Physics · Thermal radiation

Planck’s radiation law

Q: How can I check a calculation involving Planck Spectral Law?

First convert λ to metres and evaluate $x=\frac{hc}{\lambda k_{\mathrm B}T}$. This exponent must be dimensionless. For 5800 K and 500 nm, $x\approx4.96130$ and $B_\lambda\approx2.68822\cdot10^{13}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$. Next check whether the requested value is per metre, nanometre or micrometre; the per-nanometre numerical value is smaller by 10⁻⁹. For small x, an expm1 evaluation avoids subtracting nearly equal numbers inaccurately. A measured power also requires wavelength bandwidth, projected area and accepted solid angle. The factor π describes hemisphere integration and must not replace a detector’s small acceptance angle.

Planck’s law describes blackbody equilibrium spectral radiance per wavelength interval. Power additionally requires appropriate area, angle and band integration.

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Formula

Calculate

B_{λ} (T) = \frac{2 h c ^{2}}{λ ^{5}} \frac{1}{exp ( \frac{h c}{λ k _{B} T} ) - 1}

Bλ in W·m⁻²·sr⁻¹·m⁻¹, equivalently W·m⁻³·sr⁻¹; λ in m; T in K; h in J·s; c in m/s; kB in J/K. The final m⁻¹ specifies spectral bandwidth.

Variables & units

Variables & units
Symbol	Meaning	Unit
$Bλ$	Spectral radiance per vacuum wavelength interval	W·m⁻²·sr⁻¹·m⁻¹
$λ$	Positive vacuum wavelength	m
$T$	Absolute blackbody equilibrium temperature	K
$h$	Planck constant: exactly 6.62607015×10⁻³⁴ J·s	J·s
$c$	Speed of light in vacuum: exactly 299792458 m/s	m/s
$kB$	Boltzmann constant: exactly 1.380649×10⁻²³ J/K	J/K

LaTeX

B_\lambda(T)=\frac{2hc^2}{\lambda^5}\frac{1}{\exp\!\left(\frac{hc}{\lambda k_{\mathrm B}T}\right)-1}

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A qualitative Planck curve marks its peak at λmax; the normalized wavelength is λ divided by λmax. — Spectral radiance per wavelength interval has a peak; increasing temperature shifts it to shorter wavelengths.

Planck's law calculator

Find the spectral radiance of an ideal black body at a wavelength or frequency, the temperature from it, or the power in a narrow measuring channel.

Calculation model

Find

Wavelength λ

Wavelength you are looking at, e.g. 500 nm for green light.

Absolute equilibrium temperature T

Absolute temperature of the emitter, e.g. 5800 K for the surface of the Sun.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for Bλ, substitute, evaluate

1. Rearrange for Bλ
$B_\lambda=\frac{2hc^2}{\lambda^5[\exp(\frac{hc}{\lambda k_{\rm B}T})-1]}$
2. Substitute values (in SI units)
- λ = 500 nm = 5 · 10⁻⁷ m
$\frac{2\cdot 6.62607\times10^{-34}\cdot 299792458\cdot 299792458}{5\times10^{-7}^5\cdot[\exp(\frac{6.62607\times10^{-34}\cdot 299792458}{5\times10^{-7}\cdot 1.38065\times10^{-23}\cdot 5800})-1]}=26882199625900$
3. Result
Bλ = 26 880 W/(m²·sr·nm)

Result: Spectral radiance Bλ

Bλ = 26 880 W/(m²·sr·nm)

With λ = 500 nm and T = 5800 K, Bλ = 26 880 W/(m²·sr·nm).

Quick tips

Use Wien as a check: Bλ peaks at λ ≈ 2898 μm·K divided by T, so at about 500 nm for a 5800 K emitter such as the Sun.
You cannot turn Bν into Bλ just by substituting λ = c/ν: Bλ = Bν · c/λ², which is why their peaks sit at different places.

Assumptions and limits

Ideal blackbody thermal equilibrium radiator in vacuum, T>0 and positive spectral coordinate. Radiance is per projected area, solid angle and its own spectral interval, not power in watts. Wavelength and frequency densities have different Jacobians and peaks. Temperature inversion applies only to known calibrated blackbody radiance, not uncalibrated light with unknown emissivity or background.

