Physics · Secondary School · Modern physics

Modern Physics

The GPS in a mobile phone, the X-ray at the hospital, the solar panel on the roof and the colour of the stars all depend on a physics that emerged at the start of the twentieth century. We start from the photon and the photoelectric effect, go through the energy levels of the atom and the wave–particle duality, reach the nucleus, with radioactivity and nuclear energy, and finish with relativity, which changed our idea of time.

1Spectrum and photon
2Photoelectric effect
3Bohr model
4Wave and particle
5Radioactivity
6Nuclear energy
7Relativity
✓Challenges

STEP 1

From radio to gamma rays

The light we see is a narrow band of a much larger family, the electromagnetic waves, in which electric and magnetic fields oscillate and travel together. In a vacuum they all travel at the same speed, \(c = 3 \cdot 10^8\ \text{m/s}\), and what tells one from another is the wavelength or, which amounts to the same thing, the frequency, linked by \(c = \lambda\,f\).

When we order them by wavelength, we obtain the electromagnetic spectrum, with gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves and radio waves. The visible band runs from about 400 nm, in the violet, to about 700 nm, in the red. The borders between the other regions are conventions, with no exact point where one ends and the next begins.

In 1900, to explain the light given off by hot bodies, Max Planck assumed that the energy exchanged with radiation came in packets. In 1905, Einstein went further and proposed that light itself is made of these packets, which we now call photons, each with energy \(E = h\,f\), where \(h\) is Planck's constant.

\(c = \lambda\,f\)\(E = h\,f = \dfrac{h\,c}{\lambda}\)\(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\)Throughout we use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), the same as \(4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\), the energy an electron gains when it crosses a potential difference of 1 V. With these values, \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\), and the photon energy in eV comes from \(1240/\lambda\), with λ in nanometres. Since \(h\) is rounded, calculations done by different routes may differ in the second digit. In the simulation, for simplicity, we treat every photon above 3.9 eV, the edge of UV-B, as harmful.

The idea of the photon tells us why ultraviolet causes sunburn and visible light does not. A UV-B photon carries about 4 eV, an energy of the order of the one that holds together the atoms of molecules such as DNA, and so it can break these bonds. A photon of red light has less than 2 eV.

A more intense visible light brings more photons, and each one is still too weak for that damage. It can warm the skin, which is a different effect, the energy of many photons added up and turned into heat. What decides the sunburn, as we can see in the simulation, is the energy of each photon, and not the amount of light.

Let's discuss

Drag the marker from red to violet, on the ruler or on the zoom of the visible band. What happens to the frequency and to the energy of each photon?
Click 'Wi-Fi' and then 'Microwave (oven)'. Are the frequencies similar? How many times greater is the energy of a photon of green light than that of a Wi-Fi photon?
Choose green light and turn the intensity up to the maximum. Does any photon break bonds in the skin? Now choose UV-B at low intensity and compare.
Find the point on the ruler where the photon goes above 10 eV. Which region of the spectrum are we in, and why is this radiation called ionising?

Wavelength

Intensity

Wavelength

Frequency

Photon energy

Energy in joules

Region

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
An FM radio station broadcasts at a frequency of 100 MHz. What is the wavelength of these waves? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
From the relation \(c = \lambda\,f\), with \(f = 100 \cdot 10^6 = 10^8\ \text{Hz}\), we have \(\lambda = \dfrac{c}{f} = \dfrac{3 \cdot 10^8}{10^8}\).
\(\lambda = 3\ \text{m}\)
basic
Put the following radiations in order, from the lowest to the highest frequency: visible light, X-rays, microwaves, ultraviolet and radio waves.

Show solution
In a vacuum they all travel at the same speed \(c\), and, since \(f = c/\lambda\), the frequency rises as the wavelength falls.
We therefore only need to order them from the longest to the shortest wavelength.
Radio waves, microwaves, visible light, ultraviolet and X-rays.
basic
A photon of orange light has a frequency of \(5 \cdot 10^{14}\ \text{Hz}\). What is its energy, in joules and in electronvolts? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
The photon energy is \(E = h\,f = 6.6 \cdot 10^{-34} \cdot 5 \cdot 10^{14}\) \(= 3.3 \cdot 10^{-19}\ \text{J}\).
To convert to electronvolts, we divide by \(1.6 \cdot 10^{-19}\), \(\dfrac{3.3 \cdot 10^{-19}}{1.6 \cdot 10^{-19}} \approx 2.1\).
\(E = 3.3 \cdot 10^{-19}\ \text{J}\), about \(2.1\ \text{eV}\).
basic
Débora spends the afternoon under a very bright LED lamp and does not get burnt. The next day she spends half an hour in the winter sun, which seems weak, and her skin turns red. How does the idea of the photon explain the difference?

Show solution
Sunburn is caused by ultraviolet, whose photons carry enough energy to break bonds in molecules of the skin, such as DNA.
The photons of the visible light from the LED each have less energy. A more intense light brings more photons, but each photon still lacks the energy for the damage, and what decides is the energy per photon, \(E = h\,f\).
The winter sun still sends ultraviolet photons, of high frequency, and the LED, however bright, sends only visible photons, of lower energy.
intermediate
A microwave oven works with waves of 2.45 GHz. What is the wavelength of these waves and the energy, in eV, of each photon? How many times smaller is this energy than that of a 2 eV photon of visible light? Use \(c = 3 \cdot 10^8\ \text{m/s}\) and \(h = 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\).

Show solution
The wavelength is \(\lambda = \dfrac{3 \cdot 10^8}{2.45 \cdot 10^9} \approx 0.12\ \text{m}\).
The photon energy is \(E = h\,f = 4.1 \cdot 10^{-15} \cdot 2.45 \cdot 10^9\) \(\approx 1.0 \cdot 10^{-5}\ \text{eV}\), and the ratio is \(\dfrac{2}{1.0 \cdot 10^{-5}} = 2 \cdot 10^5\).
\(\lambda \approx 12\ \text{cm}\) and \(E \approx 1.0 \cdot 10^{-5}\ \text{eV}\), about 200 thousand times less than the visible photon.
intermediate
The Sun's UV-B radiation has a wavelength of about 300 nm. What is the energy of one of these photons, in joules and in eV? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
With \(E = \dfrac{h\,c}{\lambda}\) and \(\lambda = 3 \cdot 10^{-7}\ \text{m}\), we have \(E = \dfrac{6.6 \cdot 10^{-34} \cdot 3 \cdot 10^8}{3 \cdot 10^{-7}}\) \(= 6.6 \cdot 10^{-19}\ \text{J}\).
In electronvolts, \(\dfrac{6.6 \cdot 10^{-19}}{1.6 \cdot 10^{-19}} \approx 4.1\).
\(E = 6.6 \cdot 10^{-19}\ \text{J}\), about \(4.1\ \text{eV}\).
intermediate
A green laser pointer emits 5 mW of light with a wavelength of 532 nm. How many photons does it emit per second? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
Each photon has \(E = \dfrac{h\,c}{\lambda}\) \(= \dfrac{6.6 \cdot 10^{-34} \cdot 3 \cdot 10^8}{532 \cdot 10^{-9}}\) \(\approx 3.7 \cdot 10^{-19}\ \text{J}\).
The power is the energy emitted per second, and the number of photons per second is \(\dfrac{5 \cdot 10^{-3}}{3.7 \cdot 10^{-19}}\).
About \(1.3 \cdot 10^{16}\) photons per second.
intermediate
The LED of a remote control emits infrared at 940 nm. What is the frequency of this radiation and the energy of each photon, in eV? Why can we not see the light of the remote control, although the camera of many mobile phones shows it? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
The frequency is \(f = \dfrac{c}{\lambda}\) \(= \dfrac{3 \cdot 10^8}{940 \cdot 10^{-9}} \approx 3.2 \cdot 10^{14}\ \text{Hz}\), and the energy, \(E = h\,f \approx 2.1 \cdot 10^{-19}\ \text{J}\), or about 1.3 eV.
Our eyes respond to wavelengths between about 400 nm and 700 nm, and 940 nm lies outside this band. The camera sensor responds to a slightly wider band and records part of this infrared.
\(f \approx 3.2 \cdot 10^{14}\ \text{Hz}\) and \(E \approx 1.3\ \text{eV}\), with λ outside the band the eye perceives.
challenge
A 10 W LED turns about 30% of the energy it receives into light. Assuming that all this light has a wavelength of 550 nm, how many photons does it emit per second? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
The light output is \(0.30 \cdot 10 = 3\ \text{W}\), or 3 J of light per second.
Each photon has \(E = \dfrac{6.6 \cdot 10^{-34} \cdot 3 \cdot 10^8}{550 \cdot 10^{-9}}\) \(= 3.6 \cdot 10^{-19}\ \text{J}\), and the number of photons per second is \(\dfrac{3}{3.6 \cdot 10^{-19}}\).
About \(8 \cdot 10^{18}\) photons per second.
challenge
The X-rays of a radiograph have a wavelength of about 0.1 nm. Calculate the energy of one of these photons, in eV, and compare it with that of a 500 nm photon of visible light. Why can an X-ray photon ionise atoms when a visible-light photon cannot? Use \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\).

