Answer key · Unit 16

The nucleus.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 25 minutes

Nuclear fission

A uranium nucleus that tears in two: why a neutron splits it, where the energy comes from by E = mc², the multiplication factor k, control rods that hold a chain steady, and why a growing chain multiplies itself.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A nucleus that tears
Practice · Choose
Question
Why is it a neutron, not a proton, that splits a uranium-235 nucleus?

Correct answer
- Correct The neutron has no charge, so the nucleus doesn’t push it away
Setup, choices and feedback
The Scene at this step
Fissile fuel
50 %
Control rods
20 % absorbed
Count the neutrons
Off
Choices
Shown in a random order.
- Correct The neutron has no charge, so the nucleus doesn’t push it away
- Wrong The neutron is much heavier than the proton and hits harder
  Feedback
  A neutron and a proton have almost the same mass. The difference is the charge.
- Wrong The neutron is very fast and smashes the nucleus apart
  Feedback
  Slow neutrons actually split uranium-235 better, which is why reactors slow them down with water or graphite. The neutron doesn’t smash the nucleus: it is swallowed, and the nucleus wobbles and tears apart.
Feedback for a right answer
The nucleus has 92 protons that repel any positive charge coming near. The neutron walks straight in, the nucleus swallows it, wobbles like a water drop, pinches in the middle and tears into two smaller nuclei and two or three new neutrons.

Worked solution
1. The nucleus has a large positive charge, 92 protons.
2. An approaching proton is pushed away by a strong electric force; a neutron, with no charge, is not pushed at all.
3. The strong force grabs the neutron when it touches the nucleus, and the nucleus that swallowed it is unstable and splits.
Step 2 of 6
Open this step
Where the energy comes from
Practice · Calculate
Question
When one kilogram of uranium-235 is split completely, the fragments together weigh about 0.9 g less. How much energy is released? The speed of light is 3 × 10⁸ m/s.

Correct answer
81000000000000 J ± 3%
Setup, choices and feedback
The Scene at this step
Fissile fuel
50 %
Control rods
20 % absorbed
Count the neutrons
Off
Accepted within
± 3%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 81000000000000000 J ± 3%
  Misconception: Puts a mass into E = mc² in grams instead of kilograms.
  Feedback from the catalogue
  In $E = mc^2$ ⁠, with $c$ ⁠ in metres per second, the mass must be in kilograms for the energy to come out in joules. 0.9 g is $9 \times 10^{-4}$ ⁠ kilograms. Put in 0.9 and the answer is a thousand times too big.
Feedback for a right answer
$E = 9 \times 10^{-4} \times (3 \times 10^8)^2 \approx 8 \times 10^{13}$ ⁠ joules: about as much as burning three thousand tonnes of coal.

Hint
$E = mc^2$ ⁠, with $m$ ⁠ in kilograms.

Worked solution
1. The mass that vanished: 0.9 grams, $9 \times 10^{-4}$ ⁠ kilograms.
2. $c^2 = (3 \times 10^8)^2 = 9 \times 10^{16}$ ⁠ square metres per second squared.
3. $E = mc^2 = 9 \times 10^{-4} \times 9 \times 10^{16} = 8.1 \times 10^{13}$ ⁠ joules.
If you got $8.1 \times 10^{16}$ ⁠, you put the mass in grams.
Step 3 of 6
Open this step
The multiplication factor
Practice · Calculate
Question
Each split frees 2.5 neutrons on average. The Fissile fuel is at 90%, and the Control rods absorb 20% of the neutrons. What is the multiplication factor $k$ ⁠?

