Worked examples — Blackbody radiation — Planck's quantum hypothesis

2.3.1 · D3 · Physics › Modern Physics › Blackbody radiation — Planck's quantum hypothesis

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The scenario matrix

Blackbody radiation ka har problem actually ek dimensionless number ke baare mein sawaal hai: $x=\frac{h\nu }{k_{B}T}\text{("ek quantum ki cost kitni thermal units mein hai?")}$

• $x$ small ($\ll 1$): ek quantum sasta hai → classical behaviour wapas aata hai.
• $x$ large ($\gg 1$): ek quantum mehnga hai → mode freeze ho jaata hai.
• $x$ around $2$–$5$: interesting middle jahan spectrum peak karta hai.

Is page par do spectral densities aate hain, aur yeh matter karta hai ki tum kaun sa use karo:

Niche ki table har class ki situation list karti hai. Har worked example us cell(s) ke saath tagged hai jise wo cover karta hai.

Cell	Situation	Physical regime	Example
A	$h\nu \ll k_{B}T$ (low frequency)	classical / Rayleigh–Jeans limit	Ex 1
B	$h\nu \gg k_{B}T$ (high frequency)	Wien / frozen-out limit	Ex 2
C	$h\nu \approx k_{B}T$ (comparable)	exact formula, no shortcut	Ex 3
D	Degenerate: $\nu \to 0$ aur $T\to 0$	limiting values / boundaries	Ex 4
E	Peak-finding (turning point), $\lambda$-peak vs $\nu$-peak	Wien's displacement, transcendental	Ex 5 (figure)
F	Total power (integral over all $\nu$)	Stefan–Boltzmann, real object	Ex 6
G	Real-world word problem	astronomy: colour se temperature	Ex 7
H	Exam twist: "kaun si curve hotter hai?"	do curves compare karna, ratios	Ex 8 (figure)

Constants jo poore use honge (inhe haath ke paas rakho):

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Ex 1 — Cell A: low-frequency (classical) limit

Step 1 — $x=h\nu /k_{B}T$ compute karo. Ye step kyun? $x$ hi decide karta hai ki regime kaun sa hai; baaki sab us ki size se follow karta hai. $x=\frac{(6.626\times 10^{-34})(1.0\times 10^{10})}{(1.381\times 10^{-23})(300)}=\frac{6.626\times 10^{-24}}{4.143\times 10^{-21}}\approx 1.60\times 10^{-3}.$ Ye bahut chhota hai — ek quantum ek thermal unit ke thousandth se bhi kam cost karta hai. Toh hum Cell A mein deep hain.

Step 2 — Planck ki average energy. Ye step kyun? Hum honest quantum answer chahte hain compare karne ke liye. $⟨E⟩=\frac{h\nu }{e^x-1}=\frac{6.626\times 10^{-24}}{e^{0.00160}-1}.$ Kyunki $x$ small hai, $e^x-1\approx x=1.60\times 10^{-3}$, jo deta hai $⟨E⟩\approx \frac{6.626\times 10^{-24}}{1.60\times 10^{-3}}\approx 4.14\times 10^{-21}\text{J}.$

Step 3 — classical value. Ye step kyun? Puri baat hi comparison ki hai. $k_{B}T=(1.381\times 10^{-23})(300)=4.14\times 10^{-21}\text{J}.$

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Ex 2 — Cell B: high-frequency (frozen-out) limit

Step 1 — $x$ compute karo. Ye step kyun? Phir se, $x$ hi sab kuch decide karta hai. $x=\frac{(6.626\times 10^{-34})(3.0\times 10^{15})}{4.143\times 10^{-21}}=\frac{1.988\times 10^{-18}}{4.143\times 10^{-21}}\approx \mathrm{479.8.}$ Ye bahut bada hai — ek quantum ~480 thermal units cost karta hai. Cell B.