Show solution
With \(E = \dfrac{h\,c}{\lambda}\), the X-ray photon has \(E \approx \dfrac{1240}{0.1} \approx 1.2 \cdot 10^4\ \text{eV}\), and the visible-light one, \(E \approx \dfrac{1240}{500} \approx 2.5\ \text{eV}\).
The ratio of the energies equals the inverse ratio of the wavelengths, \(\dfrac{500}{0.1} = 5000\).
Pulling an electron out of an atom typically takes from a few to a few tens of eV, and a single visible photon falls short of that.
The X-ray photon has about 12,000 eV, 5000 times the energy of the visible photon, with energy to spare for ionising.

STEP 2

Light that knocks out electrons

When we shine light of high enough frequency on a metal plate, electrons escape from the surface. This is the photoelectric effect, observed by Heinrich Hertz in 1887 and studied in detail by Philipp Lenard around 1900.

The results went against the wave theory of light. According to it, a more intense light should knock out faster electrons, and any colour would eventually knock them out if we waited long enough. In the experiments, below a certain threshold frequency no electron comes out, however strong the light, and above it the electrons come out practically at once, with a maximum energy that depends only on the frequency.

In 1905, Einstein explained the data by assuming that light arrives in photons of energy \(h\,f\) and that each photon gives all its energy to a single electron. One part, at least the work function \(W\) of the metal, is spent getting the electron out of the plate, and the rest becomes kinetic energy. If we raise the intensity, more photons arrive per second and we knock out more electrons, each with the same maximum energy as before.

\(E_{k,\text{max}} = h\,f - W\)\(f_0 = \dfrac{W}{h}\)\(\lambda_0 = \dfrac{h\,c}{W}\)\(W\) is the work function, the minimum energy to knock an electron out of the metal, and \(f_0\) and \(\lambda_0\) are the threshold frequency and wavelength. The values in the simulation are approximate, with caesium 2.1 eV, sodium 2.3 eV, zinc 4.3 eV, copper 4.7 eV and platinum 5.6 eV, and the measurements vary with the state of the surface. We assume that each photon is absorbed by a single electron, and the maximum speed comes from \(E_k = m\,v^2/2\), with the mass of the electron, \(9.1 \cdot 10^{-31}\ \text{kg}\).

Millikan measured the energy of the electrons from several metals and found, on the graph of \(E_k\) against \(f\), parallel straight lines with slope equal to \(h\), as Einstein had predicted, even though he himself distrusted the idea of the photon. It was for this work that Einstein received the 1921 Nobel Prize.

We find the photoelectric effect in light sensors, such as those of automatic doors and those that switch on the street lights at dusk, almost always in the version that happens inside semiconductors, in which the electron is freed inside the material without leaving it. Solar panels and mobile-phone cameras work on the same idea.

Let's discuss

With sodium, start at 700 nm and reduce λ little by little. At what wavelength do the first electrons appear? Compare with \(\lambda_0 = h\,c/W\).
At 600 nm, turn the intensity up to the maximum. Does any electron come out? Now, at 400 nm, lower the intensity. What changes, the number of electrons or their speed?
Swap sodium for zinc without touching the light. Why do the electrons stop coming out, and what λ would be needed for them to return?
On the graph, compare the lines of the five metals. Do they have the same slope? Where does each one cross the frequency axis?

Wavelength

Intensity

Photon energy

Work function

Threshold frequency

Threshold λ

Maximum kinetic energy

Maximum speed

Electrons per second

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A sodium plate, with a work function of 2.3 eV, is lit by photons of 3.0 eV. What is the maximum kinetic energy of the electrons knocked out?

Show solution
Each photon gives all its energy to one electron, and at least 2.3 eV are spent getting it out of the plate, \(E_{k,\text{max}} = h\,f - W = 3.0 - 2.3\).
\(E_{k,\text{max}} = 0.7\ \text{eV}\)
basic
The work function of zinc is about 4.3 eV. What is the threshold frequency of zinc, and what wavelength does it correspond to? Use \(h = 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\) and \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
At the threshold frequency, the photon has just the energy needed to knock out the electron, \(h\,f_0 = W\), and so \(f_0 = \dfrac{4.3}{4.1 \cdot 10^{-15}}\) \(\approx 1.05 \cdot 10^{15}\ \text{Hz}\).
The corresponding wavelength is \(\lambda_0 = \dfrac{c}{f_0} = \dfrac{3 \cdot 10^8}{1.05 \cdot 10^{15}}\).
\(f_0 \approx 1.0 \cdot 10^{15}\ \text{Hz}\) and \(\lambda_0 \approx 2.9 \cdot 10^{-7}\ \text{m}\), about 290 nm, in the ultraviolet.
basic
Violet light knocks electrons out of a caesium plate. If we double the intensity of the light, without changing its colour, what happens to the number of electrons knocked out per second and to the maximum kinetic energy of each one?

Show solution
Doubling the intensity, at the same frequency, doubles the number of photons arriving per second, and each photon keeps the same energy \(h\,f\).
Since each photon knocks out at most one electron, the number of electrons per second doubles, and the maximum energy of each one, \(h\,f - W\), does not change.
The number of electrons per second doubles, and the maximum kinetic energy stays the same.
basic
Iuri shines a very strong red floodlight on a zinc plate for several minutes, and no electron is knocked out. Then a weak ultraviolet lamp knocks out electrons at once. Explain.

Show solution
Each photon acts on its own on one electron. A photon of red light has less than 2 eV, and zinc requires about 4.3 eV to release an electron.
Adding up weak photons does not help, because an electron does not accumulate the energy of several photons, and waiting longer only brings more weak photons.
The ultraviolet photon has more energy than the work function of zinc, and the red one does not, whatever the intensity.
intermediate
A copper plate, with a work function of 4.7 eV, is lit by 200 nm ultraviolet. What is the maximum kinetic energy of the electrons, in eV? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
The photon energy is \(E = \dfrac{h\,c}{\lambda} = \dfrac{6.6 \cdot 10^{-34} \cdot 3 \cdot 10^8}{2 \cdot 10^{-7}}\) \(= 9.9 \cdot 10^{-19}\ \text{J}\), or about 6.2 eV.
Subtracting the work function, \(E_{k,\text{max}} = 6.2 - 4.7\).
\(E_{k,\text{max}} \approx 1.5\ \text{eV}\)
intermediate
The electrons knocked out of a plate leave with a maximum kinetic energy of 2.0 eV. What is the maximum speed of these electrons? Use the mass of the electron, \(9.1 \cdot 10^{-31}\ \text{kg}\), and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
In joules, the energy is \(2.0 \cdot 1.6 \cdot 10^{-19} = 3.2 \cdot 10^{-19}\ \text{J}\).
From \(E_k = \dfrac{m\,v^2}{2}\), we have \(v = \sqrt{\dfrac{2\,E_k}{m}}\) \(= \sqrt{\dfrac{2 \cdot 3.2 \cdot 10^{-19}}{9.1 \cdot 10^{-31}}}\).
\(v \approx 8.4 \cdot 10^5\ \text{m/s}\), about 840 km/s.
intermediate
In an experiment, the graph of the maximum kinetic energy of the electrons against the frequency of the light is a straight line that crosses the frequency axis at \(5.0 \cdot 10^{14}\ \text{Hz}\) and passes through the point (\(1.0 \cdot 10^{15}\ \text{Hz}\); 2.05 eV). Find Planck's constant, in eV·s, and the work function of the metal.

Show solution
The slope of the line is \(h = \dfrac{\Delta E_k}{\Delta f}\) \(= \dfrac{2.05}{1.0 \cdot 10^{15} - 5.0 \cdot 10^{14}}\) \(= 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\).
At the crossing point, \(E_k = 0\) and \(W = h\,f_0 = 4.1 \cdot 10^{-15} \cdot 5.0 \cdot 10^{14}\).
\(h = 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\) and \(W \approx 2.1\ \text{eV}\)
intermediate
A potassium plate, with a work function of 2.3 eV, is lit first with 600 nm light and then with 400 nm light. In which case do electrons come out, and with what maximum kinetic energy? Use \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\).

Show solution
With 600 nm, each photon has \(E \approx \dfrac{1240}{600} \approx 2.1\ \text{eV}\), less than the work function, and no electron comes out.
With 400 nm, \(E \approx \dfrac{1240}{400} = 3.1\ \text{eV}\), and \(3.1 - 2.3\) is left over for the kinetic energy.
Only the 400 nm light knocks out electrons, with a maximum kinetic energy of about 0.8 eV.
challenge
With light of \(1.0 \cdot 10^{15}\ \text{Hz}\), the electrons of a metal leave with a maximum kinetic energy of 2.0 eV. With \(1.5 \cdot 10^{15}\ \text{Hz}\), they leave with 4.05 eV. Without using the value of \(h\) given in the lesson, find Planck's constant and the work function of the metal.