Derivation & background

Quantized mode energy and thermal occupation yield the Planck spectrum. This form is per wavelength interval. Frequency conversion requires the Jacobian $B_\lambda=B_\nu\frac c{\lambda^2}$ , so the two maxima do not correspond through simple coordinate substitution. Blackbody surface exitance is $M_\lambda=\pi B_\lambda$ ; integrating it over all wavelengths gives σT⁴. Real surfaces need spectral emissivity. Measured signals additionally need bandwidth, angle, projected area, transmission and background.

Exam blueprint

Validity range

Blackbody thermal equilibrium radiation in vacuum, with absolute T > 0 and λ > 0. Bλ is radiance per wavelength interval, projected area and solid angle; blackbody surface exitance is $M_\lambda=\pi B_\lambda$ . Real thermal surfaces need spectral and potentially directional emissivity, along with incident-background and instrument contributions. Nonthermal radiation does not automatically follow this law. Bν and Bλ are related by a Jacobian; their maxima must be calculated separately.

Derivation steps

Combine the thermal mean energy of quantized radiation modes with their mode density, then transform spectral coordinates correctly.

1Consider equilibrium radiation in a large ideal cavity. A mode at frequency $\nu$ has excitation energies $E_n=nh\nu$ , with $n=0,1,2,\ldots$ .
2At absolute temperature $T$ , Boltzmann weights give $Z=\sum_{n=0}^{\infty}e^{-\frac{nh\nu}{k_{\mathrm B}T}}=\frac1{1-e^{-\frac{h\nu}{k_{\mathrm B}T}}}$ .
3Mean occupation is $\bar n=\frac1{e^{\frac{h\nu}{k_{\mathrm B}T}}-1}$ , so mean thermal mode energy is $\bar E=h\nu\bar n$ .
4Mode density including two polarizations is $g_\nu=\frac{8\pi\nu^2}{c^3}$ . Its product with $\bar E$ gives spectral energy density $u_\nu$ .
5For isotropic vacuum radiation, $B_\nu=\frac{cu_\nu}{4\pi}=\frac{2h\nu^3}{c^2\left(e^{\frac{h\nu}{k_{\mathrm B}T}}-1\right)}$ . This is radiance per frequency interval.
6Transform with $\nu=\frac c\lambda$ and $B_\lambda=B_\nu\left|\frac{\mathrm d\nu}{\mathrm d\lambda}\right|=B_\nu\frac c{\lambda^2}$ . This gives $B_\lambda=\frac{2hc^2}{\lambda^5\left(e^{\frac{hc}{\lambda k_{\mathrm B}T}}-1\right)}$ . The Jacobian also changes peak location.

Rearrangements

Calculate spectral radiance

B_\lambda=\frac{2hc^2}{\lambda^5\left[\exp\!\left(\frac{hc}{\lambda k_{\mathrm B}T}\right)-1\right]}

Use wavelength in metres and the per-wavelength convention.

Task variant

Calculate blackbody spectral radiance at 5800 K and 500 nm. Express it per metre and per nanometre of wavelength interval. What additional information would detector power require?

Approach: convert $\lambda=5.00\cdot10^{-7}\,\mathrm m$ , then calculate dimensionless exponent $x=\frac{hc}{\lambda k_{\mathrm B}T}\approx4.96130$ .
The full Planck law gives $B_\lambda\approx2.68822\cdot10^{13}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ .
Per nanometre this is $2.68822\cdot10^4\,\mathrm{W\,m^{-2}\,sr^{-1}\,nm^{-1}}$ .

Check: corresponding band integrals agree. Power additionally requires area, viewing geometry or solid angle, spectral band and transmission.

Common mistakes

Reading spectral radiance directly as power.

Integrate over the detection band and projected area with the collected solid angle. Curve height alone does not include these factors.

Equating wavelength and frequency peaks by simple conversion.

Densities satisfy $B_\nu=B_\lambda\frac{\lambda^2}c$ . The variable Jacobian changes peak location.

Substituting nanometres into an SI formula or leaving density units unchanged.

