaun sa classical Rayleigh–Jeans law hai, aur doosre formula ki wo kaunsi ek cheez hai jo "quantum signature" hai?

Recall Solution

(Q) classical Rayleigh–Jeans law hai: mode density $\frac{8 π ν ^{2}}{c ^{3}}$ times har mode ki classical average energy $k_{B} T$ .

(P) Planck's law hai. Quantum signature denominator mein " $- 1$ " hai — yaani $e^{h ν / k_{B} T} - 1$ . Agar tum ye $- 1$ hata do toh Wien's high-frequency tail milta; agar $h ν ≪ k_{B} T$ ho toh (Q) wapas mil jaata hai. Ye akela $- 1$ hi ultraviolet catastrophe ko rokta hai.

Level 2 — Application

L2.1 — Ek star ka temperature uski peak se

Ek star ka spectrum $λ_{peak} = 290 nm$ par peak karta hai. Uska surface temperature nikalo.

Recall Solution

KYA: Wien's law, $T$ ke liye rearrange karke. $T = \frac{b}{λ _{peak}} = \frac{2.898 \times 1 0 ^{- 3} m\cdotK}{290 \times 1 0 ^{- 9} m} .$ nm → m convert kyun karein: $b$ metre·kelvin mein express hai, toh dono lengths metres mein honi chahiye warna units cancel nahi hongi. $T = \frac{2.898 \times 1 0 ^{- 3}}{2.90 \times 1 0 ^{- 7}} = 1.00 \times 1 0^{4} K = 10, 000 K .$ Ek hot blue-white star.

L2.2 — Ek quantum ki energy

Ek cavity mode ki frequency $ν = 5.0 \times 1 0^{14} Hz$ hai (roughly green light). Ek single quantum $h ν$ ki energy kya hai, joules aur electron-volts mein? ( $1 eV = 1.602 \times 1 0^{- 19} J$ use karo.)

Recall Solution

KYA: Planck's postulate kehta hai energy $h ν$ size ke lumps mein aati hai. $E = h ν = (6.626 \times 1 0^{- 34}) (5.0 \times 1 0^{14}) = 3.31 \times 1 0^{- 19} J .$ Electron-volts mein convert karo (joules-per-eV se divide karo): $E = \frac{3.31 \times 1 0 ^{- 19}}{1.602 \times 1 0 ^{- 19}} = 2.07 eV .$ Ye "a few eV" wali size exactly visible-light photons aur Photoelectric effect work functions ka scale hai — ye koi coincidence nahi, ye same quanta hain.

Level 3 — Analysis

L3.1 — Kya $⟨ E ⟩$ formula sach mein classical physics recover karta hai?

Dikhao ki low frequency par ( $h ν ≪ k_{B} T$ ) Planck ki average energy $⟨ E ⟩ = \frac{h ν}{e ^{h ν / k_{B} T} - 1}$ classical $k_{B} T$ mein reduce ho jaati hai. Bilkul clearly batao ki tum kaun sa mathematical tool use kar rahe ho aur kyun ye sahi tool hai.

Recall Solution

KAUN SA tool aur KYU: exponential ka Taylor expansion. Hum ise isliye use karte hain kyunki " $h ν ≪ k_{B} T$ " ka matlab hai ki exponent $x = h ν / k_{B} T$ ek zero ke paas chhota number hai, aur Taylor expansion exactly wo tool hai jo kisi chosen point ke paas function ko uske pehle kuch terms se approximate karta hai. Yahan wo point $x = 0$ hai.

Chhote $x$ ke liye: $e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots \approx 1 + x .$ Sirf $1 + x$ kyun rakhein: agla term $x^{2} /2$ ek chhote number ka square hai — $x$ ki tulna mein negligible.

Toh $e^{x} - 1 \approx x$ , isliye $⟨ E ⟩ = \frac{h ν}{e ^{x} - 1} \approx \frac{h ν}{x} = \frac{h ν}{h ν / k _{B} T} = k_{B} T . ✓$ Ye Equipartition theorem ka result hai — quantum formula smoothly classical wale ko apni low-frequency limit ke roop mein contain karta hai.

L3.2 — High frequency par freeze-out

Ab ulta regime lo $h ν ≫ k_{B} T$ (matlab $x ≫ 1$ ). Dikhao ki $⟨ E ⟩$ exponentially suppress ho jaata hai, aur physically explain karo ki ye ultraviolet catastrophe ko kaise rokta hai.

Recall Solution

KYA: jab $x$ bada ho, toh $e^{x}$ bahut bada hota hai, toh $e^{x} - 1 \approx e^{x}$ ( $- 1$ ek huge number mein tiny correction hai). $⟨ E ⟩ = \frac{h ν}{e ^{x} - 1} \approx \frac{h ν}{e ^{x}} = h ν e^{- h ν / k_{B} T} .$ Ye kyun important hai: factor $e^{- h ν / k_{B} T}$ zero ki taraf uss se kahin zyada tezi se jaata hai jitna mode count $\frac{8 π ν ^{2}}{c ^{3}}$ badhta hai. Toh chahe high-frequency modes zyada se zyada ho jaayein, har ek almost koi energy carry nahi karta.