Show solution
Both measurements obey \(E_{k,\text{max}} = h\,f - W\), with \(2.0 = h \cdot 1.0 \cdot 10^{15} - W\) and \(4.05 = h \cdot 1.5 \cdot 10^{15} - W\).
Subtracting the first from the second, \(2.05 = h \cdot 0.5 \cdot 10^{15}\), and \(h = 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\).
Going back to the first, \(W = 4.1 - 2.0\).
\(h = 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\) and \(W = 2.1\ \text{eV}\)
challenge
In the classical wave model, the energy of the light would spread evenly over the plate, and an atom could only capture the energy falling on its own area, of about \(10^{-20}\ \text{m}^2\). With a faint light, of \(10^{-6}\ \text{W/m}^2\), how long would an atom take to gather the 2 eV needed to release an electron? Compare with what is observed, in which the electrons come out in less than a billionth of a second. Use \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
The power reaching one atom, in this model, is \(10^{-6} \cdot 10^{-20} = 10^{-26}\ \text{W}\), and the energy needed is \(2 \cdot 1.6 \cdot 10^{-19} = 3.2 \cdot 10^{-19}\ \text{J}\).
The time would be \(t = \dfrac{3.2 \cdot 10^{-19}}{10^{-26}} = 3.2 \cdot 10^7\ \text{s}\).
About \(3 \cdot 10^7\ \text{s}\), close to a year, whereas the electrons come out practically at once, one of the pieces of evidence that led Einstein to propose photons.

STEP 3

Why does each element have its own colours?

When we heat a gas or pass an electric discharge through it, the gas gives off light, and a prism shows that this light contains only a few colours, in thin, separate lines. This is the emission spectrum, and each element has its own set of lines, as characteristic as a fingerprint.

In 1913, Niels Bohr proposed a model for hydrogen in which the electron can occupy only certain energy levels, \(E_n = -13.6/n^2\) eV, with \(n = 1, 2, 3, \dots\) The negative sign indicates that the electron is bound to the nucleus, and we need 13.6 eV to pull it out of the lowest level, the ground state.

When it goes from a higher level to a lower one, the electron emits a photon with energy equal to the difference between the levels, and to move up it must absorb a photon with exactly that energy. The drops to level 2 give the visible lines of hydrogen, the Balmer series, measured at 656, 486, 434 and 410 nm, and Bohr's model reproduces these values with a precision that impressed the physicists of the time.

\(E_n = -\dfrac{13.6}{n^2}\ \text{eV}\)\(E_{\text{photon}} = E_i - E_f\)\(\lambda = \dfrac{h\,c}{E_{\text{photon}}}\)\(E_i\) and \(E_f\) are the energies of the initial and final levels. The Balmer lines quoted in the text are measurements; with the rounded values of \(h\) and \(c\) we use here, the calculation gives 655, 485, 433 and 409 nm, with differences of about 1 nm that come mainly from the rounding. In the simulation, the heights of the levels and the radii of the orbits are not to scale.

We should, however, treat this model with care. The picture of the electron going round in well-defined orbits was abandoned with quantum mechanics, and the model fails for atoms with more than one electron. The idea of discrete energy levels still holds, and quantum mechanics gives for hydrogen the same levels that Bohr calculated.

We use spectra to identify elements, on Earth and beyond it. The yellowish light of sodium lamps comes from two lines near 589 nm, and the colours of fireworks come from salts of strontium, in the red, of barium, in the green, and of copper, in the blue.

The spectrum of the Sun shows dark absorption lines, because the gases of its atmosphere absorb the same colours they would emit. This is how helium was discovered in the Sun, in 1868, almost thirty years before it was found on Earth.

Let's discuss

Take the electron to level 3 and let it drop to level 2. Which colour appears in the emission spectrum, and what is the wavelength?
Repeat with the drops from 4, 5 and 6 to level 2. Do the lines get closer to one another as \(n\) increases?
With the electron at level 2, move it up to level 4. Where does the dark line appear in the absorption spectrum? Does it coincide with any emission line?
Make the electron drop from any level to level 1. Why do these lines not appear in the visible band? Then switch on the hydrogen lamp and wait for the spectrum to build up.

Electron level

Level energy

Last photon

Wavelength

Type and region

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
In Bohr's model, the energy levels of hydrogen are \(E_n = -13.6/n^2\) eV. Calculate the energy of levels 1, 2 and 3.

Show solution
Substituting \(n\), we have \(E_1 = -\dfrac{13.6}{1}\), \(E_2 = -\dfrac{13.6}{4}\) and \(E_3 = -\dfrac{13.6}{9}\).
\(E_1 = -13.6\ \text{eV}\), \(E_2 = -3.4\ \text{eV}\) and \(E_3 \approx -1.51\ \text{eV}\)
basic
The electron of a hydrogen atom drops from level 3 to level 2. What is the energy of the photon emitted? Use \(E_n = -13.6/n^2\) eV.

Show solution
The photon carries the energy difference between the levels, \(E = E_3 - E_2 = -1.51 - (-3.4)\).
\(E \approx 1.89\ \text{eV}\)
basic
Why does the light from a heated gas, such as the hydrogen in a discharge lamp, form separate lines when it passes through a prism, and not a continuous rainbow like the light from an incandescent bulb?

Show solution
The electrons of the gas occupy only certain energy levels, and each photon emitted has the energy of the difference between two of these levels.
Only some energies are possible, and therefore only some wavelengths. The filament of an incandescent bulb is a hot solid, in which the atoms interact strongly and the emission covers every colour.
The discrete levels of the isolated atom allow only certain photon energies, which appear as lines.
basic
How do astronomers know that there is hydrogen and sodium in the atmosphere of a distant star, without ever having been there?

Show solution
The light of the star, as it crosses the star's atmosphere, loses the colours that the atoms there absorb, and the spectrum shows dark lines.
Each element absorbs at the same positions at which it emits in the laboratory, and by comparing the dark lines with these known positions we identify the elements.
The dark lines in the star's spectrum coincide with the lines of hydrogen and sodium measured on Earth.
intermediate
Calculate the wavelength of the photon emitted when the electron of hydrogen drops from level 3 to level 2. Which region of the spectrum is it in? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\), \(c = 3 \cdot 10^8\ \text{m/s}\) and \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
The photon energy is \(1.89\ \text{eV} = 1.89 \cdot 1.6 \cdot 10^{-19}\) \(\approx 3.02 \cdot 10^{-19}\ \text{J}\).
With \(\lambda = \dfrac{h\,c}{E}\) \(= \dfrac{6.6 \cdot 10^{-34} \cdot 3 \cdot 10^8}{3.02 \cdot 10^{-19}}\), we arrive at \(6.55 \cdot 10^{-7}\ \text{m}\).
\(\lambda \approx 655\ \text{nm}\), red light, the same line that is measured at 656 nm.
intermediate
How much energy must we give an electron of hydrogen at level 2 to pull it out of the atom? What is the longest wavelength of a photon able to do this? Use \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\).

Show solution
To pull the electron out, we take its energy from \(E_2 = -3.4\ \text{eV}\) up to 0, which requires 3.4 eV.
The longest wavelength corresponds to the photon of minimum energy, \(\lambda = \dfrac{1240}{3.4}\).
It takes 3.4 eV, and the photon must have a λ of at most about 365 nm, in the ultraviolet.
intermediate
In a hydrogen gas, the electrons have been excited up to level 4 and go down, by every possible route, to level 1. How many different lines can appear in the spectrum? Which of them are visible?

Show solution
Each line corresponds to a pair of levels, and the possible pairs are 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1.
The visible ones are those that end at level 2, the Balmer series, and those that end at level 1 fall in the ultraviolet. The 4→3 drop falls in the infrared.
Six lines, of which only two are visible, 3→2 (red, 656 nm) and 4→2 (blue-green, 486 nm).
intermediate
A hydrogen atom in the ground state receives a 10.2 eV photon, and then another atom, also in the ground state, receives an 11 eV photon. What happens in each case? Use \(E_n = -13.6/n^2\) eV.

Show solution
With 10.2 eV, the electron would go from \(-13.6\ \text{eV}\) to \(-13.6 + 10.2 = -3.4\ \text{eV}\), which is exactly \(E_2\), and the photon is absorbed.
With 11 eV, the electron would go to \(-2.6\ \text{eV}\), and there is no level with that energy, because \(E_2 = -3.4\ \text{eV}\) and \(E_3 \approx -1.51\ \text{eV}\).
The 10.2 eV photon is absorbed and takes the electron to level 2; the 11 eV one is not absorbed and passes through the gas.
challenge
Calculate the wavelengths of the drops from level 4 to level 2 and from level 6 to level 2 in hydrogen, and compare them with the measured values, 486 nm and 410 nm. Use \(E_n = -13.6/n^2\) eV and \(h\,c = 1237.5\ \text{eV} \cdot \text{nm}\), the value that comes from the constants of this lesson.

Show solution
In the 4→2 drop, \(E = 13.6 \left(\dfrac{1}{4} - \dfrac{1}{16}\right)\) \(= 2.55\ \text{eV}\), and \(\lambda = \dfrac{1237.5}{2.55} \approx 485\ \text{nm}\).
In the 6→2 drop, \(E = 13.6 \left(\dfrac{1}{4} - \dfrac{1}{36}\right)\) \(\approx 3.022\ \text{eV}\), and \(\lambda = \dfrac{1237.5}{3.022} \approx 409.5\ \text{nm}\).
The differences of about 1 nm come mainly from the rounding of \(h\) and \(c\).
About 485 nm and 409.5 nm, practically the measured values, one of the great successes of Bohr's model.
challenge
The yellow light of the sodium lamps used in street lighting has a wavelength of about 589 nm. What is the energy difference, in eV, between the two levels involved? Why can we not use \(E_n = -13.6/n^2\) to calculate the levels of sodium? Use \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\).

