## Whoa, black body (bam-ba-lam): part two

In part one, we looked at the fact that the hotter an object then the greater the intensity of electromagnetic radiation that will be emitted. For simplicity, we looked at so-called ‘blackbodies’ — that is say, objects which are perfect absorbers (hence ‘blackbodies’) and more importantly, perfect emitters of electromagnetic radiation.

To human eyes, things look very dull in the visible part of the electromagnetic spectrum until we reach temperatures of several hundreds of degrees — however, objects at room temperature (or just above) glow brightly in the infrared part of the electromagnetic spectrum, as we can see easily if we have access to an infrared camera.

By ‘intensity’ of course, we mean the power (‘energy per second’) emitted per unit area.

This links in neatly with 4.6.3.2 of the 2015 AQA GCSE Physics specification:

#### Stretch and challenge for students (1): Is the intensity of emitted radiation directly proportional to the temperature of the object?

The short answer is no. If you doubled the temperature (measured in kelvins!) of an object then the intensity of radiation would increase by a factor of 16. In other words, the intensity I of radiation emitted by an object is directly proportional to the absolute temperature T raised to the power of 4.

This is a consequence of the Stefan-Boltzmann radiation law (covered in A-level Physics):

In part 1 we estimated the intensity of radiation emitted by two blackbodies by ‘counting squares’ to find the area underneath a graph. We can show that the values obtained are consistent with the Stefan-Boltzmann radiation law.

Since we have dealt comprehensively with the relationship between intensity of radiation and temperature, I propose to move along and look at how the wavelength distribution changes with the temperature of the body.

### How does the temperature of a blackbody affect the distribution of emitted wavelengths?

Let’s consider an object that approximates to a blackbody: the filament of an old school incandescent lamp.

The graph of the radiation produced by both objects is shown below.

First, let’s look at the visible wavelengths produced by both bulbs.

• The 1700 degree Celsius bulb produces only a very small amount of visible light and the vast majority of that is towards the red end of the spectrum: you can see the section where the left hand edge of the 1700 curve just nicks the visible light wavelengths. This means that the 1700 degree filament emits a barely perceptible reddish glow to our eyes with its peak output still firmly in the infrared.
• The 2200 degree Celsius bulb produces a much larger amount of visible light: look at the left hand side of the curve. What is more, it appears as white light to our eyes since it includes all the colours of the rainbow. However, it’s still a very reddish-tinged white. Photographs taken in artificial light with chemical films (very old school!) had to be taken using special colour balanced film stock otherwise this bias was very evident in the final print(!) Modern digital cameras have software that automatically compensates for artificial vs. daylight colour balance issues.

Second, let’s look at the position of the peak wavelength.

• The 1700 degree Celsius bulb has its peak output at a wavelength of 1.5 x 10-6 m (shown by the blue dotted line on the graph).
• The 2200 degree Celsius bulb has its peak output at a wavelength of 1.2 x 10-6 m (shown by the red dotted line on the graph.)

Assuming that you wanted to, these findings could be summarised in song (sung to the tune of ‘Black Betty’ by Ram Jam):

```Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
More heat, peak shifts left
Waves out with more zest
Wavelengths out not alike
Some hues power spike!
Whoa, black body (Bam-ba-lam)
Whoa, black body
Bam-ba-laaam, yeah yeah
```

#### Stretch and challenge for students (2): predicting the position of the peak output wavelength

The position of the peak output wavelength can be predicted using Wien’s Displacement Law (studied in A-level Physics:

As we can see, the peak output wavelength on the graph agrees well with the position as calculated by Wien’s Displacement Law.

## Whoa, black body (bam-ba-lam)

The 2015 AQA GCSE Physics specification (4.6.3.1) asks that students understand that:

In my experience, students find it natural to accept that bodies absorb electromagnetic radiation — but surely only extremely hot objects (like the filament of a light bulb) emit electromagnetic waves?

This is a consequence of the fact that our eyes can only detect the tiny slice of the vast electromagnetic spectrum.

That tiny sliver known as ‘visible light’ looks insultingly small even on a diagram with a logarithmic scale as shown above. On a linear scale, it’s even worse: if we represented the em spectrum by a line stretching from London to New York, then the range of wavelengths that human eyes can detect would be a strip two centimetres wide.

This calls to my mind some lines quoted many years ago by Arthur C. Clark in his wonderful essay ‘Things We Cannot See’: A being who hears me tapping / The five-sensed cane of mind / Amid such greater glories / That I am worse than blind.

### Seeing the unseeable

A Leslie’s cube is a cuboid with black and silver coloured faces that can filled with hot water.

In visible light, there is no difference between its appearance when at room temperature (say 15 degrees Celsius) and when filled with hot water (say 70 degrees Celsius).

However, seen through an infrared camera, things look very different: the hot sides glow brightly, emitting huge amounts of infrared em waves.

There is another effect: the black coloured side throws out more infrared than the silver side. Why? Because any object which is good at absorbing em radiation is also good at emitting radiation.

As the AQA GCSE spec puts it:

By this definition, the Sun is a good approximation of a black body since it absorbs nearly all of the radiation falling on it (from other stars! — as well as the odd photon bounced back from minuscule specks like the Earth) as well being highly effective at emitting em radiation.

4.6.3.2 of the AQA GCSE Physics spec says:

One of the ways to cover this is to look at the radiation curves of two black bodies at different temperatures (pdf here). Both of these objects are at relatively low temperatures, so they emit most of their energy in the infrared part of the em spectrum. The visible light range is shown by the coloured bar just to the right of the y-axis.

Because it is a perfect emitter — as well as absorber — of radiation, the intensity (power per unit area) of emitted radiation from a black body depends only on the temperature of the black body.

### Estimating the total power emitted per unit area

We can estimate the total power emitted per unit area by approximating the area underneath the curve. We’re going to count any square which is larger than a half square as one whole square and ignore any part squares which are smaller than a half square.

The area under the blue curve is 325 W/m^2.

The intensity of radiation emitted by the hot object (red curve) is larger than the intensity emitted by the cold object (blue curve).

We will look at the distribution of wavelengths in a later post.

### Whoa, black body (bam-ba-lam)

Physics can occasionally, go better with a song.

```Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
Black body e.m. waves (Bam-ba-lam)
Come out with peak shapes (Bam-ba-lam)
Planck said, “I’m worryin’ outta mind (Bam-ba-lam)
UV won’t align!” (Bam-ba-lam)
He said, oh black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)

Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
Planck set out to Quantise (Bam-ba-lam)
UV don’t catastrophize! (Bam-ba-lam)
“No prob now,” said he (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)

Get it!

Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
More heat, peak shifts left
Waves out with more zest
Wavelengths out not alike
Some hues power spike!
Whoa, black body (Bam-ba-lam)
Whoa, black body
Bam-ba-laaam, yeah yeah

(with apologies to Huddie Ledbetter)
```