Step 2 — Planck ki average energy. Ye step kyun? Hum actual quantum answer chahte hain, jo bahut hi chhota hona chahiye. $⟨E⟩=\frac{h\nu }{e^x-1}\approx h\nu e^{-x}=(1.988\times 10^{-18})e^{-479.8}.$ $e^{-479.8}$ ko size karne ke liye, base 10 mein convert karo: ${\log }_{10}(e^{-479.8})=-479.8\times 0.4343\approx -208.4$, isliye $e^{-479.8}\approx 10^{-208.4}\approx 4\times 10^{-209}$ — ek unimaginably chhota number. $⟨E⟩\approx (1.988\times 10^{-18})(4\times 10^{-209})\approx 8\times 10^{-227}\text{J}\approx 0.$

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Ex 3 — Cell C: comparable energies (koi shortcut allowed nahi)

Step 1 — $x$ compute karo. Ye step kyun? Ye confirm karne ke liye ki hum dangerous "middle" mein hain jahan koi bhi shortcut valid nahi. $x=\frac{(6.626\times 10^{-34})(5.45\times 10^{14})}{(1.381\times 10^{-23})(5800)}=\frac{3.611\times 10^{-19}}{8.010\times 10^{-20}}\approx \mathrm{4.51.}$ Na chhota na huge — hume pura formula use karna hi hoga.

Step 2 — full expression evaluate karo. Ye step kyun? Cell C mein "$-1$" genuinely matter karta hai; ise drop karna kuch percent galat hoga. $e^{4.51}\approx 90.9,e^{4.51}-1\approx \mathrm{89.9.}$ $⟨E⟩=\frac{h\nu }{e^x-1}=\frac{3.611\times 10^{-19}}{89.9}\approx 4.02\times 10^{-21}\text{J}.$

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Ex 4 — Cell D: degenerate / boundary inputs

Step 1 — case (a): $\nu \to 0$. Ye step kyun? Ye Cell A ki boundary hai extreme par push ki gayi; hum test karte hain ki formula sensible rehta hai. Jab $\nu \to 0$, $x=h\nu /k_{B}T\to 0$, isliye $e^x-1\to x$: $⟨E⟩=\frac{h\nu }{e^x-1}\to \frac{h\nu }{x}=\frac{h\nu k_{B}T}{h\nu }=k_{B}T.$ Isliye har mode $k_{B}T$ approach karta hai low frequency par — ek finite, non-zero value.

Step 2 — case (b): $T\to 0$. Ye step kyun? Ye Cell B ko extreme par push karta hai; ye check karta hai ki freeze-out absolute zero par complete ho jaata hai. Jab $T\to 0^{+}$, $x=h\nu /k_{B}T\to +\infty$, isliye $e^x\to \infty$ aur $⟨E⟩=\frac{h\nu }{e^x-1}\to \frac{h\nu }{\infty }=0.$ Absolute zero par koi bhi mode excited nahi hai — cavity dark hai.

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Ex 5 — Cell E: peak dhundna (Wien's displacement) + $\lambda$-peak vs $\nu$-peak subtlety

Step 1 — figure padho. Ye step kyun? Matrix se yaad karo ki $u(\lambda ,T)$ energy density per unit wavelength hai. Figure mein solid teal curve $u(\lambda ,T)$ hai; uska maximum orange dot se mark hai, aur us dot ke through chhota horizontal plum segment ek flat tangent hai (slope $=0$). Dashed orange curve wahi radiation hai re-sliced karke $u(\nu ,T)$ ke roop mein (per unit frequency, comparison ke liye wavelength ke against plot ki gayi); uska apna peak — plum dot — ek chhote wavelength par baith ta hai. Do dots ke beech woh visible gap hi part (b) ki poori subtlety hai. Horizontal tangent ka matlab slope $=0$ hai, isliye part (a) ka maths task hai: $du/d\lambda =0$ set karo.

Step 2 — $u(\lambda ,T)$ likho aur differentiate karo. Ye step kyun? Peak wahan hai jahan $du/d\lambda =0$ hai, exactly figure mein flat-tangent point (orange dot). $u(\lambda ,T)=\frac{8\pi hc}{{\lambda }^5}\frac{1}{e^{hc/\lambda k_{B}T}-1}.$ $u=C{\lambda }^{-5}g(\lambda )$ likho jahan $C=8\pi hc$ aur $g=(e^{a/\lambda }-1{)}^{-1}$, aur $a=hc/k_{B}T$. Product rule: $\frac{du}{d\lambda }=C[-5{\lambda }^{-6}g+{\lambda }^{-5}{g}^{\prime }].$

