Wien Displacement Law Simulator — Peak Wavelength of Black-Body Radiation
Compute the Wien displacement law lambda_max*T = b and the Stefan-Boltzmann law q = eps*sigma*T^4 in real time from temperature T, emissivity eps, surface area A and observation distance R, and visualise the peak wavelength, radiant exitance, total radiated power and irradiance at distance R. The Planck curve with visible-band shading and a log-log T-lambda_max line let you study the Sun, tungsten filaments, the human body and cosmic background radiation under one consistent framework.
Parameters
Temperature T
K
Emissivity eps
Surface area A
m²
Distance R
m
Defaults: T = 5800 K (solar surface), eps = 1.0 (black body), A = 1.0 m^2 (unit area), R = 10.0 m (point-source observation). The relation E = P/(4*pi*R^2) holds when R is much larger than the radiator size.
Results
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Peak wavelength lambda_max
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Radiant exitance q
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Total radiated power P
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Irradiance E at distance R
Planck spectrum and peak wavelength
Wavelength lambda (nm) on the x-axis (0-3000), spectral radiance (relative) on the y-axis. The orange curve is the Planck law and the yellow marker is lambda_max. The pale band marks the visible range (380-780 nm). Higher T shifts the curve to the left and upward.
T - lambda_max relation (log-log)
Temperature T (K) on the x-axis (100-10000), lambda_max (nm) on the y-axis, both logarithmic. The blue line is the Wien displacement law lambda_max = b/T (slope -1). The yellow marker is the current operating point (T, lambda_max).
Theory & Key Formulas
The peak wavelength and total radiated energy of a black body are described by the Wien displacement law and the Stefan-Boltzmann law respectively.
Wien displacement law:
$$\lambda_{\max}\,T = b,\quad b = 2.898\times10^{-3}\ \mathrm{m\cdot K}$$
Stefan-Boltzmann law (radiant exitance per unit area):
Total radiated power and point-source irradiance at distance R:
$$P = q\,A,\qquad E = \frac{P}{4\pi R^{2}}$$
$T$ is the absolute temperature [K], $\varepsilon$ is the emissivity (1 for a black body), $A$ is the radiating surface area [m^2], $R$ is the observation distance [m], $b$ is the Wien displacement constant and $\sigma$ is the Stefan-Boltzmann constant. $\lambda_{\max}$ is the wavelength at which the spectrum of a black body at temperature $T$ peaks; higher $T$ shifts it to shorter wavelengths.
What is the Wien Displacement Law Simulator?
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With the default T = 5800 K the tool says lambda_max = 500 nm. Is that the Sun? And why does temperature alone fix the peak wavelength?
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Spot on. The Wien displacement law lambda_max*T = b with b = 2.898e-3 m*K says that the wavelength at which a black-body spectrum peaks depends only on temperature. It comes out of differentiating the Planck law and setting the derivative to zero, and is independent of size or material - emissivity eps only changes the height of the whole spectrum, not the peak position. The Sun at 5800 K peaks at about 500 nm (green), a tungsten bulb at 2800 K at 1035 nm (near infrared), and the human body at 310 K at 9.4 um (mid infrared). Thermal cameras see people because the human peak sits right in the infrared band their detectors are tuned to.
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The right log-log chart looks like a clean straight line. Is the slope -1 also a consequence of this law?
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Yes. Taking the log of lambda_max = b/T gives log(lambda_max) = log(b) - log(T), which is a straight line of slope -1 on log-log axes. Click Sweep T and watch the yellow marker glide along that line as T runs from 100 to 10000 K. From the intercept you can even read off log(b) = -2.54 in m*K (or +6.46 with lambda in nm), so the chart is essentially a graphical recipe for measuring the Wien constant from a series of black-body spectra at different temperatures.
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What is the pale band under the Planck curve? The Sun seems to peak right inside it.
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That is the visible band 380-780 nm. The Sun at 5800 K peaks right inside it, which is often given as an evolutionary explanation for why life on Earth is built around visible light. Drop T to 2000 K and the peak escapes far to the right (infrared), with almost no visible component left. Push T to 10000 K (a hot blue-giant star) and the peak is in the ultraviolet, with the visible band catching only the long-wavelength tail. A tungsten bulb looks dim and yellowish for the same reason: its peak is in the infrared, so most of its energy is wasted as heat. White LEDs sidestep the law entirely by exciting narrow visible bands directly with electricity, which is why their luminous efficacy is so much higher.
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The stat cards show P = 64.2 MW and E = 51.1 kW/m^2 at R = 10 m. That is enormous for one square metre of solar surface seen from 10 metres away.
