About 99% of the solar power reaching Earth's surface has a wavelength between \(\SI{300}{\nano\meter}\) and \(\SI{2.5}{\micro\meter}\). This is a wide range of wavelengths: solar photons can differ in wavelength by a factor of 10.

This wide range of available wavelengths, as well as their distribution (i.e. which wavelengths are most common), is important to many solar energy conversion technologies.

Which wavelengths make up the peak intensity of solar radiation?

**Hint**: keep in mind that life on Earth evolved to be sensitive to electromagnetic radiation with wavelengths it was most exposed to (i.e., the wavelengths with the highest intensity). If a particular type of energy or radiation isn't available on Earth, it can't influence what happens here.

The solar spectrum at Earth’s surface is shown in the plot below. Higher values mean more solar energy comes in the form of photons with that wavelength. The peak intensity of solar radiation is in the visible spectrum, but there is still a significant amount of solar energy in ultraviolet and infrared wavelengths.

Approximately \(4\%\) of the solar energy reaching Earth's surface is in the ultraviolet spectrum, \(44\%\) is in the visible spectrum, and \(52\%\) is in the infrared spectrum. There is more energy contained in the infrared spectrum despite it having lower intensity because the infrared portion of the solar spectrum has a much wider range of wavelengths than the visible spectrum.

Suppose we have a solar collector which is only able to utilize photons that have an energy of at least \(\SI{1.1}{\electronvolt}\). (We will learn about a collector that works like this in Chapter 3.)

Approximately what portion of solar energy is carried by photons that have an energy of at least \(\SI{1.1}{\electronvolt}\)?

Recall that the Planck relation, which relates photon energy \(E\) and wavelength \(\lambda\), is \(E = hc/\lambda,\) where the product of the Planck constant \(h\) and the speed of light \(c\) is \(\SI{1240}{\electronvolt\nano\meter}\).

The wide range of wavelengths in the solar spectrum is explained by the fact that sunlight originates as thermal radiation from the Sun. Everything emits thermal radiation, but the intensity and wavelength distribution of that radiation depends on its temperature.

A glowing coal is hot enough to emit some visible radiation, which is why it appears to glow, whereas most objects we encounter in everyday life only emit infrared radiation, which we cannot see.

A

black bodyis an object that absorbs all incident radiation and emits the maximum amount of thermal radiation that any object can emit (for its given temperature). Objects at very high temperature tend to act like black bodies.

The spectral distribution of thermal radiation for a black body emitter is given by **Planck’s law**, and depends on the temperature \(T\) (in Kelvin) of the black body a la

\[\displaystyle B_{\lambda}(\lambda,T) = \frac{2\pi h c^2}{ \lambda^5}\frac{1}{e^{h c/\lambda k_B T}-1},\]

where \(B_{\lambda}\) is the spectral irradiance of the body in terms of wavelength \(\lambda\), typically reported in units of \(\si[per-mode=symbol]{\watt\per\meter\squared\per\nano\meter}\). Here \(c = \SI[per-mode=symbol]{3e8}{\meter\per\second}\) is the speed of light, \(h\) is the Planck constant (for this equation it is more useful to use standard SI units, in which case \(h = \SI{6.626e-34}{\joule\second}),\) and \(k_B = \SI[per-mode=symbol]{1.38e-23}{\joule\per\kelvin}\) is the Boltzmann constant.

From Planck's law, we can see that a black body emitter will have a peak emission wavelength (a wavelength which corresponds to the highest spectral radiance for that emitter), which will be a function of the temperature of that emitter.

Recall Planck's law: \(\displaystyle B_{\lambda}(\lambda,T) = \frac{2\pi h c^2}{\lambda^5}\frac{1}{e^{hc/\lambda k_B T}-1}\)

You can find the relationship between the peak emission wavelength and temperature by taking the derivative of spectral irradiance with respect to wavelength and setting it to zero. This finds the maximum spectral irradiance value because it finds the point on the curve that corresponds to zero slope.

The peak solar wavelength is around \(\SI{500}{\nano\meter}\). What temperature (in Kelvin) corresponds to a black body with a peak emission wavelength of \(\SI{500}{\nano\meter}\)?

**Remember**:

- \(c = \SI[per-mode=symbol]{3e8}{\meter\per\second}\)
- \(h = \SI{6.626e-34}{\joule\second}\)
- \(k_B = \SI[per-mode=symbol]{1.38e-23}{\joule\per\kelvin}\)

**Hint**:

The solution to \(\frac{x e^x}{e^x-1}-5=0\) is \(x \approx 4.965\). (This should come in handy when you take the derivative of Planck's law with respect to wavelength and set it equal to zero)

Solar photons have a wide range of wavelengths, and the solar spectral irradiance curve is characteristic of thermal radiation from a black body at \(\SI{5800}{\kelvin}\). This is because Sunlight is thermal radiation from the Sun, which has a surface temperature of around \(\SI{5800}{\kelvin}\).

The actual spectrum which reaches Earth’s surface looks a bit different from black body radiation from a \(\SI{5800}{\kelvin}\) emitter, primarily due to absorption in the Earth's atmosphere. Certain gas species in the atmosphere absorb over narrow wavelength ranges, which leads to "missing" portions of the solar spectrum.

This spectrum represents the distribution of wavelengths available in solar radiation. Many solar energy systems are optimized to convert this particular spectrum to electricity, since it represents the energy flow available to us on Earth. We'll learn about some of these systems in future chapters of this exploration.

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