Use metres in the exponent and prefactor. Changing to radiance per nanometre additionally multiplies the numerical value by 10⁻⁹.

Exam context

Model thermal laboratory and stellar spectra, infer temperature from suitable spectral measurements and calculate detector-band power using appropriate geometry.

Formula cluster

Thermal radiation

Related relations: stefan-boltzmann-strahlung, bohr-energieniveaus-wasserstoff.

P = σ·A·T⁴

Stefan-Boltzmann Law

En = −13,6 eV/n²; hν = |Ef − Ei|

Bohr Energy Levels and Transitions in Hydrogen

Worked example

At $T=5800\,\mathrm K$ and $\lambda=500\,\mathrm{nm}=5.00\cdot10^{-7}\,\mathrm m$ , $B_\lambda\approx2.68822\cdot10^{13}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ . Check: $\frac{hc}{\lambda k_{\mathrm B}T}\approx4.96130$ is dimensionless. Per nanometre this is 2.68822×10⁴ W·m⁻²·sr⁻¹·nm⁻¹; the factor 10⁻⁹ applies to spectral bandwidth.

Applications

Model thermal laboratory and stellar spectra, infer temperature from suitable spectral measurements and calculate detector-band power using appropriate geometry.

How do you calculate with Planck Spectral Law?

Here is how to work through a typical Planck Spectral Law (Bλ = (2hc²/λ⁵)/(e^(hc/λkBT)−1)) task step by step:

1
Task
Calculate blackbody spectral radiance at 5800 K and 500 nm. Express it per metre and per nanometre of wavelength interval. What additional information would detector power require?
Solution path
1. Approach: convert $\lambda=5.00\cdot10^{-7}\,\mathrm m$ , then calculate dimensionless exponent $x=\frac{hc}{\lambda k_{\mathrm B}T}\approx4.96130$ .
2. The full Planck law gives $B_\lambda\approx2.68822\cdot10^{13}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ .
3. Per nanometre this is $2.68822\cdot10^4\,\mathrm{W\,m^{-2}\,sr^{-1}\,nm^{-1}}$ .
Check: corresponding band integrals agree. Power additionally requires area, viewing geometry or solid angle, spectral band and transmission.

Understand and apply the equation

Thermal radiation spans many wavelengths. Planck’s law describes how much a blackbody emits in each direction at a given temperature. Measured power additionally depends on bandwidth, area and viewing angle. The calculations therefore distinguish a spectrum, one detection channel and temperature inferred from two channels.

What quantity does Planck’s radiation law describe?

B_\lambda(T)=\frac{2hc^2}{\lambda^5}\frac1{\exp[\frac{hc}{\lambda k_BT}]-1}

λ is vacuum wavelength here; T is absolute temperature.[2]

Spectral radiance counts power per projected source area, solid angle and wavelength interval. In W·m⁻²·sr⁻¹·m⁻¹, the final inverse metre describes a spectral interval rather than volume density. One Bλ value is therefore not power in watts. Measurement also needs bandwidth, accepted area, viewing direction and solid angle; optical losses can further reduce received power.[2]

The ideal black cavity radiator is in thermal equilibrium. Radiance does not depend on wall color or material when the blackbody assumptions hold. A real emitting surface needs appropriate spectral and possibly directional emissivity. Reflected radiation and nonthermal sources cannot simply be represented by substituting another temperature.[1]

Compare spectra: 1400 and 2000 kelvin

Temperature changes spectrum height and peak position

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Both curves use the same density per micrometre and the same scale. A detection band corresponds to an area under the curve; height alone is not power.[1][2]

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Explore further: mean energy of a quantised mode

Model an electromagnetic mode with occupations $n=0,1,2,\ldots$ and thermal excitation energies $nh\nu$ . Equilibrium weights each occupation by $e^{-nx}$ , with $x=\frac{h\nu}{k_{\mathrm B}T}$ . A common zero-point offset does not change mean thermal excitation. A geometric series yields mean occupation and hence mean thermally transported energy per mode.[1]

\begin{aligned}Z&=\sum_{n=0}^\infty e^{-nx}=\frac1{1-e^{-x}}\\{} \langle n\rangle&=-\frac{d\ln Z}{dx}=\frac1{e^x-1}\\{} \overline E_\nu&=\frac{h\nu}{e^{\frac{h\nu}{k_BT}}-1}\end{aligned}

Quantised energy hν and thermal weighting are model assumptions; subsequent algebra uses a geometric series.