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Yes - the Stefan-Boltzmann law q = eps*sigma*T^4 is fourth-power in T, so 5800 K already gives q = 64.2 MW/m^2. With A = 1 m^2 the total radiated power is P = 64.2 MW, and the point-source approximation E = P/(4*pi*R^2) gives 51.1 kW/m^2 at 10 m. The real Sun has radius 6.96e8 m and radiates exactly that q from its photosphere; stretch R out to the Earth orbit, 1.496e11 m, and you recover about 1.4 kW/m^2, the solar constant. Within the same framework you can size tungsten filaments, LED substrates, photovoltaic modules and even human-body radiation, which is why the same equations sit under almost every CAE thermal-radiation boundary condition.
Physical model and key equations
The Wien displacement law follows from differentiating the Planck law $B_{\lambda}(\lambda,T) = (2hc^2/\lambda^5) / (\exp(hc/\lambda k_{B}T) - 1)$ with respect to $\lambda$ and setting the derivative to zero.
$$\lambda_{\max}\,T = b,\qquad b = \frac{hc}{k_{B}\,x_{W}} = 2.898\times10^{-3}\ \mathrm{m\cdot K}$$
Here $x_{W} = 4.965114\ldots$ is the numerical solution of $x e^{x} - 5(e^{x}-1) = 0$, $h$ is the Planck constant, $c$ is the speed of light and $k_{B}$ is the Boltzmann constant. Integrating the same Planck spectrum over all wavelengths returns the Stefan-Boltzmann law: $q = \int_{0}^{\infty} \pi B_{\lambda}\,d\lambda = \sigma T^{4}$ with $\sigma = 2\pi^{5}k_{B}^{4}/(15h^{3}c^{2}) = 5.670\times10^{-8}\ \mathrm{W/(m^{2}\cdot K^{4})}$. A real surface is treated as a gray body with $\varepsilon \le 1$ and gives $q = \varepsilon\sigma T^{4}$.
For a radiator of surface area $A$, the total radiated power is $P = q A$, and assuming an isotropic point source the irradiance on a sphere of radius $R$ is $E = P/(4\pi R^{2})$. The point-source approximation requires $R$ to be much larger than the radiator size; in the near field a view factor must be used and the net exchange between two surfaces becomes $Q_{\mathrm{net}} = \varepsilon\sigma A F_{12}(T_{1}^{4}-T_{2}^{4})$.
Internally the simulator draws the spectrum in the dimensionless self-similar form $u(x) = (1/x^{5})/(\exp(x_{W}/x)-1)$ with $x = \lambda/\lambda_{\max}$, so changing $T$ rescales the curve horizontally without changing its shape - the famous Wien-Planck similarity property of black-body radiation.
Real-world applications
Solar irradiance and photovoltaic design: the Sun (T = 5800 K) peaks at lambda_max = 500 nm and concentrates about 43% of its radiated energy in the visible band 380-780 nm. Plug the defaults into the tool and you get q = 64.2 MW/m^2; stretch R to the Earth orbit (1.496e11 m) and the irradiance drops to 1.36 kW/m^2 - the solar constant. This sets the Shockley-Queisser limit (about 33%) for silicon cells with a 1.12 eV band gap and a long-wavelength cut-off near 1100 nm, and directly informs the spectral splitting in tandem and perovskite photovoltaics.
Incandescent bulbs and LED lighting: a tungsten filament at 2800 K peaks at 1035 nm in the near infrared, so only the long-wavelength tail of its spectrum reaches the visible band, giving a luminous efficacy of just 10-15 lm/W. Pushing T to 3300 K shortens lambda_max to 878 nm and almost doubles q (factor 1.94), but Arrhenius-driven evaporation collapses the lifetime. White LEDs avoid the constraint entirely by directly exciting narrow visible bands, reaching efficacies above 100 lm/W.
Infrared thermography and ear thermometers: the human body (T = 310 K) peaks at lambda_max = 9.4 um, well into the mid infrared, so VOx microbolometers and InSb/HgCdTe quantum detectors can image people at room temperature. Set T = 310 K in the tool and lambda_max falls off the 0-3000 nm chart to the right, showing why detector and lens designs must be tuned to the 8-14 um atmospheric window. Applications run from power-plant pipe diagnostics through building-envelope inspection to non-contact fever screening.
Stellar classification and colour temperature: astronomers use the colour of starlight to infer surface temperature. Red giants such as Betelgeuse (T = 3500 K, lambda_max = 828 nm), the Sun (G2V, 5800 K, 500 nm), white dwarfs such as Sirius B (25000 K, 116 nm) and blue giants such as Rigel (11000 K, 263 nm) span ultraviolet to infrared peaks. Sweeping T from 100 to 10000 K shows visually that only a narrow window (about 3700-7600 K) places the peak inside the visible band, highlighting how special the Sun is. The Hertzsprung-Russell diagram is essentially a plot of colour temperature against luminosity sigma*T^4 * 4*pi*R_star^2.