Show solution
The energy difference is the photon energy, \(E = \dfrac{1240}{589} \approx 2.1\ \text{eV}\).
Bohr's formula holds for a single electron around a nucleus with one proton. Sodium has 11 electrons, which repel one another and screen part of the charge of the nucleus, and Bohr's model cannot account for this, even though the levels are still discrete.
About 2.1 eV; the levels of sodium exist, but they do not follow the hydrogen formula, because the atom has many electrons.

STEP 4

Wave or particle?

If light, which we treated as a wave, also behaves as a particle, perhaps matter, which we treat as particles, also behaves as a wave. This was the hypothesis of Louis de Broglie, in 1924, who associated with every particle of mass \(m\) and speed \(v\) a wavelength \(\lambda = h/(m\,v)\).

In 1927, Davisson and Germer, in the United States, and G. P. Thomson, in England, observed the diffraction of electrons by crystals, with the wavelength de Broglie had predicted. Electron microscopes use this behaviour, and the short wavelength of electrons reveals details that visible light cannot reach.

If we calculate \(\lambda\) for a 0.43 kg football at 20 m/s, we find about \(8 \cdot 10^{-35}\ \text{m}\), a value so small next to any slit or obstacle that no wave effect can be noticed. For an electron at \(10^6\ \text{m/s}\), \(\lambda\) is close to 0.7 nm, comparable to the distance between the atoms of a crystal.

\(\lambda = \dfrac{h}{m\,v}\)\(\Delta y \approx \dfrac{\lambda\,L}{d}\)\(\Delta y\) is the distance between neighbouring bright fringes on a screen at a distance \(L\) from two slits a distance \(d\) apart, when \(\Delta y\) is much smaller than \(L\). In the simulation, the slits are 1 µm apart, each 0.2 µm wide, and the screen is 1 m away, values chosen so that the fringes have a visible size. The mass of the electron is \(9.1 \cdot 10^{-31}\ \text{kg}\).

In the double-slit experiment with electrons, carried out with electrons arriving one at a time in Italy, in the 1970s, and repeated more clearly in Japan, in 1989, each electron leaves a single dot on the screen, like a particle. With thousands of dots, bright and dark fringes appear, the interference pattern of a wave that went through both slits, and if we close one of the slits the fringes disappear.

What we can state with confidence is what is measured. Quantum mechanics predicts the probability of the electron arriving at each point on the screen, and this prediction agrees with the experiments. The question of which slit the electron went through, when we do not measure it, has no answer accepted by everyone, and it is precisely there that the interpretations of the theory diverge.

Let's discuss

Stop the firing, clear the screen and fire the electrons one by one. Can we predict where the next one will land?
Let a few thousand electrons arrive. Where are the bands with the most dots, and where does hardly any arrive? Compare with the dashed curve.
Close the lower slit and wait again. Are the fringes still there? Why is the pattern not simply half of the previous one?
Increase the speed of the electrons. What happens to \(\lambda\) and to the distance between the fringes?

Speed (m/s)

Electrons on the screen

Speed

De Broglie λ

Fringe spacing

Open slits

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the de Broglie wavelength of an electron with a speed of \(2 \cdot 10^6\ \text{m/s}\)? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\) and the mass of the electron, \(9.1 \cdot 10^{-31}\ \text{kg}\).

Show solution
From the de Broglie relation, \(\lambda = \dfrac{h}{m\,v}\) \(= \dfrac{6.6 \cdot 10^{-34}}{9.1 \cdot 10^{-31} \cdot 2 \cdot 10^6}\).
\(\lambda \approx 3.6 \cdot 10^{-10}\ \text{m}\), about 0.36 nm.
basic
What is the de Broglie wavelength of a 60 g tennis ball at 50 m/s? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\).

Show solution
With the mass in kilograms, \(m = 0.06\ \text{kg}\), we have \(\lambda = \dfrac{h}{m\,v} = \dfrac{6.6 \cdot 10^{-34}}{0.06 \cdot 50}\).
\(\lambda = 2.2 \cdot 10^{-34}\ \text{m}\)
basic
If every particle has an associated wavelength, why do we not see a football produce interference when it passes through the gap between two players in the wall?

Show solution
Wave effects, such as interference and diffraction, only become visible when the wavelength is comparable to the size of the slits or obstacles.
For the ball, \(\lambda\) is close to \(10^{-34}\ \text{m}\), trillions of trillions of times smaller than a gap of half a metre, and the fringes would be so close together that no instrument could separate them.
The wavelength of the ball is far too small, next to any gap, for any wave effect to appear.
basic
In a double-slit experiment, electrons are fired one at a time. What does each electron leave on the screen? What appears after thousands of electrons have arrived?

Show solution
Each electron is detected at a single point on the screen, like a particle, and we cannot predict at which point.
With many electrons, the dots build up in bright and dark bands, the interference pattern we would expect from a wave passing through both slits.
One dot per electron and, taken together, interference fringes.
intermediate
An electron and a proton have the same speed. Which of the two has the longer de Broglie wavelength, and how many times longer? Use \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\) and \(m_p = 1.67 \cdot 10^{-27}\ \text{kg}\).

Show solution
At the same speed, \(\lambda = \dfrac{h}{m\,v}\) is inversely proportional to the mass, and \(\dfrac{\lambda_e}{\lambda_p} = \dfrac{m_p}{m_e}\) \(= \dfrac{1.67 \cdot 10^{-27}}{9.1 \cdot 10^{-31}}\).
The electron, with a wavelength about 1800 times longer.
intermediate
What speed must an electron have for its de Broglie wavelength to be 0.1 nm, similar to the distance between the atoms of a crystal? Use \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\) and \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\).

Show solution
Isolating the speed, \(v = \dfrac{h}{m\,\lambda}\) \(= \dfrac{6.6 \cdot 10^{-34}}{9.1 \cdot 10^{-31} \cdot 10^{-10}}\).
\(v \approx 7.3 \cdot 10^6\ \text{m/s}\), about 2% of the speed of light.
intermediate
Electrons with a wavelength of 0.5 nm pass through two slits 1 µm apart and reach a screen 1 m away. What is the distance between neighbouring bright fringes? And if the speed of the electrons doubles?

Show solution
For closely spaced fringes, \(\Delta y = \dfrac{\lambda\,L}{d}\) \(= \dfrac{0.5 \cdot 10^{-9} \cdot 1}{10^{-6}}\) \(= 5 \cdot 10^{-4}\ \text{m}\).
Doubling the speed, \(\lambda = h/(m\,v)\) falls by half, and so does \(\Delta y\).
\(\Delta y = 0.5\ \text{mm}\), and with twice the speed, 0.25 mm.
intermediate
In the double-slit experiment with electrons, we close one of the slits. Is the pattern on the screen half of the pattern with both slits open? What appears?

Show solution
No. With both slits open, the dark fringes are places where almost no electron arrives, and with a single slit some of these places start to receive electrons.
What appears is a broad band, the diffraction by a single slit, without the fine interference fringes.
The fringes disappear and a broad band is left, which shows that the pattern with two slits is not the sum of what each slit would do on its own.
challenge
An electron starts from rest and is accelerated through a potential difference of 100 V. Calculate its speed and its de Broglie wavelength. Why are electrons like these useful for studying the structure of crystals? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\), \(m_e = 9.1 \cdot 10^{-31}\ \text{kg}\) and \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\).

Show solution
The kinetic energy is \(E_k = q\,U = 1.6 \cdot 10^{-19} \cdot 100\) \(= 1.6 \cdot 10^{-17}\ \text{J}\), and \(v = \sqrt{\dfrac{2\,E_k}{m}}\) \(\approx 5.9 \cdot 10^6\ \text{m/s}\).
The wavelength is \(\lambda = \dfrac{h}{m\,v}\) \(= \dfrac{6.6 \cdot 10^{-34}}{9.1 \cdot 10^{-31} \cdot 5.9 \cdot 10^6}\) \(\approx 1.2 \cdot 10^{-10}\ \text{m}\).
This value is similar to the distance between the atoms of a crystal, and the electrons are diffracted by the lattice, like those of Davisson and Germer in 1927.
\(v \approx 5.9 \cdot 10^6\ \text{m/s}\) and \(\lambda \approx 0.12\ \text{nm}\), of the order of the atomic spacing.
challenge
Two classmates are discussing the double-slit experiment with electrons. For Lara, each electron is a wave that turns into a particle when it hits the screen. For Sônia, each electron is a particle that goes through one of the slits, only we do not know which. What does the experiment establish, and what remains open?

