The atmosphere, the ozone layer and the ionosphere

Sunlight on its way down

The Sun sends out waves of every band. Not all of them reach the ground. The air has a window for visible light and for part of the radio waves, and it blocks much of the rest. That is lucky for us. This lesson looks at three effects.

The ozone layer and ultraviolet

Ultraviolet light is divided into three parts by its wavelength. We give the photon energies.

• UV-A, $315$ nm to $400$ nm, photon energy $3.1$ eV to $3.9$ eV. Most of it reaches the ground.
• UV-B, $280$ nm to $315$ nm, photon energy $3.9$ eV to $4.4$ eV. Ozone takes in most of it. What remains causes sunburn.
• UV-C, $100$ nm to $280$ nm, photon energy $4.4$ eV to $12.4$ eV. The air and the ozone layer take in all of it.

A typical chemical bond in a living cell, such as the C-C bond, needs about $3.6$ eV to break. A photon of UV-B or UV-C has more than this energy, so it can break bonds in the molecules of DNA and damage a cell. A photon of visible light, at most $3.1$ eV, cannot. That is why ultraviolet is harmful and visible light is not.

Ozone is a molecule of three oxygen atoms, $\text{O}_3$. A layer of it lies in the stratosphere, roughly $15$ km to $35$ km above the ground. Ultraviolet light splits $\text{O}_2$ molecules into atoms, the atoms join other $\text{O}_2$ molecules to make ozone, and ozone in turn takes in ultraviolet. This cycle takes in the harmful part of the Sun's ultraviolet. Chlorine from certain gases (CFCs) destroys ozone. An international agreement, the Montreal Protocol of 1987, ended their use.

The greenhouse effect and infrared

The Sun's light warms the ground. The ground, in turn, gives out infrared. Wien's law puts its peak at $10.1\ \mu\text{m}$.

The gases carbon dioxide, water vapour and methane take in infrared. They send part of it back to the ground. This extra warming is the greenhouse effect. It is natural, and without it the Earth would be frozen. Adding more of these gases to the air increases it.

A balance of power shows the size of the effect. The Earth takes in sunlight on a disc of area $\pi R^2$ and gives out heat from a sphere of area $4\pi R^2$. With the solar constant $S = 1361$ W/m² and a reflected fraction (the albedo) of $0.30$,

A worked example. Find the temperature of the Earth with no greenhouse effect. Use $\sigma = 5.67 \times 10^{-8}\ \text{W/(m}^2\ \text{K}^4\text{)}$.

1. Power absorbed in each square metre of surface. $\dfrac{1361 \times 0.70}{4} = 238\ \text{W/m}^2$.
2. The temperature. We want $T$, so $T^4 = \dfrac{238\ \text{W/m}^2}{\sigma} = 4.20 \times 10^{9}\ \text{K}^4$. The fourth root gives $T = 255$ K.
3. Compare. The real average is $288$ K. The greenhouse effect warms the Earth by about $33$ K.

Radio waves and the ionosphere

High in the atmosphere, between about $60$ km and $400$ km, sunlight and cosmic rays strip electrons from atoms. The result is a layer of free electrons and ions, the ionosphere. Its electron density $N$ is largest in the F layer, a few hundred kilometres up, with about $10^{12}$ electrons in each cubic metre by day. The lower E layer, near $100$ km, has about $10^{11}$.

The free electrons react to a radio wave. A wave with a frequency below the critical frequency $f_c$ is turned back. A wave above it passes through. The critical frequency for a wave going straight up is

with $f_c$ in Hz and $N$ in m⁻³. For a wave that meets the layer at an angle of incidence $i$, the highest frequency that returns is $f_c/\cos i$.

Set the frequency to $12$ MHz and the electron density to $6 \times 10^{11}$. At what angle of incidence does the line meet the curve? Below that angle the signal goes through into space. At larger angles it returns.