Physical picture: ek single quantum ki cost $h ν$ hai. High $ν$ par wo lump thermally available energy ( $\sim k_{B} T$ ) se zyada cost karta hai. Bath simply afford nahi kar sakta ek bhi quantum excite karna, toh wo mode apni ground state mein rehta hai — "freeze out" ho jaata hai. Koi infinity nahi. Quantisation ka yahi poora point hai.

Level 4 — Synthesis

L4.1 — Frequency spectrum ka peak derive karo

Wo frequency nikalo jis par frequency-form Planck spectrum $u (ν, T) = \frac{8 π h}{c ^{3}} \frac{ν ^{3}}{e ^{h ν / k_{B} T} - 1}$ maximum ho. Dikhao ki ye transcendental equation $3 (1 - e^{- x}) = x$ deta hai jahan $x = h ν / k_{B} T$ , aur $x$ ke liye numerically solve karo. Har step ka purpose explain karo.

Recall Solution

KAUN SA tool aur KYU: hum derivative $\frac{\partial u}{\partial ν} = 0$ set karte hain. Derivative curve ka slope measure karta hai; peak par curve momentarily flat hoti hai, toh uska slope zero hota hai. Mathematically "peak nikalna" iska yahi matlab hai.

$x = h ν / k_{B} T$ likho, toh $ν \propto x$ aur $u \propto \frac{x ^{3}}{e ^{x} - 1} \equiv f (x)$ . $ν$ ke upar maximize karna $f (x)$ ko maximize karne jaisa hai.

Step — quotient rule. Numerator $x^{3}$ aur denominator $g = e^{x} - 1$ ke saath: $f^{'} (x) = \frac{3 x ^{2} ( e ^{x} - 1 ) - x ^{3} e ^{x}}{( e ^{x} - 1 ) ^{2}} .$ Numerator zero set karo (fraction tab zero hota hai jab uska upar zero ho): $3 x^{2} (e^{x} - 1) - x^{3} e^{x} = 0.$ $x^{2}$ se divide karo (valid hai kyunki peak par $x \neq = 0$ ): $3 (e^{x} - 1) - x e^{x} = 0.$ $e^{x}$ se divide karo aur rearrange karo: $3 (1 - e^{- x}) = x . ✓$ Numerically solve karo: root $x \approx 2.821$ hai. (Tum figure mein dekh sakte ho: do curves $y = x$ aur $y = 3 (1 - e^{- x})$ $x \approx 2.82$ ke paas milti hain.)

Toh frequency peak $ν_{peak} = \frac{2.821 k _{B} T}{h}$ par hai — $T$ ke proportional.

L4.2 — Wavelength peak aur frequency peak kyun alag hote hain

Wavelength-form peak $5 (1 - e^{- y}) = y$ se milti hai jisme $y \approx 4.965$ hai, lekin upar frequency-form peak $x \approx 2.821$ deta hai. Frequency peak ko $λ = c / ν$ se wavelength mein convert karo aur dikhao ki ye wavelength peak ke BARABAR nahi hai. Explain karo kyun nahi.

Recall Solution

Naively convert karna: $λ_{ν -peak} = c / ν_{peak} = \frac{h c}{2.821 k _{B} T}$ , jabki true wavelength peak $λ_{peak} = \frac{h c}{4.965 k _{B} T}$ hai. $\frac{λ _{peak}}{λ _{ν -peak}} = \frac{2.821}{4.965} = 0.568 \neq = 1.$ Ye almost $43%$ se alag hain.

Kyun alag hain: "per unit frequency" aur "per unit wavelength" mein measure kiya gaya spectrum alag functions hain, kyunki convert karne ke liye factor $∣ d ν / d λ ∣ = c / λ^{2}$ aata hai. Ye extra $λ$ -dependent factor curve ko reshape karta hai aur maximum kahin aur shift kar deta hai. Nonlinear change of variable ke under maximum preserve nahi hota. $u_{ν}$ ka peak aur $u_{λ}$ ka peak genuinely alag physical statements hain — dono correct hain, bas alag sawaalon ke jawab de rahe hain ("per unit frequency sabse bright" vs "per unit wavelength").

Level 5 — Mastery

L5.1 — Planck's law se Stefan–Boltzmann constant

$U = \frac{8 π h}{c ^{3}} (\frac{k _{B} T}{h})^{4} \int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x$ se shuru karke jisme $\int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x = \frac{π ^{4}}{15}$ hai, radiated power per area $P = σ T^{4}$ hai jahan $σ = \frac{2 π ^{5} k _{B}^{4}}{15 h ^{3} c ^{2}}$ . $σ$ numerically evaluate karo aur known value se check karo.

