# Blackbody Radiation - Peak wavelength

• teme92
In summary, the peak of the Planck spectrum for a blackbody at a temperature T occurs at the wavelength λ_{max}T=0.29 where T is in Kelvin.
teme92

## Homework Statement

The Planck blackbody spectrum is given by

$$u(ω,t)=\frac{ħω^3}{π^2c^3(e^{βħω}-1)}$$

Show that the peak of the Planck spectrum for a blackbody at a temperature T occurs at the wavelength

$$λ_{max}T=0.29$$

where T is in Kelvin and λmax is in cm.

## Homework Equations

$$\frac{d(ω,T)}{dω}=0$$

## The Attempt at a Solution

So using the equation above I get down to:

$$(3-βħω)(e^{βħω})-3=0$$

This is a transcendental equation. I said:

$$x=βħω$$

I've tried using the bisection method to solve but my answer doesn't match up with what it should as from the Wien Displacement Law it should be 0.29 for λ in cm. If anyone could help explain the solving for x, I would greatly appreciate it.

This one is somewhat complicated by the fact that you first need to convert the Planck function to a wavelength spectral distribution before you find the maximum in the wavelength spectral distribution (density) function. You get a different answer if you try to work the problem (finding the maximum) with the distribution (density) function in frequency space.

If its converted to the wavelength I got.

$$x=\frac{hc}{λkT}$$

$$5(e^x-1)=xe^x$$

Which is a similar equation to solve. Where would I go from here?

## e^{-x}=1-x/5 ##. Numeric solution with x not equal to zero. x=4.95 approximately.

I don't understand how you solved for x. I know normally with an exponential, a natural log would be involved.

## e^{-x} ## is going to be nearly zero. On a first iteration, ## x=5 ##. Taylor series can get you about 2 or 3 decimal places in one or two iterations. Otherwise iterate ##x=5(1-e^{-x} ) ## without any Taylor series. That's probably much quicker.

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Additional comment: You already did the hardest part. Using iterative method (see post #6), you can start with ##x=5 ## on the right side and I think you get ## x=4.95 ## on the left. Take this 4.95 and insert it into the right side, and I think you get close to 4 decimal place accuracy for your answer. This is the number you need to write out Wien's law with extreme precision. (## h ## ,## c ##, and ## k_b ## are already known. Plug it in and you should have the right answer.)

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## 1. What is blackbody radiation?

Blackbody radiation refers to the electromagnetic radiation emitted by an object that absorbs all wavelengths of light and reflects none. It is a fundamental concept in physics and is used to describe the thermal radiation emitted by objects at a certain temperature.

## 2. What is the peak wavelength of blackbody radiation?

The peak wavelength of blackbody radiation refers to the wavelength at which the intensity of the radiation is the highest. This wavelength is determined by the temperature of the object and follows the Wien's displacement law, which states that the peak wavelength is inversely proportional to the temperature of the object.

## 3. How is the peak wavelength of blackbody radiation calculated?

The peak wavelength of blackbody radiation can be calculated using the Wien's displacement law, which states that the peak wavelength (in meters) is equal to the constant 2.898 × 10^-3 divided by the temperature (in Kelvin).

## 4. What is the significance of the peak wavelength in blackbody radiation?

The peak wavelength in blackbody radiation is significant because it provides information about the temperature of the object. The higher the temperature, the shorter the peak wavelength will be. This information is useful in various fields such as astronomy and thermodynamics.

## 5. How does the peak wavelength of blackbody radiation change as the temperature increases?

As the temperature of the object increases, the peak wavelength of blackbody radiation decreases. This means that the radiation shifts towards shorter wavelengths, such as from infrared to visible light. This phenomenon is known as the "red shift" and is observed in various astronomical objects, such as stars and galaxies.

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