For hν≪kBT, eˣ−1≈x makes mean energy approach kBT. For hν≫kBT, the exponential denominator suppresses high-frequency excitation. Classical equipartition omits that suppression and yields a divergent high-frequency integral. Quantisation is therefore physical content rather than a correction to units. Equipartition of energy therefore describes only the classical limit of the quantised mode used here.[1]

From mode counting to radiance

Counting both electromagnetic polarizations in a large cavity gives 8πν²/c³ modes per volume per frequency interval. Multiply by mean thermal energy to obtain spectral energy density uν. For isotropic radiation, radiance relates to uν through c/(4π); flux through a surface additionally requires angular projection.

\begin{aligned}u_\nu&=\frac{8\pi h\nu^3}{c^3(e^{\frac{h\nu}{k_BT}}-1)}\\{} B_\nu&=\frac{c}{4\pi}u_\nu=\frac{2h\nu^3}{c^2(e^{\frac{h\nu}{k_BT}}-1)}\end{aligned}

Bν has W·m⁻²·sr⁻¹·Hz⁻¹; uν has J·m⁻³·Hz⁻¹.[2]

\begin{aligned}B_\lambda&=B_\nu\left|\frac{\mathrm d\nu}{\mathrm d\lambda}\right|=B_\nu\frac c{\lambda^2}\\\nu&=\frac c\lambda\end{aligned}

Equal physical band energy requires the Jacobian, not merely replacing ν with c/λ.[2]

Equal wavelength widths correspond to different frequency widths at different spectral positions. Numerical density values and curve shape therefore change with the spectral coordinate. Integrals over matching bands remain equal when frequency limits are reversed appropriately. Choosing a spectral axis does not change the actual radiation.[2]

Which units and numerical limits apply to the Planck spectrum?

Use $h=6.62607015\cdot10^{-34}\,\mathrm{J\,s}$ , $c=299792458\,\mathrm{m\,s^{-1}}$ and $k_{\mathrm B}=1.380649\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ , exact SI defining values. Convert wavelength to metres before substitution. The exponent $\frac{hc}{\lambda k_{\mathrm B}T}$ is dimensionless. Values per nanometre are $10^{-9}$ times the numerical density per metre; per micrometre they are $10^{-6}$ times that density. Source area remains in square metres.[3]

\begin{aligned}x\ll1:\quad B_\lambda&\approx\frac{2ck_{\mathrm B}T}{\lambda^4}\\x\gg1:\quad B_\lambda&\approx\frac{2hc^2}{\lambda^5}e^{-x}\end{aligned}

Rayleigh–Jeans requires small x; the Wien approximation requires large x.

Numerically, expm1(x) computes small eˣ−1 more reliably than directly subtracting exp(x)−1 when x is small. At very large x, use exponential suppression instead of evaluating an overflowing exponential. These techniques do not change the physical model. Observed signals additionally include spectral filtering and instrument calibration.[2]

Example: turn radiance into narrow-channel power

An ideal narrow infrared channel

Given: Black model surface at 1400 K, $\lambda_0=2.00\,\mathrm{\mu m}$ , symmetric 20.0 nm band. Accepted source area $A=2.00\cdot10^{-4}\,\mathrm{m^2}$ , small $\Omega=0.00300\,\mathrm{sr}$ near the surface normal; ideal transmission and no background subtraction.
Find: Find Bλ and approximate channel power.