Correct answer
1.8 ± 0.02
Setup, choices and feedback
The Scene at this step
Fissile fuel
90 %fixed
Control rods
20 % absorbedfixed
Count the neutrons
Off
Accepted within
± 0.02
Computed from the Scene at this step’s setting
K, new neutrons each neutron makes on average, ν · enrichment · (1 − absorption)/nodes/core/outputs/multiplication
Wrong answers it recognises
- 2.5 ± 0.02
  Misconception: Counts every neutron a split frees as causing another split, forgetting those the rods absorb and those that hit non-fissile nuclei.
  Feedback from the catalogue
  Only some of the new neutrons carry the chain on. Some are absorbed by the control rods, and some hit non-fissile nuclei and are swallowed without a split. The multiplication factor is $k = \nu \times (\text{fissile share}) \times (1 - \text{absorbed share})$ ⁠: the neutrons per split, times the chance that each causes another.
- 2.25 ± 0.02
  Misconception: Counts every neutron a split frees as causing another split, forgetting those the rods absorb and those that hit non-fissile nuclei.
  Feedback from the catalogue
  Only some of the new neutrons carry the chain on. Some are absorbed by the control rods, and some hit non-fissile nuclei and are swallowed without a split. The multiplication factor is $k = \nu \times (\text{fissile share}) \times (1 - \text{absorbed share})$ ⁠: the neutrons per split, times the chance that each causes another.
Feedback for a right answer
$k = 2.5 \times 0.9 \times 0.8 = 1.8$ ⁠: each generation is 1.8 times the last, and the chain grows fast.

Hint
$k$ ⁠ is the neutrons per split, times the chance a neutron misses the rods, times the chance it hits a fissile nucleus.

Worked solution
1. The chance a neutron is not absorbed by the rods: $1 - 0.2 = 0.8$ ⁠.
2. The chance it then hits a fissile nucleus: $0.9$ ⁠.
3. $k = 2.5 \times 0.8 \times 0.9 = 1.8$ ⁠.
If you answered 2.5 or 2.25, you forgot some of the neutrons that are lost: in the rods or in non-fissile nuclei.
Step 4 of 6
Open this step
A steady chain
Practice · Reach the target
Question
The Fissile fuel is fixed at 90%. Push the Control rods in until the chain holds steady: $k = 1$ ⁠.

Goal
- 1 ± 0.02
  At the start of the step: 1.8
Setup, choices and feedback
The Scene at this step
Fissile fuel
90 %fixed
Control rods
20 % absorbed
Count the neutrons
On
Read from the Scene at
K, new neutrons each neutron makes on average, ν · enrichment · (1 − absorption)/nodes/core/outputs/multiplication
One setting that reaches it
Control rods
55 % absorbed
Feedback for a right answer
The rods must absorb about 56% of the neutrons: $1 - \dfrac{1}{2.25} \approx 0.56$ ⁠. A reactor runs exactly like this, at $k = 1$ ⁠: a steady chain whose heat boils water for a turbine.

Hint
$2.5 \times 0.9 \times (1 - a) = 1$ ⁠. Find $a$ ⁠.

Worked solution
1. $k = 2.5 \times 0.9 \times (1 - a) = 2.25 \times (1 - a)$ ⁠.
2. $k = 1$ ⁠ means $1 - a = \dfrac{1}{2.25} \approx 0.44$ ⁠.
3. $a \approx 0.56$ ⁠: the rods absorb about 56%.
Step 5 of 6
Open this step
Pulling the rods out
Practice · Predict, then look
Question
The Fissile fuel is at 50% and the Control rods at 20% absorbed, so $k = 1$ ⁠: the number of neutrons holds steady. The rods are now pulled all the way out, to 0%. Predict what will happen.

Correct answer
- Correct The number of neutrons will grow from generation to generation, and the core will heat up
Setup, choices and feedback
The Scene at this step
Fissile fuel
50 %fixed
Control rods
20 % absorbed
Count the neutrons
Off
Choices
- Correct The number of neutrons will grow from generation to generation, and the core will heat up
- Wrong The reactor will explode like a nuclear bomb
  Misconception: Thinks a power reactor whose chain starts to grow will explode like a nuclear bomb.
  Feedback from the catalogue
  In a reactor the fuel is enriched only to a few per cent, and some neutrons arrive late, so a growing chain heats the core fast but does not blow up like a bomb. In a bomb the fuel is almost pure, and the mass is assembled in an instant far beyond the critical mass. A reactor accident, as at Chernobyl, is overheating and melting, not a nuclear explosion.
- Wrong Nothing: the fuel hasn’t changed
  Feedback
  The fuel is the same, but now no neutron is absorbed by the rods, and more of them carry the chain on.
Once the prediction is locked in, the Scene changes
Control rods
0 % absorbed
Count the neutrons
On
The Scene then shows: 1.25 · K, new neutrons each neutron makes on average, ν · enrichment · (1 − absorption)/nodes/core/outputs/multiplication
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With no rods $k = 2.5 \times 0.5 = 1.25$ ⁠, and each generation is a quarter bigger than the last. The line on the graph climbs, on a log scale, in a straight line: exponential growth.