A worked example. An F layer at noon has $N = 1 \times 10^{12}$ electrons in each cubic metre, and an E layer has $1 \times 10^{11}$. Find the critical frequency of each, and the highest frequency the E layer returns at an angle of incidence of $60^\circ$.

1. The F layer. $f_c = 9\sqrt{1 \times 10^{12}}\ \text{Hz} = 9.0$ MHz.
2. The E layer. $f_c = 9\sqrt{1 \times 10^{11}}\ \text{Hz} = 2.85$ MHz.
3. The oblique wave. $f = \dfrac{f_c}{\cos i} = \dfrac{2.85\ \text{MHz}}{\cos 60^\circ} = 5.69$ MHz.

So the layer returns the waves of short-wave radio, $3$ MHz to $30$ MHz, over long distances by bouncing them between the layer and the ground. Medium-wave (AM) broadcast radio also travels along the ground. The ionosphere does not return waves of FM radio, television and mobile phones, whose frequencies are above $30$ MHz. They go through it.

Three ways to reach a distant place

• Ground wave. It follows the curve of the Earth. It works for low frequencies, such as AM radio below a few MHz.
• Sky wave. The ionosphere bends it back. It works for the short-wave band, and it reaches far.
• Space wave. It goes in a straight line from antenna to antenna, and it is stopped by the horizon. It is used by FM, television and mobile links, which is why those antennas stand on tall towers. For a tower of height $h$, the distance to the horizon is $d = \sqrt{2Rh}$, with $R = 6.4 \times 10^{6}$ m. Satellite links use frequencies that go through the ionosphere.

A worked example. A tower is $150$ m tall, and the receiving antenna is $25$ m above the ground. Then $d_1 = \sqrt{2 \times 6.4 \times 10^{6}\ \text{m} \times 150\ \text{m}} = 43.8$ km, and $d_2 = \sqrt{2 \times 6.4 \times 10^{6}\ \text{m} \times 25\ \text{m}} = 17.9$ km. They can be at most $d_1 + d_2 = 61.7$ km apart.

Practise with new numbers

Try it in Python

The program first balances the sunlight that the Earth takes in against the heat that it gives out, to find its temperature with no greenhouse effect. Then it finds the critical frequency for three electron densities.

Common mistakes

• Saying that the greenhouse effect is the same as global warming. The effect is natural and keeps the Earth warm. The concern is the increase of it from added gases.
• Believing that ozone takes in infrared. The ozone layer takes in ultraviolet. The greenhouse gases take in infrared.
• Using the density in cm⁻³ in $f_c = 9\sqrt{N}$. The formula needs $N$ in m⁻³.
• Thinking that all radio waves bounce off the ionosphere. Only the waves below the critical frequency (for a vertical wave) return. FM, TV and microwaves pass through.

Remember these

• UV-B and UV-C photons are more energetic than the bonds of a cell, so they harm it. The ozone layer takes in most of them.
• The greenhouse gases take in the infrared of the ground. Without them the Earth would be about $33$ K colder.
• The ionosphere returns waves below $f_c = 9\sqrt{N}$ (Hz, with $N$ in m⁻³), and $f_c/\cos i$ at an angle $i$.
• Ground wave, sky wave, space wave. Horizon: $d = \sqrt{2Rh}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why can a photon of ultraviolet light harm a living cell when a photon of visible light cannot?

2. An ionosphere layer has $4.0 \times 10^{11}$ electrons in each cubic metre. Find the critical frequency. Will a $1$ MHz wave sent straight up return?

3. How far is the horizon from the top of a tower that is $100$ m tall? How tall must the tower be to double that distance?

4. A student says that the greenhouse effect is a mistake of the atmosphere. What is wrong with that?

References

• OpenStax, University Physics Volume 2, Chapter 16 (Electromagnetic Waves): the electromagnetic spectrum.
• NCERT Physics Class 12, Chapter 8 (Electromagnetic Waves): ozone, greenhouse effect and the uses of the spectrum.
• NCERT Physics Class 12, the chapter on communication systems: ground, sky and space waves, and the critical frequency, in the older edition.