Approach: integrate radiance over the wavelength band and small nearly normal solid angle. With constant transmission and little variation across the band, approximate the integral by the midpoint product.
$\begin{aligned}P_{\mathrm{band}}&=A\int_{\Delta\Omega}\int_{\Delta\lambda} B_\lambda(T)\cos\theta\,\mathrm d\lambda\,\mathrm d\Omega\\&\approx A\Omega\Delta\lambda\,B_{\lambda_0}(T)\\x&=\frac{hc}{\lambda_0 k_{\mathrm B}T}\end{aligned}$
Evaluate Planck radiance per metre of wavelength.
$B_{\lambda_0}=2.19642012\times10^{10}\,\mathrm{W\,m^{-2}sr^{-1}m^{-1}}$
Multiply factors for the narrow band and small nearly normal angular acceptance.
$P_{\mathrm{band}}\approx B_{\lambda_0}A\Omega\Delta\lambda=2.63570414\times10^{-4}\,\mathrm W$

Result: The ideal channel receives about 0.26357 mW, not all power emitted by the source.

Check: The units cancel with area, solid angle and wavelength width. Independently integrating Bλ from 1.99 to 2.01 μm checks the band approximation; difference is below 0.003%. cosθ is approximated by 1 for small normal acceptance, not replaced by a hemisphere factor.

Example: two spectral peaks of the same radiator

Wavelength and frequency spectra at 3600 K

Given: Ideal blackbody $T=3600\,\mathrm K$ . Bλ and Bν are each densities per equal interval in their own coordinate.
Find: Find both peaks and express frequency-peak position as a wavelength.

Approach: Find the wavelength-density maximum; frequency and wavelength densities peak differently. $5(1-e^{-x_\lambda})=x_\lambda,\quad\lambda_{\max}=\frac{hc}{k_{\mathrm B}Tx_\lambda}$ . Differentiate Bλ proportional to $\frac{x^5}{e^x-1}$ ; its positive interior root satisfies $5(1-e^{-x})=x$ .
$x_\lambda=4.96511423,\quad\lambda_{\mathrm{peak}}=\frac{hc}{k_BT x_\lambda}=0.80493665\,\mathrm{\mu m}$
Bν proportional to $\frac{y^3}{e^y-1}$ gives $3(1-e^{-y})=y$ .
$y_\nu=2.82143937,\quad\nu_{\mathrm{peak}}=\frac{k_BT y_\nu}{h}=2.11641327\times10^{14}\,\mathrm{Hz}$
Translate frequency coordinate to wavelength without calling it the Bλ peak.
$\frac{c}{\nu_{\mathrm{peak}}}=1.41651190\,\mathrm{\mu m}$

Result: Bλ peaks at 0.80494 μm and Bν at 211.641 THz. That frequency corresponds to 1.41651 μm, not the Bλ peak.

Check: Both dimensionless roots satisfy their respective peak equations. The ratio $\frac{1.41651}{0.80494}\approx1.75978$ is independent of T. Total band energy remains consistent despite different peak coordinates.

Example: temperature from two spectral densities

Two ideal spectral channels infer one temperature

Given: A blackbody is measured in ideally narrow channels at 2.00 and 3.00 μm. Both readings are calibrated as radiance per the same wavelength unit. Their ratio is 1.63005. There is no unknown spectral filter or reflected background.
Find: Infer temperature from the density ratio and check the Wien approximation.

Approach: retain the Planck prefactor and both exponential denominators in the ratio. A ratio of raw detector powers with different bandwidths would not be this quantity.
$R(T)=\frac{B_{\lambda_1}(T)}{B_{\lambda_2}(T)}=\left(\frac{\lambda_2}{\lambda_1}\right)^5\frac{e^{\frac{hc}{\lambda_2k_{\mathrm B}T}}-1}{e^{\frac{hc}{\lambda_1k_{\mathrm B}T}}-1}$
Thermal occupation at 2 μm is more suppressed at lower temperature. A numerical root search over 1200–2000 K equates the full ratio to the observed value.
$R(1200\,\mathrm K)\approx1.01305,\quad R(2000\,\mathrm K)\approx2.14005,\quad T\approx1600\,\mathrm K$
Recalculate both radiances, then compare the short-wavelength approximation.
$B_{\lambda_1}\approx41.9742\,\mathrm{kW\,m^{-2}\,sr^{-1}\,\mu m^{-1}},\quad B_{\lambda_2}\approx25.7503\,\mathrm{kW\,m^{-2}\,sr^{-1}\,\mu m^{-1}}$

Result: The calibrated blackbody model gives approximately 1600 K. An unknown wavelength-dependent emissivity makes this inference ambiguous without additional information.