Worked solution
1. With the rods: $k = 2.5 \times 0.5 \times 0.8 = 1$ ⁠.
2. Without them: $k = 2.5 \times 0.5 \times 1 = 1.25$ ⁠.
3. $k > 1$ ⁠, so each generation is bigger than the last, and the chain grows. In a real reactor the safety systems drive the rods back in before the core gets too hot.
Step 6 of 6
Open this step
Ten generations
Practice · Calculate
Question
The Control rods absorb 12% of the neutrons, and $k = 1.1$ ⁠. For every neutron now, how many will there be ten generations on?

Correct answer
2.594 ± 0.05
Setup, choices and feedback
The Scene at this step
Fissile fuel
50 %fixed
Control rods
12 % absorbedfixed
Count the neutrons
On
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Neutrons ten generations on for every neutron now, k¹⁰/nodes/core/outputs/tenGenerations
Wrong answers it recognises
- 2 ± 0.05
  Misconception: Treats exponential growth or decay as linear: adds the same amount each generation or each half-life instead of multiplying.
  Feedback from the catalogue
  In a fission chain each generation multiplies the number of neutrons by $k$ ⁠, so after $n$ ⁠ generations there are $k^n$ ⁠ of them, not $1 + n(k - 1)$ ⁠. Decay works the same way: each half-life leaves half of what there was, so the uranium doesn’t fall by the same amount every year, and the ratio of lead to uranium grows as $2^n - 1$ ⁠.
Feedback for a right answer
$1.1^{10} \approx 2.6$ ⁠. A generation in a reactor lasts less than a thousandth of a second, so without control even a $k$ ⁠ close to 1 multiplies the power very quickly.

Hint
Each generation multiplies by $k$ ⁠.

Worked solution
1. After one generation: 1.1 neutrons. After two: $1.1^2 = 1.21$ ⁠.
2. After ten: $1.1^{10} \approx 2.59$ ⁠.
If you answered 2, you added 0.1 each generation instead of multiplying by 1.1.

Lesson 2 of 4 · About 25 minutes

Nuclear fusion

Light nuclei that join into heavier ones: the repulsion to overcome, heat that is speed, a rate that soars with temperature and with the square of the fuel density, the neutron that escapes the plasma, and why the Sun gets away with less.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
Repulsion
Practice · Choose
Question
Why is it so hard to make two heavy-hydrogen nuclei fuse?

Correct answer
- Correct Both are positively charged and repel each other, and the strong force acts only when they almost touch
Setup, choices and feedback
The Scene at this step
Temperature
15 million K
Fuel density
80 nuclei
Cut into the plasma
Machine
Choices
Shown in a random order.
- Correct Both are positively charged and repel each other, and the strong force acts only when they almost touch
- Wrong They are neutral, so they don’t attract each other
  Feedback
  In the hot plasma the nuclei have lost their electrons, and are bare, positively charged nuclei.
- Wrong Deuterium and tritium are very rare
  Feedback
  Deuterium is in all water, and tritium can be made. The difficulty is getting the nuclei close, not getting hold of them.
Feedback for a right answer
The closer two nuclei come, the harder they push apart. Only at about the width of a nucleus does the strong force win and pull them together: a tiny target behind a steep wall.

Worked solution
1. Hydrogen nuclei are positively charged, and like charges repel.
2. The repulsion grows as they come closer.
3. The strong force, which pulls them together, acts only within about a nucleus’s width. Getting there takes enormous speed, which means enormous temperature.
Step 2 of 7
Open this step
Heat is speed
Practice · Predict, then look
Question
The Temperature is at Sun’s core, 15 million kelvin, and almost nothing fuses. It is raised tenfold, to Tokamak, 150 million kelvin. Predict: how many times faster will each pair of nuclei fuse?