Step 3 — $g$ differentiate karo aur simplify karo. Ye step kyun? Hume $g^{\prime }$ chahiye bracket explicit karne ke liye, phir factor karo. $g^{\prime }=\frac{d}{d\lambda }(e^{a/\lambda }-1{)}^{-1}=-(e^{a/\lambda }-1{)}^{-2}\cdot e^{a/\lambda }\cdot (-\frac{a}{{\lambda }^2})=\frac{a}{{\lambda }^2}\frac{e^{a/\lambda }}{(e^{a/\lambda }-1{)}^2}.$ Wapas substitute karo aur $du/d\lambda =0$ set karo. Common non-zero factor $C{\lambda }^{-6}(e^{a/\lambda }-1{)}^{-1}$ divide out karne par bacha $-5+\frac{a}{\lambda }\frac{e^{a/\lambda }}{e^{a/\lambda }-1}=0.$

Step 4 — $y=a/\lambda =hc/\lambda k_{B}T$ substitute karo. Ye step kyun? $y$ saare constants ko ek dimensionless variable mein pack karta hai, mess ko ek clean equation mein collapse karta hai. $-5+y\frac{e^y}{e^y-1}=0\Rightarrow y\frac{e^y}{e^y-1}=5.$ Fraction ke top aur bottom ko $e^y$ se divide karo: $\frac{e^y}{e^y-1}=\frac{1}{1-e^{-y}}$, isliye $\frac{y}{1-e^{-y}}=5\Rightarrow \boxed{5(1-e^{-y})=y}.$

Step 5 — numerically solve karo. Ye step kyun? Transcendental equations iteration/graphing se solve hote hain.

• $y=5$ try karo: LHS $=5(1-e^{-5})=5(0.9933)=4.966$, RHS $=5$. Bahut close.
• Refine karo: crossing $y\approx 4.965$ par hai.

Step 6 — $b$ nikalo. Ye step kyun? $y=hc/{\lambda }_{\text{peak}}{k}_{B}T$ ko rearrange karo ${\lambda }_{\text{peak}}T=hc/(y{k}_B)$ mein, jo ek constant hai. $b=\frac{hc}{4.965k_B}=\frac{(6.626\times 10^{-34})(3.00\times 10^8)}{(4.965)(1.381\times 10^{-23})}\approx 2.898\times 10^{-3}\text{m⋅K}.$

Step 7 — part (b): frequency peak ek alag equation hai. Ye step kyun? Ye promised subtlety hai — students galti se assume karte hain ${\lambda }_{\text{peak}}=c/{\nu }_{\text{peak}}$. $u(\nu ,T)=\frac{8\pi {\nu }^2}{c^3}\frac{h\nu }{e^{h\nu /k_{B}T}-1}=\frac{8\pi h}{c^3}\frac{{\nu }^3}{e^{h\nu /k_{B}T}-1}$ par maximization repeat karo. Aage ka power ab ${\nu }^3$ hai (na ${\lambda }^{-5}$), isliye "$5$" ban jaata hai "$3$": $\boxed{3(1-e^{-z})=z},z=\frac{h\nu }{k_{B}T},z\approx \mathrm{2.821.}$ Kyunki $y=4.965\ne z=2.821$, do peaks alag colours pick karte hain. $\nu$-peak ko wavelength mein convert karte hue: ${\lambda }^{\prime }=c/{\nu }_{\text{peak}}=hc/(z{k}_{B}T)$, isliye ${\lambda }^{\prime }T=hc/(2.821k_B)\approx 5.10\times 10^{-3}\text{m⋅K}$ — $b=2.898\times 10^{-3}\text{m⋅K}$ se ek longer wavelength.

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Ex 6 — Cell F: total emitted power (Stefan–Boltzmann, real object)

Step 1 — surface area. Ye step kyun? Stefan–Boltzmann per unit area power deta hai; total chahiye, isliye $A$ se multiply karo. $A=4\pi r^2=4\pi (0.10{)}^2=4\pi (0.01)\approx 0.1257{\text{m}}^2.$

Step 2 — $P=\sigma A{T}^4$ apply karo. Ye step kyun? Yahi Stefan–Boltzmann law hai saari frequencies par integrated — $T^4$ un chaar $T$ factors se aata hai jo parent note ne substitution ke through track kiya. $T^4=(1000{)}^4=1.0\times 10^{12}{\text{K}}^4.$ $P=(5.67\times 10^{-8})(0.1257)(1.0\times 10^{12})\approx 7.13\times 10^3\text{W}.$ Toh lagbhag 7.1 kW.