Check: The full recalculated ratio is 1.63005. The Wien expression drops both −1 terms and gives a slightly higher ratio at 1600 K. It is a checkable approximation here, not an equivalent exact rearrangement.

Connect radiance, exitance and total power

\begin{aligned}M_\lambda&=\int_{\mathrm{hem}}B_\lambda\cos\theta\,\mathrm d\Omega=\pi B_\lambda\\M&=\pi\int_0^\infty B_\lambda\,\mathrm d\lambda=\sigma T^4\end{aligned}

The π factor applies to direction-independent blackbody or diffuse radiance.[1][2]

A detector accepting a small solid angle does not automatically use π. That factor comes only from the complete outgoing hemisphere with projection. A sphere’s emitted power likewise uses surface area 4πR²; image projection πR² is a different geometric quantity. A continuous spectrum is consistent with discrete photon energies because frequency and wavelength themselves can vary continuously. The Stefan–Boltzmann law summarizes the complete blackbody surface and spectral evaluation, rather than power in one narrow detector channel.

Inferring temperature from light requires a model

For the comparison $T=5800\,\mathrm K$ at 500 nm gives $B_\lambda\approx2.68822\cdot10^{13}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ and $x\approx4.9613$ . This describes a black model radiator. Real stellar observations include spectral lines and absorption; real surfaces require emissivity and reflected surroundings. A single band value therefore determines T only with additional known factors.

The thermal expression is not a full model of a laser, LED or externally excited fluorescent material. Media with refractive index different from 1 need explicit wavelength and radiance conventions. These equations use vacuum and do not silently change refractive-index conventions.[2]

First choose the spectral quantity and its unit. For channel power, then integrate bandwidth and viewing angle; temperature inference additionally requires emissivity and calibration. Spectral peak, total emission and one channel’s power are different evaluations of the same radiation model.

Five problems on spectral density and measurement

Calculate · Foundation

$B_\lambda=2.0\cdot10^{11}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ . What is its value per nanometre?

Show hint

One nanometre interval is 10⁻⁹m wide.

Learning goal: Convert wavelength density units.

Show solution and calculation

Approach: a nanometre interval has width $10^{-9}\,\mathrm m$ , so density per nanometre has the SI numerical value multiplied by $10^{-9}$ .
This gives $200\,\mathrm{W\,m^{-2}\,sr^{-1}\,nm^{-1}}$ .

Check: both units give the same integrated radiance of $2000\,\mathrm{W\,m^{-2}\,sr^{-1}}$ across a constant 10 nm band.

B_{\lambda,\mathrm{per\ nm}}=200\,\mathrm{W\,m^{-2}sr^{-1}nm^{-1}}

Calculate · Application

A blackbody’s Bλ peak is at 1.20 μm. Find T.

Show hint

Use $b=\frac{hc}{k_{\mathrm B}x_\lambda}\approx2.897771955\cdot10^{-3}\,\mathrm{m\,K}$ .

Learning goal: Infer temperature from a wavelength peak.

Show solution and calculation

Approach: solve Wien’s displacement law for absolute temperature.
$T=\frac{b}{1.20\cdot10^{-6}\,\mathrm m}\approx2414.81\,\mathrm K$ .
This uses a wavelength-density peak, not a converted frequency peak.

Check: $\lambda_{\max}T$ recovers $b$ .

T\approx2414.81\,\mathrm K

Calculate · Application

At $\lambda=1.00\,\mathrm{\mu m}$ , $B_\lambda=2.0\cdot10^{11}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ . Find Bν at that spectral position.

Show hint

Use $B_\nu=\frac{B_\lambda\lambda^2}{c}$ .

Learning goal: Apply the spectral-density Jacobian.

Show solution and calculation

Approach: $B_\nu=\frac{B_\lambda\lambda^2}{c}$ .
Bν≈6.67128×10⁻¹⁰ W·m⁻²·sr⁻¹·Hz⁻¹ at ν≈2.99792×10¹⁴ Hz.
The Jacobian ensures equal energy in matching small bands.