Correct answer
- Correct Thousands of times
Setup, choices and feedback
The Scene at this step
Temperature
15 million K
Fuel density
80 nucleifixed
Cut into the plasma
Machine
Choices
- Correct Thousands of times
- Wrong Ten times, like the temperature
  Misconception: Assumes the fusion rate grows in step with the temperature, or with the fuel density.
  Feedback from the catalogue
  The fusion rate depends on two things, and neither is linear. It grows as the square of the fuel density, because each deuterium nucleus needs a tritium partner, and with twice as many of each there are four times as many pairs. And with temperature it soars: only the fastest nuclei get close enough to tunnel through the repulsion, and their number climbs steeply with heat.
- Wrong A hundred times, like the temperature squared
  Misconception: Assumes the fusion rate grows in step with the temperature, or with the fuel density.
  Feedback from the catalogue
  The fusion rate depends on two things, and neither is linear. It grows as the square of the fuel density, because each deuterium nucleus needs a tritium partner, and with twice as many of each there are four times as many pairs. And with temperature it soars: only the fastest nuclei get close enough to tunnel through the repulsion, and their number climbs steeply with heat.
Once the prediction is locked in, the Scene changes
Temperature
150 million K
Cut into the plasma
Plasma
The Scene then shows: 6664 · How many times faster each D–T pair fuses than at the Sun’s core temperature, 15 million K/nodes/plasma/outputs/vsSunCore
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Almost 6700 times. Even at 150 million kelvin a typical nucleus has only a few per cent of the energy needed to climb the wall of repulsion. Fusion happens thanks to the fastest few and to quantum tunnelling through the wall, and both grow steeply with heat.

Worked solution
1. At ten times the temperature, the nuclei’s typical speed grows only $\sqrt{10} \approx 3$ ⁠ times.
2. But only nuclei in the fast tail get close enough to tunnel through the wall, and their number grows far more.
3. The chance of tunnelling through the wall also climbs steeply with speed.
4. Together: almost 6700 times for each pair.
Step 3 of 7
Open this step
A thousandfold
Practice · Reach the target
Question
Find the Temperature at which each pair of nuclei fuses 1000 times faster than in the Sun’s core.

Goal
- 1000 ± 12%
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Temperature
15 million K
Fuel density
80 nucleifixed
Cut into the plasma
Plasma
Read from the Scene at
How many times faster each D–T pair fuses than at the Sun’s core temperature, 15 million K/nodes/plasma/outputs/vsSunCore
One setting that reaches it
Temperature
75 million K
Feedback for a right answer
About 75 million kelvin: five times as hot as the Sun’s core, and a thousand times the rate.

Hint
At 150 million kelvin it is already 6700 times. Go down.

Worked solution
1. The rate climbs steeply with temperature, but not in a straight line, so you search by trying.
2. At 50 million kelvin it is about 260 times, at 70 about 850.
3. About 75 million kelvin gives about 1000 times.
Step 4 of 7
Open this step
Fuel density
Practice · Predict, then look
Question
The Fuel density is doubled from 80 to 160 nuclei, at the same temperature. Predict: what will happen to the fusion rate in the whole plasma?

Correct answer
- Correct It will grow fourfold
Setup, choices and feedback
The Scene at this step
Temperature
150 million Kfixed
Fuel density
80 nuclei
Cut into the plasma
Plasma
Choices
- Correct It will grow fourfold
- Wrong It will double, like the fuel
  Misconception: Assumes the fusion rate grows in step with the temperature, or with the fuel density.
  Feedback from the catalogue
  The fusion rate depends on two things, and neither is linear. It grows as the square of the fuel density, because each deuterium nucleus needs a tritium partner, and with twice as many of each there are four times as many pairs. And with temperature it soars: only the fastest nuclei get close enough to tunnel through the repulsion, and their number climbs steeply with heat.
- Wrong It won’t change: the temperature decides
  Feedback
  The temperature sets how fast each pair fuses. With more nuclei there are more pairs.
Once the prediction is locked in, the Scene changes
Fuel density
160 nuclei
Cut into the plasma
Plasma
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Each deuterium nucleus needs a tritium partner. Twice as many of each make four times as many possible pairs, so four times the fusions. That is why fusion needs a dense plasma, not just a hot one.