Show solution
The experiment establishes that each electron is detected at a single point, that the dots are distributed with the interference fringes and that the fringes disappear when we close a slit or record which slit each electron went through.
A particle that went through one slit with no influence from the other would not produce fringes. Some interpretations keep the trajectory of the particle, but add a wave that goes through both slits and guides it.
Quantum mechanics predicts the arrival probabilities precisely, and all the interpretations agree with these predictions. It is the description of what happens between the source and the screen that changes from one interpretation to another.
The measurements are the same for everyone; what remains open is the interpretation, and the two statements, as they were put, oversimplify what is known.

STEP 5

Nuclei that change on their own

Some atomic nuclei are unstable and, sooner or later, turn into others, emitting the radiation we detect with a Geiger counter. This phenomenon, radioactivity, was discovered by Henri Becquerel in 1896 and studied by Marie and Pierre Curie, who isolated polonium and radium.

The most common emissions are of three kinds. Alpha radiation (α) is a helium nucleus, with two protons and two neutrons, and it is stopped by a sheet of paper. Far more penetrating, beta (β) is a fast electron or positron, which goes through paper and is stopped by a few millimetres of aluminium. Gamma (γ), a photon of very high energy, is only well attenuated by thick layers of lead or concrete.

There is no way to predict when a particular nucleus will decay, but in a sample with many nuclei the fraction that decays per unit of time is constant. We call the time for half of the nuclei to decay the half-life \(T\), and after a time \(t\) there remain \(N = N_0/2^{t/T}\). Iodine-131 has a half-life of about 8 days, carbon-14, of about 5730 years, and uranium-238, of 4.5 billion years.

\(N = \dfrac{N_0}{2^{t/T}}\)\(A = \dfrac{A_0}{2^{t/T}}\)\(N\) is the number of nuclei that have not yet decayed, \(T\) is the half-life, and the activity \(A\), the number of decays per second, falls in the same proportion. The dose values in the text are approximate and vary with the place and the equipment. In the simulation, the half-life is in seconds, so that it fits on the screen, and the penetration panel shows the trends, without the exact proportions.

Living things exchange carbon with their surroundings and keep the same proportion of carbon-14 as the air. After death, the carbon-14 is no longer replaced and decays, and, by measuring the fraction that remains, we know how long ago the organism died. Carbon-14 dating works well up to a few tens of thousands of years.

In medicine, ionising radiation produces images, as in X-rays and nuclear scans, and treats tumours in radiotherapy. The effect on the body is measured by the dose, in sieverts (Sv). We receive, on average, a few millisieverts a year from natural sources, a chest X-ray stays below a tenth of a millisievert and a CT scan gives a few millisieverts.

Doses of several sieverts in a short time cause serious harm, and so radioactive sources and equipment are controlled, and the professionals who work with them use protection and dose monitors. At the doses of the examinations we have, the benefit of the diagnosis tends to far outweigh the risk, which is small.

Let's discuss

With 400 nuclei, pause when the time reaches one half-life. How many nuclei remain? Is it exactly half?
Repeat with 100 and with 1600 nuclei. In which case does the count stay closer to the theoretical curve, and why?
Look at the sample after three half-lives. Does the fraction that remains match \(1/2^3\)?
In the lower panel, choose each radiation separately. Which barrier stops each one?

Half-life

Time

Half-lives

Nuclei remaining

Prediction N₀/2^(t/T)

Difference

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Iodine-131, used in the treatment of the thyroid, has a half-life of 8 days. Of an 80 g sample, how much remains after 24 days?

Show solution
In 24 days, \(\dfrac{24}{8} = 3\) half-lives go by, and \(m = \dfrac{m_0}{2^3} = \dfrac{80}{8}\).
10 g of iodine-131 remain.
basic
Put alpha, beta and gamma radiation in order of penetrating power and say what is enough to stop each of them.

Show solution
Alpha, a helium nucleus, is heavy and charged, and it is stopped by a sheet of paper or a few centimetres of air.
Beta, a fast electron, goes through paper and stops in a few millimetres of aluminium. Gamma, a high-energy photon, has no charge and needs thick layers of lead or concrete to be well attenuated.
From the least to the most penetrating, alpha (paper), beta (aluminium) and gamma (thick lead).
basic
A sample has 1600 radioactive nuclei, with a half-life of 5 minutes. How many nuclei, on average, have not yet decayed after 20 minutes?

Show solution
In 20 minutes, 4 half-lives go by, and \(N = \dfrac{N_0}{2^4} = \dfrac{1600}{16}\).
About 100 nuclei.
basic
A classmate says that, if in one half-life the sample loses half its nuclei, in two half-lives it loses all of them. Is he right?

Show solution
In each half-life, half of the nuclei present at the start of that interval decay, and not half of the original sample.
After the first half-life, \(\dfrac{1}{2}\) remains, and after the second, half of that, \(\dfrac{1}{4}\).
No. After two half-lives, a quarter of the sample remains.
intermediate
In a fossil, the proportion of carbon-14 is 1/8 of that found in a living thing. How long ago did the organism die? The half-life of carbon-14 is about 5730 years.

Show solution
For the proportion to fall to \(\dfrac{1}{8} = \dfrac{1}{2^3}\), 3 half-lives must go by, and \(t = 3 \cdot 5730\).
About 17,190 years ago.
intermediate
Technetium-99m, used in nuclear scans, has a half-life of about 6 h. A patient receives the dose at 8 a.m. What fraction of the initial activity remains at 8 a.m. the next day?

Show solution
Between 8 a.m. on one day and 8 a.m. the next, 24 h go by, or \(\dfrac{24}{6} = 4\) half-lives.
The activity falls to \(\dfrac{1}{2^4} = \dfrac{1}{16}\) of the initial value.
1/16 of the activity remains, about 6%.
intermediate
How long does it take for the activity of a sample of iodine-131, with a half-life of 8 days, to fall to 1/32 of its initial value?

Show solution
Since \(32 = 2^5\), the activity must halve 5 times, and \(t = 5 \cdot 8\).
40 days.
intermediate
A sample has only 4 radioactive nuclei. After one half-life, must exactly 2 remain? What is the probability that exactly 2 remain?

Show solution
Each nucleus, independently of the others, has a probability \(\dfrac{1}{2}\) of not having decayed after one half-life. The half-life describes the average behaviour, and with few nuclei the result varies a great deal.
There are \(2^4 = 16\) equally likely combinations, and in 6 of them exactly 2 nuclei are left.
No; the probability that exactly 2 remain is \(\dfrac{6}{16} = 37.5\%\).
challenge
The caesium-137 involved in the radiological accident in Goiânia, Brazil, in 1987, has a half-life of about 30 years. What fraction of the activity remains after 1 year? In what year will the activity of that material fall to 1/8 of its 1987 value?

Show solution
After 1 year, the fraction is \(\dfrac{1}{2^{1/30}} \approx 0.977\), or about 97.7%, and the activity hardly changes in a year.
To fall to \(\dfrac{1}{8}\) takes 3 half-lives, \(3 \cdot 30 = 90\) years, and \(1987 + 90 = 2077\).
About 97.7% after one year, and 1/8 only around 2077.
challenge
Suppose, as approximate values, that a person receives about 3 mSv a year from natural sources, that a chest X-ray gives about 0.02 mSv and that a chest CT scan gives about 7 mSv. How many days of natural radiation is each examination equivalent to?

Show solution
Per day, natural radiation gives \(\dfrac{3}{365}\) mSv. The X-ray is equivalent to \(\dfrac{0.02}{3} \cdot 365 \approx 2.4\) days.
The CT scan is equivalent to \(\dfrac{7}{3} \approx 2.3\) years, or about 850 days.
The X-ray is equivalent to some 2 or 3 days of natural radiation, and the CT scan to a little over 2 years, which explains why it is only requested when the benefit for the diagnosis makes up for it.

STEP 6

How much energy fits in mass?

In 1905, Einstein showed that mass and energy are equivalent, \(E = m\,c^2\). Since \(c^2\) is enormous, \(9 \cdot 10^{16}\ \text{m}^2/\text{s}^2\), we see that a small mass corresponds to a great deal of energy, and one gram is equivalent to \(9 \cdot 10^{13}\ \text{J}\), roughly the energy of twenty thousand tonnes of TNT.

The mass of a nucleus is slightly less than the sum of the masses of its separate protons and neutrons, and the difference corresponds to the energy that holds it together. When a heavy nucleus splits into two smaller ones, or when two light ones join into a larger one, the total mass falls slightly and the difference appears as energy.

In fission, a uranium-235 nucleus absorbs a neutron and splits into two fragments, releasing about 200 MeV and two or three more neutrons, which can cause new fissions in a chain reaction. We call the average number of neutrons from each fission that cause another one \(k\). With \(k < 1\) the reaction dies out, with \(k = 1\) it keeps going and with \(k > 1\) it grows.

In a nuclear power station, such as Angra 1 and Angra 2, the two that Brazil runs at Angra dos Reis, on the coast of Rio de Janeiro state, the reactor operates with \(k = 1\). Control rods, made of materials that absorb neutrons, go into or come out of the reactor core to adjust \(k\), and the heat of the fissions heats water, whose steam drives the turbines. The uranium of the power station is only slightly enriched, with 3% to 5% uranium-235, and it does not explode like a bomb, which requires much more highly enriched uranium and an assembly of its own.