Check: $B_\nu\frac c{\lambda^2}$ recovers the original wavelength density.

B_\nu=\frac{B_\lambda\lambda^2}{c}\approx6.67128\times10^{-10}\,\mathrm{W\,m^{-2}sr^{-1}Hz^{-1}}

Interpret · Application

A blackbody has approximately constant Bλ over two narrow bands: 100 units per nm across 2 nm and 60 of the same units per nm across 5 nm. Area, viewing direction, projected-area factor and solid angle are shared. Which band contributes more power, and why is the taller curve insufficient?

Show hint

Compare density times bandwidth.

Learning goal: Compare band areas rather than peak heights.

Show solution and calculation

Approach: Density times width approximates each flat narrow-band integral.
Contributions are 200 and 300 common integrated units, so the second band supplies 1.5 times as much.
Greater bandwidth outweighs lower density.
This uses the stated constant-density approximation and common geometry, and is not valid for arbitrary broad bands.

\begin{aligned}J_1&\approx100\cdot2=200\\{}J_2&\approx60\cdot5=300\\{}\frac{J_2}{J_1}&=1.5\end{aligned}

[2]

Error analysis · Application

Someone locates the Bλ maximum and claims that c/λmax must also maximize Bν because the same photons are described. Diagnose the error without calculating peak values; assume vacuum and the same temperature.

Show hint

Changing density coordinates preserves band integrals, not curve height.

Learning goal: Recognize coordinate-dependent spectral peaks.

Show solution and calculation

Approach: Transform the density with its Jacobian.
Frequency density additionally contains wavelength squared divided by c.
This factor varies across the spectrum and changes the location of a maximum.
Corresponding small bands contain the same radiation, but their densities and peak locations need not agree.
The claimed peak identity is therefore false.

B_\nu=B_\lambda\left|\frac{\mathrm d\lambda}{\mathrm d\nu}\right|=B_\lambda\frac{\lambda^2}{c}

[2]

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
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Frequently asked questions about Planck Spectral Law

What does Planck Spectral Law describe physically?+

Bλ describes how thermal radiation from an ideal blackbody in equilibrium is distributed over wavelength. It counts power per projected area, solid angle and wavelength interval, rather than total power. Temperature determines both spectral height and distribution. The denominator comes from mean occupation of quantized radiation modes; hν is excitation energy per quantum. Real surfaces can have wavelength-dependent emission and reflect incident radiation. A bright laser or LED therefore does not follow a blackbody thermal spectrum simply because it is luminous. To obtain blackbody exitance over a complete hemisphere, integrate the projected angular contribution; this supplies the factor π.

How can I check a calculation involving Planck Spectral Law?+

First convert λ to metres and evaluate $x=\frac{hc}{\lambda k_{\mathrm B}T}$ . This exponent must be dimensionless. For 5800 K and 500 nm, $x\approx4.96130$ and $B_\lambda\approx2.68822\cdot10^{13}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ . Next check whether the requested value is per metre, nanometre or micrometre; the per-nanometre numerical value is smaller by 10⁻⁹. For small x, an expm1 evaluation avoids subtracting nearly equal numbers inaccurately. A measured power also requires wavelength bandwidth, projected area and accepted solid angle. The factor π describes hemisphere integration and must not replace a detector’s small acceptance angle.

What assumptions and limits apply to Planck Spectral Law?+

The displayed form applies to equilibrium thermal radiation from an ideal blackbody in vacuum. T is positive and absolute, and λ is a positive vacuum wavelength. For a real thermal surface, suitable spectral and possibly directional emissivity multiplies blackbody radiance. Reflected background and transmission along the measurement path contribute to the observed signal. One constant ε is consequently insufficient for every spectrum. Nonthermal sources such as lasers require their own model. Even an ideal blackbody has no representation-independent “peak wavelength”: maxima of Bλ and Bν differ because wavelength and frequency intervals weight spectral density differently. Specify the convention before interpreting a peak.