Worked solution
1. The fusion rate grows with the number of deuterium nuclei times the number of tritium nuclei: $n_D \times n_T$ ⁠.
2. Double both, and $2 \times 2 = 4$ ⁠.
3. The rate grows fourfold.
Step 5 of 7
Open this step
The neutron that escapes
Practice · Calculate
Question
Each deuterium–tritium fusion frees 17.6 MeV: a helium nucleus and a fast neutron, which carries 14.1 MeV of it. What share of the energy leaves with the neutron? Write it as a decimal.

Correct answer
0.8011 ± 0.01
Setup, choices and feedback
The Scene at this step
Temperature
150 million K
Fuel density
80 nuclei
Cut into the plasma
Plasma
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Share of each fusion’s 17.6 MeV the neutron carries off, 14.1 / 17.6/nodes/plasma/outputs/neutronShare
Feedback for a right answer
$\dfrac{14.1}{17.6} \approx 0.8$ ⁠: four fifths of the energy escapes the plasma with the neutron, and only the fifth in the helium stays and heats it.

Worked solution
1. The neutron’s share: $\dfrac{14.1}{17.6} \approx 0.80$ ⁠.
2. The helium’s share: $\dfrac{3.5}{17.6} \approx 0.20$ ⁠.
3. The neutron has no charge, so the magnetic field doesn’t hold it: it flies into the walls, where its energy is collected as heat.
Step 6 of 7
Open this step
Why the Sun gets away with less
Practice · Choose
Question
The Sun’s core is only at 15 million kelvin, yet it fuses about 600 million tonnes of hydrogen a second. How?

Correct answer
- Correct Gravity squeezes an enormous number of nuclei together and holds them there for billions of years
Setup, choices and feedback
The Scene at this step
Temperature
15 million K
Fuel density
80 nuclei
Cut into the plasma
Machine
Choices
Shown in a random order.
- Correct Gravity squeezes an enormous number of nuclei together and holds them there for billions of years
- Wrong In fact the Sun is hotter than any tokamak
  Misconception: Thinks the Sun's core must be hotter than a fusion reactor, since the Sun fuses so much.
  Feedback from the catalogue
  The Sun’s core is only about 15 million kelvin, ten times less than a fusion machine needs. The Sun wins by size and gravity: its weight squeezes the core to over a hundred times the density of water and holds it there for billions of years, with an enormous number of nuclei. Each nucleus waits billions of years on average, but together they are enough.
- Wrong It fuses deuterium and tritium, the easiest fuel
  Feedback
  The Sun mostly fuses ordinary hydrogen, through a far slower chain of steps. Deuterium and tritium are our choice on Earth.
Feedback for a right answer
Each nucleus in the Sun’s core waits billions of years on average to fuse, but there are so many that the total is huge. On Earth we have no such gravity, so we make up for it with heat: ten times hotter.

Worked solution
1. At 15 million kelvin each pair fuses very slowly.
2. But the Sun’s gravity squeezes the core to over a hundred times the density of water, with an enormous number of nuclei.
3. The huge number of nuclei makes up for each one’s slow rate.
Step 7 of 7
Open this step
How many fusions a second
Practice · Calculate
Question
The Temperature is locked at Tokamak, 150 million kelvin. Each cubic metre of the plasma holds 5 × 10¹⁹ deuterium nuclei and as many tritium nuclei. Each pair meets and fuses at a rate $\langle \sigma v \rangle \approx 2.05 \times 10^{-22}$ ⁠ cubic metres a second, and the number of fusions in each cubic metre per second is $n_D \, n_T \, \langle \sigma v \rangle$ ⁠. How many fusions happen in each cubic metre every second?

Correct answer
512300000000000000 ± 5%
Setup, choices and feedback
The Scene at this step
Temperature
150 million Kfixed
Fuel density
80 nuclei
Cut into the plasma
Plasma
Accepted within
± 5%
Computed from the Scene at this step’s setting
D–T reactivity ⟨σv⟩, m³/s (Bosch–Hale)/nodes/plasma/outputs/reactivity
Wrong answers it recognises
- 0.01025 ± 5%
  Misconception: Assumes the fusion rate grows in step with the temperature, or with the fuel density.
  Feedback from the catalogue
  The fusion rate depends on two things, and neither is linear. It grows as the square of the fuel density, because each deuterium nucleus needs a tritium partner, and with twice as many of each there are four times as many pairs. And with temperature it soars: only the fastest nuclei get close enough to tunnel through the repulsion, and their number climbs steeply with heat.
Feedback for a right answer
About 5 × 10¹⁷ a second, a megawatt and a half from every cubic metre. A reactor of hundreds of cubic metres gives hundreds of megawatts.