Fusion keeps the Sun shining. At its centre, at about 15 million kelvins, hydrogen nuclei fuse into helium, and the Sun converts about 4 million tonnes of mass into energy every second. We have not yet managed to use controlled fusion to generate electricity, and this is one of the great challenges of energy research.

\(E = m\,c^2\)\(\Delta E = \Delta m\,c^2\)\(1\ \text{MeV} = 1.6 \cdot 10^{-13}\ \text{J}\)Approximate values of energy per kilogram: burnt petrol, \(4.5 \cdot 10^7\ \text{J}\); uranium-235 that undergoes complete fission, \(8 \cdot 10^{13}\ \text{J}\); hydrogen that turns into helium in the Sun, \(6 \cdot 10^{14}\ \text{J}\). In the simulation, each fission releases 2 or 3 neutrons, each split nucleus is replaced after a while, as if the fuel never ran out, and a weak start-up source releases a neutron every now and then, if there are few in flight.

Let's discuss

Insert the rods well in, leaving \(k < 1\), and watch the neutron population. What happens to the chains that the source starts?
Adjust the rods until \(k = 1\). Does the number of neutrons stay stable, even with the fluctuations?
Withdraw the rods, with \(k > 1\), and follow the graph. How long does the population take to double? Then reinsert the rods and try to stabilise the reaction.
Switch off the source with \(k < 1\) and fire a neutron. How many fissions does the chain produce before it dies out?

Rods inserted

Factor k

Neutrons in flight

Fissions

Energy released

Regime

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
By the relation \(E = m\,c^2\), how much energy corresponds to a mass of 1 g? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
With the mass in kilograms, \(m = 10^{-3}\ \text{kg}\), we have \(E = m\,c^2 = 10^{-3} \cdot (3 \cdot 10^8)^2\).
\(E = 9 \cdot 10^{13}\ \text{J}\)
basic
What is the difference between nuclear fission and fusion? Which of them happens in the Angra power stations and which happens in the Sun?

Show solution
In fission, a heavy nucleus, such as that of uranium-235, splits into two smaller nuclei. In fusion, light nuclei, such as those of hydrogen, join together and form a larger nucleus.
In both cases the final mass is slightly less than the initial mass, and the difference appears as energy.
The Angra power stations use the fission of uranium, and the Sun shines through the fusion of hydrogen into helium.
basic
What are the control rods of a nuclear reactor for? What happens to the chain reaction when they are inserted further in?

Show solution
The rods are made of materials that absorb neutrons. Inserted further in, they capture more neutrons, which no longer cause new fissions, and the multiplication factor \(k\) decreases.
They regulate the reaction; inserted further in, they reduce \(k\) and the power of the reactor, and they can shut it down.
basic
In a chain reaction, \(k\) is the average number of neutrons from each fission that cause a new fission. What happens to the reaction when \(k < 1\), \(k = 1\) and \(k > 1\)? At what \(k\) should a power station operate?

Show solution
With \(k < 1\), each generation has fewer fissions than the previous one, and the reaction dies out. With \(k = 1\), the number of fissions per generation stays stable, and with \(k > 1\) it grows with each generation.
The power station operates with \(k = 1\), in the critical regime, in which the power stays constant.
intermediate
Each fission of a uranium-235 nucleus releases about 200 MeV. How much is that in joules? How many fissions per second are needed to produce a power of 1 W? Use \(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\).

Show solution
In joules, \(200 \cdot 10^6 \cdot 1.6 \cdot 10^{-19}\) \(= 3.2 \cdot 10^{-11}\ \text{J}\).
For 1 J per second, the number of fissions per second is \(\dfrac{1}{3.2 \cdot 10^{-11}}\).
About \(3.2 \cdot 10^{-11}\ \text{J}\) per fission, and some \(3 \cdot 10^{10}\) fissions per second for each watt.
intermediate
The reactor of a power station releases 3000 MW of heat. How much mass is converted into energy in one day of operation? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
In one day, of 86,400 s, the energy released is \(E = 3 \cdot 10^9 \cdot 86\,400\) \(\approx 2.6 \cdot 10^{14}\ \text{J}\).
The corresponding mass is \(m = \dfrac{E}{c^2} = \dfrac{2.6 \cdot 10^{14}}{9 \cdot 10^{16}}\).
About \(2.9 \cdot 10^{-3}\ \text{kg}\), some 3 g a day.
intermediate
A kilogram of uranium-235 that undergoes complete fission releases about \(8 \cdot 10^{13}\ \text{J}\), and a kilogram of petrol, when it burns, about \(4.5 \cdot 10^7\ \text{J}\). How many kilograms of petrol release the same energy as 1 kg of uranium-235?

Show solution
The mass of petrol is the ratio of the energies, \(\dfrac{8 \cdot 10^{13}}{4.5 \cdot 10^7}\).
About \(1.8 \cdot 10^6\ \text{kg}\), close to 1800 tonnes of petrol.
intermediate
In a chain reaction with \(k = 1.1\), roughly how many generations does it take for the number of fissions to double?

Show solution
After \(n\) generations, the number of fissions is multiplied by \(1.1^n\), and we need \(1.1^n \approx 2\).
By trial, \(1.1^7 \approx 1.95\) and \(1.1^8 \approx 2.14\).
About 7 generations.
challenge
The Sun emits about \(3.8 \cdot 10^{26}\ \text{W}\). How much mass does it convert into energy every second? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
In one second, the Sun emits \(3.8 \cdot 10^{26}\ \text{J}\), and the corresponding mass is \(m = \dfrac{E}{c^2} = \dfrac{3.8 \cdot 10^{26}}{9 \cdot 10^{16}}\).
About \(4.2 \cdot 10^9\ \text{kg}\) per second, some 4 million tonnes.
challenge
In the fusion of a deuterium nucleus (2.0141 u) with a tritium nucleus (3.0160 u), a helium-4 nucleus (4.0026 u) and a neutron (1.0087 u) are formed. How much mass disappears, and how much energy is released, in MeV? One atomic mass unit (1 u) corresponds to about 931 MeV.

Show solution
The initial mass is \(2.0141 + 3.0160 = 5.0301\ \text{u}\), and the final mass, \(4.0026 + 1.0087 = 5.0113\ \text{u}\).
The difference is \(\Delta m = 0.0188\ \text{u}\), and the energy released is \(0.0188 \cdot 931\).
0.0188 u disappears, which becomes about 17.5 MeV.

STEP 7

Does time pass the same for everyone?

In 1905, Einstein started from two postulates. The laws of physics are the same for all observers moving at constant velocity relative to one another, and the speed of light in a vacuum, \(c\), is also the same for all of them, whatever the motion of the source or of the observer.

The second postulate goes against intuition, since we would expect to add the speeds of the source and of the light, and it has a surprising consequence. In a light clock, a pulse of light bounces back and forth between two mirrors. For someone who sees the clock go past, the light follows a slanted, longer path, at the same speed \(c\), and each tick takes longer. This is time dilation, \(\Delta t = \gamma\,\Delta t_0\), where \(\Delta t_0\) is the interval measured in the clock's own frame of reference.

At everyday speeds, \(\gamma\) is so close to 1 that we do not notice the effect. Near the speed of light, it grows quickly, and with \(v = 0.8\,c\) it equals 5/3, while with \(v = 0.995\,c\) it reaches about 10.

The muons created by cosmic rays about 10 km up live, on average, 2.2 µs in their own frame of reference. Even close to the speed of light, they would travel some 660 m in that time, and hardly any would reach the ground. We do, however, measure many muons at sea level, because, for us, their lifetime is dilated.

The GPS we use on our mobile phones depends on this. The satellites of the system travel at about 3.9 km/s, and because of time dilation their clocks run about 7 µs a day slow compared with clocks on the ground. There is also an effect of gravity, predicted by general relativity, that makes these clocks run fast. The system corrects for both, and without the correction the position error would grow by several kilometres a day.

\(\Delta t = \gamma\,\Delta t_0\)\(\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}\)\(\Delta t_0\) is the proper time, measured by a clock that travels with the object, and \(\Delta t\) is the time measured by someone who sees the object go past at speed \(v\). In the simulation, the two clocks are identical, and the size of the screen and the speed of light on the screen are not to scale.

Let's discuss

With \(v = 0\), reset the ticks and compare the two clocks. Do they tick together?
Take \(v\) to 0.6 c, reset the ticks and wait for some twenty ticks of the clock at rest. How many times slower is the moving clock? Check against \(\gamma\).
Click 'Muon' and read the distance it covers in 2.2 µs, with and without dilation.
Choose the GPS. Why, at this speed, do the ticks look the same on the screen, and why does the delay over a day still matter?

Speed v/c

Speed

Factor γ

Ticks at rest / moving

Muon without dilation

Muon with dilation

Delay per day

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Calculate the factor \(\gamma = 1/\sqrt{1 - v^2/c^2}\) for a spaceship with \(v = 0.6\,c\).

Show solution
With \(v/c = 0.6\), we have \(v^2/c^2 = 0.36\) and \(\gamma = \dfrac{1}{\sqrt{1 - 0.36}}\) \(= \dfrac{1}{\sqrt{0.64}}\) \(= \dfrac{1}{0.8}\).
\(\gamma = 1.25\)
basic
On a train travelling at half the speed of light, a passenger switches on a torch pointing forwards. At what speed does the light move away from the torch, for the passenger? And for a person standing still at the station?