How do I check units and scale for Planck Spectral Law?+

The unit W·m⁻³·sr⁻¹ is more transparent as W·m⁻²·sr⁻¹·m⁻¹. Two inverse metres describe area and the remaining inverse metre describes wavelength bandwidth; this is not volumetric power density. A radiance of 2.0×10¹¹ per metre equals 200 per nanometre. Across a 10 nm band, a constant approximation then yields 2000 W·m⁻²·sr⁻¹. This still is not power in watts: suitable area and angle integration are needed. A scale check is that π∫Bλdλ equals σT⁴ for a black surface. A finite measurement band captures only part of that integral, determined by bandwidth and spectral variation.

When should I combine Planck Spectral Law with another relation?+

Complete integration gives the Stefan–Boltzmann law after using $M_\lambda=\pi B_\lambda$ . Maximizing Bλ instead gives $\lambda_{\max}T\approx2.89777\cdot10^{-3}\,\mathrm{m\,K}$ . At 3600 K this maximum is about 0.80494 µm. The Bν maximum occurs near 211.641 THz; its converted wavelength, 1.41651 µm, is not the Bλ maximum. Changing representation requires $B_\lambda=B_\nu\frac c{\lambda^2}$ , rather than only $\nu=\frac c\lambda$ . Detector calculations combine the spectrum with bandwidth, area, angular acceptance and transmission. Stellar calculations additionally need emitting surface, distance and potentially absorption; a spectrum alone does not automatically determine total luminosity.

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Curated exam set for "Planck Spectral Law":

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Which formula describes Planck Spectral Law?

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How do you rearrange Planck Spectral Law for Calculate spectral radiance?

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Which common mistake happens with Planck Spectral Law?

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Related formulas

\lambda_{\max} T = b_W

Wien displacement law

More formulas: Heat and thermodynamics

\eta_{Carnot} = 1 - \frac{T_{kalt}}{T_{warm}}

Carnot Efficiency

Q = m \cdot c \cdot \Delta T

Heat Capacity: Q = m·c·ΔT

\Delta U = Q - W

First Law of Thermodynamics

\Delta L = \alpha L_0 \Delta T

Linear thermal expansion

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Planck’s radiation law

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How do you calculate with Planck Spectral Law?

What quantity does Planck’s radiation law describe?

Temperature changes spectrum height and peak position

Explore further: mean energy of a quantised mode

From mode counting to radiance

Which units and numerical limits apply to the Planck spectrum?

Example: turn radiance into narrow-channel power

An ideal narrow infrared channel

Example: two spectral peaks of the same radiator

Wavelength and frequency spectra at 3600 K

Example: temperature from two spectral densities

Two ideal spectral channels infer one temperature

Connect radiance, exitance and total power

Inferring temperature from light requires a model

Five problems on spectral density and measurement

Bλ=2.0⋅1011 W m−2 sr−1 m−1B_\lambda=2.0\cdot10^{11}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}Bλ​=2.0⋅1011Wm−2sr−1m−1. What is its value per nanometre?

A blackbody’s Bλ peak is at 1.20 μm. Find T.

At λ=1.00 μm\lambda=1.00\,\mathrm{\mu m}λ=1.00μm, Bλ=2.0⋅1011 W m−2 sr−1 m−1B_\lambda=2.0\cdot10^{11}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}Bλ​=2.0⋅1011Wm−2sr−1m−1. Find Bν at that spectral position.

A blackbody has approximately constant Bλ over two narrow bands: 100 units per nm across 2 nm and 60 of the same units per nm across 5 nm. Area, viewing direction, projected-area factor and solid angle are shared. Which band contributes more power, and why is the taller curve insufficient?

Someone locates the Bλ maximum and claims that c/λmax must also maximize Bν because the same photons are described. Diagnose the error without calculating peak values; assume vacuum and the same temperature.

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More formulas: Heat and thermodynamics

Retain Planck Spectral Law for exams

$B_\lambda=2.0\cdot10^{11}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ . What is its value per nanometre?

At $\lambda=1.00\,\mathrm{\mu m}$ , $B_\lambda=2.0\cdot10^{11}\,\mathrm{W\,m^{-2}\,sr^{-1}\,m^{-1}}$ . Find Bν at that spectral position.