Common misconceptions and pitfalls
The most common mistake is to assume that the wavelength peak and the frequency peak are simply linked by c = lambda * nu. The wavelength distribution $B_{\lambda}$ and the frequency distribution $B_{\nu}$ are linked by a Jacobian $|d\lambda| = |c/\nu^{2}| d\nu$, and therefore peak at different wavelengths even at the same temperature. The wavelength form gives $\lambda_{\max} T = 2898\ \mu\mathrm{m\cdot K}$ (with $x_{W} = 4.965$), while the frequency form gives $\nu_{\max}/T = 58.79\ \mathrm{GHz/K}$ (with $x_{W}' = 2.821$); the corresponding wavelength $c/\nu_{\max} = 5.099\ \mathrm{mm\cdot K}/T$ is 1.76 times longer than the wavelength peak. Always check whether a paper is using the wavelength or frequency form. This simulator uses the wavelength form throughout.
Another pitfall is to claim that real surfaces are not black bodies, so the Wien law does not apply. In the gray-body approximation (eps weakly dependent on wavelength), the peak position is still b/T - emissivity only lowers the height of the whole spectrum. Only for selective emitters with strongly wavelength-dependent eps - cavity emitters, photonic crystals, two-dimensional materials - does the effective peak deviate from the black-body prediction. The simulator treats eps as wavelength-independent and is therefore correct for gray surfaces; selective emitters require integrating $B_{\lambda}(\lambda,T)\,\varepsilon(\lambda)$ explicitly.
Finally, the point-source formula E = P/(4*pi*R^2) does not work at all distances. It is a far-field approximation valid only when the radiator size sqrt(A) is much smaller than R; in the near field the view factor F is much larger than 1/(4*pi*R^2). For a sphere of diameter D evaluated at its surface (R = D/2) the true F is 0.5, several times larger than the point-source prediction 1/(pi*D^2). CAE thermal analyses use the net exchange $Q_{\mathrm{net}} = \varepsilon\sigma A F_{12}(T_{1}^{4}-T_{2}^{4})$ with view factors obtained from geometric integration or closed-form expressions for cuboids, coaxial cylinders and parallel plates. The simulator is appropriate for far-field estimates such as Sun-Earth and for q and P at the radiating surface itself.
FAQ
The Wien displacement law states that the peak wavelength lambda_max of a black-body spectrum and its absolute temperature T are linked by lambda_max*T = b, with b = 2.898e-3 m*K (the Wien displacement constant). Higher temperatures shift the peak to shorter wavelengths in the visible and ultraviolet. With the default T = 5800 K (solar surface) the tool returns lambda_max = 2.898e-3 / 5800 = 5.0e-7 m = 500 nm, in the green part of the visible band, which is why sunlight peaks within human vision.
The Stefan-Boltzmann law q = eps*sigma*T^4 gives the total radiant exitance per unit area of a black-body or a gray body of emissivity eps and depends strongly on the fourth power of T. The Wien displacement law lambda_max*T = b gives the peak wavelength of the same spectrum and fixes the spectral shape. Both follow from the Planck law B_lambda(lambda,T): integrating over all wavelengths gives sigma*T^4, while differentiating and setting the derivative to zero yields b/T. The simulator displays both on one screen.
The total radiated power is P = q*A = eps*sigma*T^4*A, and assuming an isotropic point source the irradiance on a sphere of radius R is uniform: E = P / (4*pi*R^2). With T = 5800 K, eps = 1.0, A = 1.0 m^2 and R = 10.0 m the tool returns q = 6.42e7 W/m^2, P = 64.2 MW and E = 51.1 kW/m^2. Stretching R to the Earth orbit (1.496e11 m) recovers about 1.4 kW/m^2, the solar constant. The point-source approximation requires R to be much larger than the radiator size.
The visible band spans roughly lambda = 380 to 780 nm, so T = b/lambda_max gives a temperature window of T = 2.898e-3 / 780e-9 = 3715 K (red end) to T = 2.898e-3 / 380e-9 = 7626 K (violet end). Tungsten filaments at about 2800 K peak at 1035 nm in the near infrared and waste most energy as heat, while the Sun at 5800 K places the peak right in the middle of the visible band. Sweeping T from 100 to 10000 K with the play button visualises the visible-band crossing for white-LED and solar-cell wavelength design.