Hint
Write a large number like this: 5.1e17. Multiply the two densities together, then by $\langle \sigma v \rangle$ ⁠.

Worked solution
1. $n_D \, n_T = (5 \times 10^{19})^2 = 2.5 \times 10^{39}$ ⁠.
2. $2.5 \times 10^{39} \times 2.05 \times 10^{-22} \approx 5.1 \times 10^{17}$ ⁠ fusions in each cubic metre a second.
3. Each fusion frees 17.6 MeV, about 2.8 × 10⁻¹² J, so each cubic metre gives about 1.4 MW.
The rate goes with the product of the two densities: twice the fuel, four times the fusions.

Lesson 3 of 4 · About 30 minutes

Radiometric dating and the age of the Earth

A zircon crystal that sets its own clock to zero: the ratio of lead to uranium after a half-life, reading the age from the ratio, a clock that is spent and a clock that doesn't fit, lead that leaks and clocks that stop agreeing, and how we know the age of the Earth.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
The clock starts at zero
Practice · Choose
Question
Why does a zircon crystal that has just formed hold uranium but almost no lead?

Correct answer
- Correct Uranium atoms fit places in its lattice and lead doesn’t, so the growing crystal shuts it out
Setup, choices and feedback
The Scene at this step
Clock
Uranium-238
Time since the crystal formed
0 billion years
Lead lost later
0 %
Zoom to the atoms
Atoms inside
Choices
Shown in a random order.
- Correct Uranium atoms fit places in its lattice and lead doesn’t, so the growing crystal shuts it out
- Wrong There is no lead in the magma at all
  Feedback
  There is lead in the magma. Zircon is picky: it just doesn’t let it into its lattice.
- Wrong The lead decays fast and turns into uranium
  Feedback
  It’s the other way round: uranium decays to lead, and lead-206 is stable.
Feedback for a right answer
So the clock starts at zero. From then on every lead atom in the crystal was made by a decay, right where its uranium atom sat.

Worked solution
1. Uranium and thorium ions have the right size and charge for zirconium’s places in the lattice.
2. Lead ions have the wrong size and charge, and the growing crystal shuts them out.
3. A new crystal holds uranium and almost no lead: the clock is set to zero.
Step 2 of 7
Open this step
A half-life
Practice · Predict, then look
Question
The Clock is on Uranium-238, whose half-life is 4.47 billion years. Time since the crystal formed is moved to 4.47 billion years. Predict: how many lead atoms will the crystal hold for each uranium atom left?

Correct answer
1.001 ± 0.05
A number prediction counts as right within ± 0.05
Setup, choices and feedback
The Scene at this step
Clock
Uranium-238fixed
Time since the crystal formed
0 billion years
Lead lost later
0 %fixed
Zoom to the atoms
Atoms inside
Once the prediction is locked in, the Scene changes
Time since the crystal formed
4.47 billion years
The Scene then shows: 1.001 · Lead atoms per parent atom as measured, after any loss/nodes/zircon/outputs/ratio
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
One for one: after one half-life half the uranium has become lead. After two there will be three for each one, and after three, seven.

Worked solution
1. After one half-life half the uranium is left.
2. Each atom that decayed became a lead atom, so the other half is lead.
3. The ratio $\dfrac{D}{P} = \dfrac{1/2}{1/2} = 1$ ⁠.
Step 3 of 7
Open this step
Reading the age
Practice · Calculate
Question
A zircon crystal holds about 7 lead-207 atoms for each Uranium-235 atom left. Uranium-235’s half-life is 0.704 billion years. How old is the crystal, in billions of years?