Recall Solution

KYA: bas constants ko closed form mein substitute karo. $σ = \frac{2 π ^{5} k _{B}^{4}}{15 h ^{3} c ^{2}} .$ Piece by piece:

$k_{B}^{4} = (1.381 \times 1 0^{- 23})^{4} = 3.638 \times 1 0^{- 92}$
$h^{3} = (6.626 \times 1 0^{- 34})^{3} = 2.909 \times 1 0^{- 100}$
$c^{2} = (3.00 \times 1 0^{8})^{2} = 9.00 \times 1 0^{16}$
$2 π^{5} = 612.0$

Combine karo: $σ = \frac{612.0 \times 3.638 \times 1 0 ^{- 92}}{15 \times 2.909 \times 1 0 ^{- 100} \times 9.00 \times 1 0 ^{16}} \approx 5.67 \times 1 0^{- 8} W\cdotm^{- 2} K^{- 4} . ✓$ Ye measured $σ = 5.670 \times 1 0^{- 8}$ se match karta hai. Stefan–Boltzmann law koi independent empirical law nahi hai — ye seedha Planck's spectrum se nikalta hai, aur $σ$ poori tarah $h, k_{B}, c$ se bana hai.

L5.2 — Cosmic Microwave Background temperature

Cosmic Microwave Background ek near-perfect blackbody hai jo (apne wavelength spectrum mein) $λ_{peak} \approx 1.06 mm$ par peak karta hai. Uska temperature nikalo. Phir, frequency-peak result $ν_{peak} = 2.821 k_{B} T / h$ use karte hue, peak frequency nikalo.

Recall Solution

Wien se Temperature: $T = \frac{b}{λ _{peak}} = \frac{2.898 \times 1 0 ^{- 3}}{1.06 \times 1 0 ^{- 3}} = 2.73 K .$ CMB ka famous $2.7 K$ — poora sky almost absolute zero par ek blackbody hai.

Peak frequency (frequency spectrum): $ν_{peak} = \frac{2.821 k _{B} T}{h} = \frac{2.821 \times 1.381 \times 1 0 ^{- 23} \times 2.73}{6.626 \times 1 0 ^{- 34}} = 1.60 \times 1 0^{11} Hz \approx 160 GHz .$ Dhyan do ki ye wavelength wale peak se genuinely alag peak hai (L4.2 yaad karo): $c / ν_{peak} = 1.87 mm \neq = 1.06 mm$ — Jacobian shift ek real experiment mein action mein.

L5.3 — Photon occupation number

Planck ki average energy $⟨ E ⟩ = \frac{h ν}{e ^{h ν / k_{B} T} - 1}$ hai. Kyunki ek mode mein har photon energy $h ν$ carry karta hai, us mode mein photons ki average number $\overset{n}{ˉ} = ⟨ E ⟩ / h ν = \frac{1}{e ^{h ν / k_{B} T} - 1}$ hai. Ye Bose–Einstein statistics distribution hai. Ek mode ke liye jisme $h ν = k_{B} T$ ho (yaani $x = 1$ ), $\overset{n}{ˉ}$ compute karo.

Recall Solution

$x = h ν / k_{B} T = 1$ set karo: $\overset{n}{ˉ} = \frac{1}{e ^{1} - 1} = \frac{1}{2.71828 - 1} = \frac{1}{1.71828} = 0.582.$ Toh ek mode jiska quantum exactly ek $k_{B} T$ cost karta hai, average mein roughly 0.58 photons rakhta hai — ek se bhi kam. Jinme $h ν ≫ k_{B} T$ hota hai unme bahut kam hote hain (L3.2 wala freeze-out); jinme $h ν ≪ k_{B} T$ hota hai unme $\overset{n}{ˉ} \approx k_{B} T / h ν ≫ 1$ hota hai (bahut saare photons, classical crowd). Har aisa mode thermal equilibrium mein ek Quantum harmonic oscillator ki tarah behave karta hai.

Recall Jaane se pehle ek-line self-test

Peak $T$ ke saath kaise shift karta hai? ::: Inversely — $λ_{peak} = b / T$ , zyada hot matlab bluer. Planck's law ki kaun si ek feature UV catastrophe ko khatam karti hai? ::: $e^{h ν / k_{B} T} - 1$ mein $- 1$ , high $ν$ par exponential freeze-out karti hai. $ν$ -peak aur $λ$ -peak kyun alag hain? ::: Spectral density ek Jacobian $∣ d ν / d λ ∣$ pick up karta hai; peaks nonlinear variable change ke under preserve nahi hote. $σ$ teen kaun se constants se bana hai? ::: $h$ , $k_{B}$ , $c$ (via $σ = 2 π^{5} k_{B}^{4} /15 h^{3} c^{2}$ ).