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Ex 7 — Cell G: real-world word problem (colour se temperature)

Step 1 — Wien's law apply karo. Ye step kyun? Wien directly measurable peak colour ko temperature se jodta hai — exactly wahi jo astronomers exploit karte hain, kyunki tum star mein thermometer nahi daal sakte. (Hum yahan wavelength peak $b$ use karte hain, jo per unit wavelength measure kiye gaye spectrum ke saath consistent hai — Ex 5 ki caution yaad karo.) ${\lambda }_{\text{peak}}T=b\Rightarrow T=\frac{b}{{\lambda }_{\text{peak}}}.$

Step 2 — numbers plug karo. Ye step kyun? Pehle nm ko metres mein convert karo, phir divide karo. $T=\frac{2.898\times 10^{-3}}{966\times 10^{-9}}=\frac{2.898\times 10^{-3}}{9.66\times 10^{-7}}\approx 3000\text{K}.$

Step 3 — colour interpret karo. Ye step kyun? Koi number answer nahi hota jab tak kuch mean na kare. $3000\text{K}$ par peak near-infrared mein hai, lekin visible tail red end par strongest hai — isliye star distinctly reddish-orange dikhta hai. Cooler blackbodies hamesha longer wavelengths par peak karte hain (yahi Wien's displacement law relationship hai, ulte padha gaya).

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Ex 8 — Cell H: exam twist ("kaun si curve hotter hai, aur kitni?")

Step 1 — (a) kaun si hotter hai? Ye step kyun? Wien kehta hai peak shorter wavelength ki taraf shift hoti hai jab $T$ badhti hai. Figure mein teal curve 500 nm par peak karti hai (bluer) aur uska poora body orange curve ke upar towering hai jo 1000 nm par peak karti hai. Dono facts same direction point karte hain: Curve 2 (teal) hotter hai.

Step 2 — (b) temperature ratio. Ye step kyun? Wien ka constant $b$ dono bodies ke liye same hai, isliye $\lambda T$ ek fixed number hai — temperatures ka ratio sirf peaks ka inverse ratio hai. ${\lambda }_1{T}_1={\lambda }_2{T}_2=b\Rightarrow \frac{T_2}{T_1}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{1000}{500}=2.$ Curve 2 exactly double hot hai. (Ye figure mein visible hai: do dotted peak-lines 1000 nm aur 500 nm par baithe hain, factor of 2 apart.)

Step 3 — (c) power ratio. Ye step kyun? Total power $T^4$ ke anusaar scale hoti hai (Stefan–Boltzmann), isliye factor-2 temperature ban jaata hai factor-$2^4$ power mein. Figure mein isliye teal curve ka area orange wale ko dwarf karta hai — height ka gap temperature gap ke factor-2 se kahin zyada bada hai. $\frac{P_2}{P_1}={\left(\frac{T_2}{T_1}\right)}^4=2^4=16.$

Answers: (a) Curve 2 hotter hai. (b) $T_2/T_1=2$. (c) $P_2/P_1=16$.

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Recall Self-test: har mini-scenario ke liye cell name karo

Ek warm oven mein microwave photon ::: Cell A (low frequency, classical limit, $⟨E⟩\to k_{B}T$). Room-temperature cavity mein ek X-ray mode ::: Cell B (high frequency, frozen out, $⟨E⟩\approx 0$). Exact colour jahan Sun ki curve peak karti hai ::: Cell C / Cell E (comparable energies, $x\approx 4.965$). Frequency-peak ek alag colour kyun name karta hai wavelength-peak se? ::: Cell E part (b): prefactor ${\lambda }^{-5}$ se ${\nu }^3$ mein change hota hai, $z\approx 2.821$ deta hai $y\approx 4.965$ ki jagah. Peak colour se star ka temperature compute karna ::: Cell G (Wien word problem). "Peak half ho gayi — kitni zyada power?" ::: Cell H ($T$ double, power $\times 16$).