Show solution
By Einstein's second postulate, the speed of light in a vacuum is the same for all observers moving at constant velocity, whatever the motion of the source.
The sum \(c + 0.5\,c\), which intuition suggests, does not hold for light.
Both people measure the same speed, \(c\).
basic
A clock on a spaceship shows 2 s between two events that happen on the spaceship itself. For observers on Earth, the spaceship travels with \(\gamma = 2\). How much time passes between the events, for them?

Show solution
The 2 s of the spaceship are the proper time \(\Delta t_0\), and on Earth \(\Delta t = \gamma\,\Delta t_0 = 2 \cdot 2\).
\(\Delta t = 4\ \text{s}\)
basic
Why do we not notice time dilation when we travel by car or by plane?

Show solution
The factor \(\gamma\) depends on \(v^2/c^2\), and even the speed of a plane, about 250 m/s, is less than a millionth of the speed of light.
So \(\gamma\) stays at 1 + something of the order of \(10^{-13}\), and the time difference over a whole journey is billionths of a second, which only atomic clocks can measure.
At everyday speeds, \(\gamma\) is practically 1, and the effect is far too small to be noticed.
intermediate
A muon has an average lifetime of 2.2 µs in its own frame of reference and travels at \(0.995\,c\). How far would it travel in that time without dilation? And with dilation, for an observer on Earth? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
Without dilation, \(d = v\,\Delta t_0\) \(= 0.995 \cdot 3 \cdot 10^8 \cdot 2.2 \cdot 10^{-6}\) \(\approx 657\ \text{m}\).
With \(v = 0.995\,c\), \(\gamma = \dfrac{1}{\sqrt{1 - 0.990}} \approx 10\), and for us the muon lives \(\gamma\,\Delta t_0 \approx 22\ \mu\text{s}\), travelling about ten times further.
About 660 m without dilation, and about 6.6 km with it.
intermediate
A spaceship travels at \(0.8\,c\) relative to Earth. The on-board clock shows 30 min of travel. How much time passes for those who stayed on Earth?

Show solution
With \(v/c = 0.8\), \(\gamma = \dfrac{1}{\sqrt{1 - 0.64}} = \dfrac{1}{0.6} = \dfrac{5}{3}\).
The time on Earth is \(\Delta t = \gamma\,\Delta t_0 = \dfrac{5}{3} \cdot 30\).
\(\Delta t = 50\ \text{min}\)
intermediate
In a light clock, the mirrors are 1.5 m apart, and one tick is a round trip of the light. How long does a tick last with the clock at rest? And for someone who sees the clock go past at \(0.6\,c\)? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
At rest, the light covers 3 m per tick, and \(\Delta t_0 = \dfrac{3}{3 \cdot 10^8} = 10^{-8}\ \text{s}\).
Seen going past at \(0.6\,c\), with \(\gamma = 1.25\), each tick lasts \(\Delta t = 1.25 \cdot 10^{-8}\ \text{s}\).
\(10^{-8}\ \text{s}\) at rest and \(1.25 \cdot 10^{-8}\ \text{s}\) in motion.
intermediate
An unstable particle lives, on average, 1 µs in its own frame of reference. In an accelerator, it travels at \(0.99\,c\). How long does it live, on average, for the physicists in the laboratory?

Show solution
With \(v/c = 0.99\), \(\gamma = \dfrac{1}{\sqrt{1 - 0.9801}}\) \(= \dfrac{1}{\sqrt{0.0199}} \approx 7.1\).
In the laboratory, \(\Delta t = \gamma\,\Delta t_0 \approx 7.1 \cdot 1\ \mu\text{s}\).
About 7.1 µs.
challenge
GPS satellites travel at about 3.9 km/s. For low speeds, we can use \(\gamma \approx 1 + \dfrac{v^2}{2\,c^2}\). How much does a satellite's clock fall behind per day, through time dilation alone? What distance error would this cause, multiplied by \(c\)? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
We have \(\dfrac{v^2}{2\,c^2} = \dfrac{(3.9 \cdot 10^3)^2}{2 \cdot 9 \cdot 10^{16}}\) \(\approx 8.5 \cdot 10^{-11}\), the fraction of time that the satellite's clock loses.
In one day, \(8.5 \cdot 10^{-11} \cdot 86\,400 \approx 7.3 \cdot 10^{-6}\ \text{s}\), and multiplying by \(c\), \(7.3 \cdot 10^{-6} \cdot 3 \cdot 10^8 \approx 2.2 \cdot 10^3\ \text{m}\).
The effect of gravity, predicted by general relativity, acts in the opposite direction and is larger, and the system corrects for both together.
About 7 µs per day, which would give an error of some 2 km per day if it were not corrected.
challenge
At what speed must a spaceship travel for its clock, seen from Earth, to run at half the rate, that is, for \(\gamma = 2\)? Use \(c = 3 \cdot 10^8\ \text{m/s}\).

Show solution
From \(\gamma = 2\), we have \(\sqrt{1 - v^2/c^2} = \dfrac{1}{2}\), \(1 - \dfrac{v^2}{c^2} = \dfrac{1}{4}\) and \(\dfrac{v^2}{c^2} = \dfrac{3}{4}\).
So \(v = \dfrac{\sqrt{3}}{2}\,c \approx 0.866 \cdot 3 \cdot 10^8\).
\(v \approx 0.87\,c\), about \(2.6 \cdot 10^8\ \text{m/s}\).

WRAP-UP

Challenges

Spectrum and photon

A 4G mobile phone transmits at 1.8 GHz. What is the wavelength of these waves and the energy of each photon, in eV? (\(h = 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\))

Show solution

The wavelength is \(\lambda = \dfrac{3 \cdot 10^8}{1.8 \cdot 10^9} \approx 0.17\ \text{m}\), some 17 cm.
Each photon has \(E = h\,f = 4.1 \cdot 10^{-15} \cdot 1.8 \cdot 10^9\) \(\approx 7.4 \cdot 10^{-6}\ \text{eV}\), far from the electronvolts needed to break bonds.

Photoelectric effect

A caesium plate, with \(W = 2.1\ \text{eV}\), is lit with blue light of 450 nm. What is the maximum kinetic energy of the electrons? (\(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\))

Show solution

The photon has \(E \approx \dfrac{1240}{450} \approx 2.76\ \text{eV}\), and \(E_{k,\text{max}} = 2.76 - 2.1\).
The electrons come out with up to about 0.66 eV.

Bohr model

What is the wavelength of the photon emitted when the electron of hydrogen drops from level 2 to level 1? Which region of the spectrum is it in? (\(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\))

Show solution

The energy is \(E = -3.4 - (-13.6) = 10.2\ \text{eV}\), and \(\lambda \approx \dfrac{1240}{10.2} \approx 122\ \text{nm}\).
It is the first line of the Lyman series, in the ultraviolet.

Duality

'Thermal' neutrons, used to study materials, travel at about 2200 m/s. What is their de Broglie wavelength? (\(m_n = 1.67 \cdot 10^{-27}\ \text{kg}\), \(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\))

Show solution

From the de Broglie relation, \(\lambda = \dfrac{6.6 \cdot 10^{-34}}{1.67 \cdot 10^{-27} \cdot 2200}\) \(\approx 1.8 \cdot 10^{-10}\ \text{m}\).
That is 0.18 nm, of the order of the distance between the atoms of a crystal, and so the neutrons are diffracted by materials.

Half-life

A counter records 800 counts per minute from a sample and, 12 h later, 50 counts per minute. What is the half-life of the material?

Show solution

The activity fell to \(\dfrac{50}{800} = \dfrac{1}{16} = \dfrac{1}{2^4}\), which corresponds to 4 half-lives.
So \(T = \dfrac{12}{4}\), 3 h.

Nuclear energy

A power station produces 1000 MW of electricity and turns about a third of the reactor's heat into electricity. How many kilograms of uranium-235 does it fission per day? (1 kg of uranium-235 that undergoes fission releases about \(8 \cdot 10^{13}\ \text{J}\))

Show solution

The reactor releases about 3000 MW of heat, and in one day, \(3 \cdot 10^9 \cdot 86\,400\) \(\approx 2.6 \cdot 10^{14}\ \text{J}\).
The mass that undergoes fission is \(\dfrac{2.6 \cdot 10^{14}}{8 \cdot 10^{13}}\), about 3.2 kg per day.

Relativity

A spaceship travels at 0.8 c to a star that, measured from Earth, is 4 light-years away. How long does the journey last for those who stayed on Earth and for those on the spaceship?

Show solution

On Earth, the journey lasts \(\dfrac{4\ \text{light-years}}{0.8\,c} = 5\) years.
With \(\gamma = \dfrac{5}{3}\), the on-board clock shows \(\Delta t_0 = \dfrac{\Delta t}{\gamma} = \dfrac{5}{5/3}\), or 3 years.

Thinking, no calculation

Under a 'black light' lamp, of weak ultraviolet, certain minerals and white fabrics glow blue. Under a very strong red lamp, nothing glows. Why?