Correct answer
2.11 ± 0.04
Setup, choices and feedback
The Scene at this step
Clock
Uranium-235fixed
Time since the crystal formed
2.11 billion yearshidden
Lead lost later
0 %fixed
Zoom to the atoms
In the rockfixed
Accepted within
± 0.04
Computed from the Scene at this step’s setting
The age the chosen clock reads, billions of years/nodes/zircon/outputs/ageRead
Wrong answers it recognises
- 4.93 ± 0.04
  Misconception: Treats exponential growth or decay as linear: adds the same amount each generation or each half-life instead of multiplying.
  Feedback from the catalogue
  In a fission chain each generation multiplies the number of neutrons by $k$ ⁠, so after $n$ ⁠ generations there are $k^n$ ⁠ of them, not $1 + n(k - 1)$ ⁠. Decay works the same way: each half-life leaves half of what there was, so the uranium doesn’t fall by the same amount every year, and the ratio of lead to uranium grows as $2^n - 1$ ⁠.
Feedback for a right answer
$1 + 7 = 8 = 2^3$ ⁠: three half-lives, $3 \times 0.704 \approx 2.11$ ⁠ billion years. An eighth of the uranium is left, and the other seven eighths are lead.

Hint
$t = t_{1/2} \cdot \log_2\left(1 + \dfrac{D}{P}\right)$ ⁠.

Worked solution
1. $1 + \dfrac{D}{P} = 1 + 7 = 8$ ⁠.
2. $\log_2 8 = 3$ ⁠: three half-lives have passed.
3. $t = 3 \times 0.704 \approx 2.11$ ⁠ billion years.
If you got about 4.93, you multiplied the half-life by 7, as if each half-life added one lead atom. But each half-life halves the uranium that is left.
Step 4 of 7
Open this step
A spent clock
Practice · Predict, then look
Question
The crystal is 4.4 billion years old, and the Clock is on Uranium-238: about half the atoms are still uranium. It is switched to Uranium-235, with a half-life of 0.704 billion years. Predict: what share of the uranium-235 is left?

Correct answer
- Correct About one per cent
Setup, choices and feedback
The Scene at this step
Clock
Uranium-238
Time since the crystal formed
4.4 billion yearsfixed
Lead lost later
0 %fixed
Zoom to the atoms
Atoms inside
Choices
- Correct About one per cent
- Wrong About half as well: same crystal, same age
  Feedback
  The age is the same, but uranium-235 decays six times as fast. More than six of its half-lives have passed.
- Wrong Nothing is left
  Feedback
  Decay never quite reaches zero: each half-life leaves half of what there was.
Once the prediction is locked in, the Scene changes
Clock
Uranium-235
Zoom to the atoms
Atoms inside
The Scene then shows: 0.01312 · Share of the starting parent atoms still there, 0..1/nodes/zircon/outputs/parentLeft
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$\dfrac{4.4}{0.704} \approx 6.25$ ⁠ half-lives, and $2^{-6.25} \approx 0.013$ ⁠: about one per cent. The clock is nearly spent, and reads less well: a small difference in the uranium left changes the age a lot.

Worked solution
1. The number of half-lives: $\dfrac{4.4}{0.704} \approx 6.25$ ⁠.
2. What is left: $2^{-6.25} \approx 0.013$ ⁠.
3. About 1% of the uranium-235 is left.
Step 5 of 7
Open this step
Which clock
Practice · Choose
Question
Which clock can not date a 4.4-billion-year-old rock?

Correct answer
- Correct Carbon-14, with a half-life of 5730 years
Setup, choices and feedback
The Scene at this step
Clock
Uranium-238
Time since the crystal formed
4.4 billion years
Lead lost later
0 %
Zoom to the atoms
Atoms inside
Choices
Shown in a random order.
- Correct Carbon-14, with a half-life of 5730 years
- Wrong Uranium-238, with a half-life of 4.47 billion years
  Feedback
  This is actually the best clock here: its half-life is close to the age being measured.
- Wrong Thorium-232, with a half-life of 14 billion years
  Feedback
  Thorium-232 is slow, but in a 4.4-billion-year-old crystal it has made enough lead-208 to measure.
- Wrong All of them can: any radioactive clock dates any age
  Misconception: Thinks carbon-14 dating can date any old object, rocks included.
  Feedback from the catalogue
  Carbon-14’s half-life is only 5730 years. After about 50,000 years too little is left to measure, so it dates bones, wood and cloth, never rocks millions or billions of years old. A clock reads best when its half-life is not far from the age being measured.
Feedback for a right answer
After about 50,000 years almost no carbon-14 is left to measure, so it dates bones and wood. Rocks are dated with slow clocks: uranium, thorium and potassium-40.