Show solution

Fluorescent materials absorb a photon, move up a level and come back down emitting photons of slightly lower energy, in the visible band.
The ultraviolet photon has the energy for this, and the red photon, of lower energy, cannot take the material to a level from which it would emit blue light, of higher energy, however many photons arrive.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with a base text taken from everyday life, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on modern physics, which may well be the best practice once these are done.

Spectrum and photon · Step 1

For a poster on skin protection, the pharmacist Teresa put together the table below, with everyday radiations and their approximate wavelengths. For simplicity, she assumed that only photons with more than 3.5 eV can break the bonds in the molecules of the skin that lead to sunburn.

Everyday radiations
Radiation	Wavelength
FM radio	3 m
Wi-Fi	12 cm
Remote-control infrared	940 nm
Green light	550 nm
Ultraviolet B	300 nm

Using \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\), which radiations in the table have photons able to cause this damage?

All of them, as long as the intensity is high enough.
Only ultraviolet B.
Only green light and ultraviolet B.
Only FM radio and Wi-Fi, which have the longest wavelengths.
None, because a single photon never has the energy to break a bond.

Show solution

Answer: B.
The energy of each photon is \(E = \dfrac{h\,c}{\lambda}\), and with λ in nanometres, \(E \approx \dfrac{1240}{\lambda}\) eV.
Ultraviolet B gives \(\dfrac{1240}{300} \approx 4.1\ \text{eV}\), above 3.5 eV, and green light gives \(\dfrac{1240}{550} \approx 2.3\ \text{eV}\), below it. The infrared is at about 1.3 eV, and the Wi-Fi and radio photons have energies of \(10^{-5}\ \text{eV}\) or less.
Option A confuses intensity with energy per photon, and C treats visible light as if it had the energy of ultraviolet. Option D reverses the relation between wavelength and energy, and E contradicts the calculation itself.
The single threshold of 3.5 eV is a simplification; in real skin, the damage increases gradually across the ultraviolet.

Photoelectric effect · Step 2
At a science fair, Fábio builds an alarm with a photoelectric cell whose cathode is made of a metal with a work function of 2.3 eV. He has two lasers available, a red one, of 650 nm and 50 mW, and a violet one, of 405 nm and only 5 mW. The alarm works when the light knocks electrons out of the cathode. Take \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\).

Which of the lasers makes the cell work?
1. Only the red one, because its power is ten times greater.
2. Only the violet one, because each of its photons has more energy than the work function of the metal.
3. Both, because any light knocks out electrons if it is left on for long enough.
4. Only the red one, because longer wavelengths carry more energy per photon.
5. Neither, because photons of visible light never exceed 2.3 eV.
Show solution

Answer: B.
The red photon has \(E \approx \dfrac{1240}{650} \approx 1.9\ \text{eV}\), less than the 2.3 eV of the work function, and it knocks out no electron, whatever the power.
The violet photon has \(E \approx \dfrac{1240}{405} \approx 3.1\ \text{eV}\), and the electrons come out with up to \(3.1 - 2.3 = 0.8\ \text{eV}\) of kinetic energy.
Options A and C use the wave idea that more power or more time would make up for the low frequency, which is precisely what the photoelectric effect disproves. Option D reverses the relation between λ and energy, and E forgets that violet goes above 3 eV.
The lower power of the violet laser only reduces the number of electrons knocked out per second, and for an alarm this is likely to be enough.
Bohr model · Step 3
Analysing the light from a nebula, the astronomer Bruna finds an intense line at 486 nm, in the blue-green band. She suspects that the line comes from hydrogen, whose energy levels, in Bohr's model, are \(E_n = -13.6/n^2\) eV. Take \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\).

Which transition of the electron in the hydrogen atom emits a photon with this wavelength?
1. From level 2 to level 1.
2. From level 3 to level 2.
3. From level 4 to level 2.
4. From level 4 to level 3.
5. From level 2 to level 4.
Show solution

Answer: C.
The 486 nm photon has \(E \approx \dfrac{1240}{486} \approx 2.55\ \text{eV}\).
In the drop from level 4 to level 2, \(E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 2.55\ \text{eV}\), exactly the energy we are looking for.
The 2→1 drop gives 10.2 eV, in the ultraviolet, and the 3→2 drop gives 1.89 eV, the red line at 656 nm. The 4→3 drop gives about 0.66 eV, in the infrared, and E describes a move upwards, in which the atom absorbs a photon instead of emitting one.
Lines like this one, measured in nebulae and stars, are one of the ways of knowing what the universe far from Earth is made of.
Half-life · Step 5
Technetium-99m is used in nuclear scans. The graph shows the activity of a dose of this material, as a percentage of the activity at the moment it was given, over the following hours.

Gustavo received a dose at 7 a.m. The hospital only considers the activity negligible for the next examination when it has fallen to 1/8 of its initial value.

From what time does the activity of Gustavo's dose reach this value?
1. 1 p.m. on the same day.
2. 7 p.m. on the same day.
3. 1 a.m. the following day.
4. 7 a.m. the following day.
5. 7 a.m. two days later.
Show solution

Answer: C.
From the graph, the activity halves every 6 h, and this is the half-life of technetium-99m.
To reach \(\dfrac{1}{8} = \dfrac{1}{2^3}\), 3 half-lives go by, \(3 \cdot 6 = 18\ \text{h}\), and \(7\ \text{h} + 18\ \text{h}\) takes us to 1 a.m. the following day.
Option A corresponds to a single half-life and B to two. Option D counts 24 h, when the activity has already fallen to 1/16, and E takes 1/8 as if it were eight half-lives.
The short half-life is one of the reasons technetium is so widely used, because the examination is ready in a few hours and the dose in the body falls quickly.
Time dilation · Step 7
Cosmic rays that strike the upper atmosphere create muons at an altitude of about 10 km. In its own frame of reference, a muon lives, on average, 2.2 µs before it decays. The muons come down at about 0.995 of the speed of light, for which the factor \(\gamma = 1/\sqrt{1 - v^2/c^2}\) is approximately 10. Take \(c = 3 \cdot 10^8\ \text{m/s}\).

For an observer on the Earth's surface, what is the approximate average distance that one of these muons travels before it decays?
1. 66 m
2. 660 m
3. 6.6 km
4. 66 km
5. 10 km, because every muon created in the upper atmosphere reaches the ground.
Show solution

Answer: C.
For us, the muon's lifetime is dilated, \(\Delta t = \gamma\,\Delta t_0 \approx 10 \cdot 2.2\) \(= 22\ \mu\text{s}\).
In this time it travels \(d = v\,\Delta t \approx 0.995 \cdot 3 \cdot 10^8 \cdot 22 \cdot 10^{-6}\), about 6.6 km.
Option B ignores time dilation, and A divides by \(\gamma\) instead of multiplying. Option D applies the factor of 10 twice, and E treats the mean lifetime as if it held for every muon, when it is only an average.
Without dilation, hardly any muon would get much further than 1 km down, and the large number of muons measured at sea level is one of the classic tests of relativity.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2013, Day 1, blue booklet, question 49. Glucose labelled with carbon-11, with a half-life of about 20 minutes, is used in positron emission tomography, and we calculate how much of 1 g of the nuclide remains after five half-lives.
ENEM 2015, Day 1, blue booklet, question 50. A figure gives the frequencies that separate UV-A, UV-B and UV-C, and we need to convert them into wavelengths to choose, from five absorption spectra, the sun cream that absorbs most in the UV-B.
ENEM 2017, Day 2, blue booklet, question 104. The carbon-14 dating of a fossil appears through the counting of beta emissions, and the age comes from the number of half-lives the activity took to fall to the measured value.
ENEM 2020, Day 2, blue booklet, question 126. The question asks which nuclear process takes place when an atomic bomb is detonated, and the options compare fission, fusion and other processes with the particles that trigger them.
ENEM 2022, Day 2, blue booklet, question 130. The text says that the cones of the eye respond to the energy of the photons and the rods to the number of photons, and we need to work out how an animal with more sensitive rods sees.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Wave\(c = \lambda\,f\), \(c = 3 \cdot 10^8\ \text{m/s}\)

Photon energy\(E = h\,f = h\,c/\lambda\)

Planck's constant\(h = 6.6 \cdot 10^{-34}\ \text{J} \cdot \text{s}\)
\(h = 4.1 \cdot 10^{-15}\ \text{eV} \cdot \text{s}\)

Electronvolt\(1\ \text{eV} = 1.6 \cdot 10^{-19}\ \text{J}\), \(h\,c \approx 1240\ \text{eV} \cdot \text{nm}\)

Photoelectric effect\(E_{k,\text{max}} = h\,f - W\)

Threshold frequency\(f_0 = W/h\)

Hydrogen levels\(E_n = -13.6/n^2\ \text{eV}\)

Photon emitted or absorbed\(E = E_i - E_f\)

De Broglie\(\lambda = h/(m\,v)\)

Half-life\(N = N_0/2^{t/T}\)

Mass and energy\(E = m\,c^2\)

Time dilation\(\Delta t = \gamma\,\Delta t_0\), \(\gamma = 1/\sqrt{1 - v^2/c^2}\)