Worked solution
1. 4.4 billion years is about 770,000 half-lives of carbon-14.
2. What is left, $2^{-770\,000}$ ⁠, is zero for every purpose.
3. A clock reads best when its half-life is not far from the age, so ancient rocks suit uranium and thorium.
Step 6 of 7
Open this step
Lead that leaks
Practice · Predict, then look
Question
The crystal is 4.4 billion years old, and its clocks all agree. Lead lost later is raised to 30%: about a third of the lead escapes. Predict: what will the clocks read?

Correct answer
- Correct Ages that are too young, and differ from each other
Setup, choices and feedback
The Scene at this step
Clock
Uranium-238
Time since the crystal formed
4.4 billion yearsfixed
Lead lost later
0 %
Zoom to the atoms
Atoms inside
Choices
- Correct Ages that are too young, and differ from each other
- Wrong They will all read the same age, too young
  Misconception: Thinks that if a crystal loses lead, every clock in it is thrown off by the same amount, so they still agree.
  Feedback from the catalogue
  Losing lead does not shift every clock by the same amount. Uranium-235 has made about 75 lead atoms for each one left, uranium-238 only about one. Losing a third of the lead takes a big bite out of the uranium-238 age and a small one out of the uranium-235 age, and the clocks stop agreeing. That disagreement is what gives the leak away.
- Wrong Ages that are too old
  Feedback
  Less lead for each uranium atom looks like less time has passed, so the age comes out younger.
Once the prediction is locked in, the Scene changes
Lead lost later
30 %
Zoom to the atoms
Atoms inside
The Scene then shows: 3.364 · The age the U-238 clock reads, billions of years/nodes/zircon/outputs/ageU238
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Uranium-238 now reads about 3.36 billion years, and uranium-235 about 4.04. Neither is right, and the gap between them gives the leak away. Ages are trusted only when the clocks agree.

Worked solution
1. Losing lead cuts each clock’s $\dfrac{D}{P}$ ⁠ by 30%.
2. For uranium-238 $\dfrac{D}{P} \approx 1$ ⁠, and dropping to 0.7 lowers $\log_2(1 + \dfrac{D}{P})$ ⁠ a lot: from 1 to 0.77.
3. For uranium-235 $\dfrac{D}{P} \approx 75$ ⁠, and dropping to 53 hardly changes the logarithm.
4. So the two clocks read ages that are too young, and different.
Step 7 of 7
Open this step
The age of the Earth
Practice · Choose
Question
The oldest crystals found on Earth are about 4.4 billion years old. How do we know the Earth itself is 4.54 billion years old?

Correct answer
- Correct From dating meteorites, which formed with the planets from the same cloud and have hardly changed since
Setup, choices and feedback
The Scene at this step
Clock
Uranium-238
Time since the crystal formed
4.54 billion years
Lead lost later
0 %
Zoom to the atoms
Atoms inside
Choices
Shown in a random order.
- Correct From dating meteorites, which formed with the planets from the same cloud and have hardly changed since
- Wrong From the oldest rock found on Earth
  Feedback
  The oldest whole rock is only about 4.0 billion years old. The Earth’s surface is endlessly recycled, and nothing from the first days survives.
- Wrong From working out how long the Earth took to cool
  Feedback
  That is how Lord Kelvin worked it out, and he got only tens of millions of years. He didn’t know that radioactive decay keeps heating the Earth from inside.
Feedback for a right answer
In 1956 Clair Patterson measured the lead in meteorites and in sediment from the ocean floor, and found they all fit a single age: about 4.55 billion years.

Worked solution
1. The Earth’s surface is melted, worn down and rebuilt by plate tectonics and erosion, so the first rocks did not survive.
2. Meteorites are pieces of asteroids that formed at the same time and from the same cloud as the Earth, and never went through that cycle.
3. Their uranium–lead clocks read about 4.55 billion years.

Lesson 4 of 4 · About 25 minutes

Particle physics and the Standard Model

What is inside a proton: quarks and fractional charges, mass that is mostly energy, a gluon string that pulls with the weight of fifteen tonnes and snaps into a new particle, and the three generations of the Standard Model.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Quarks
Practice · Predict, then look
Question
An up quark carries a charge of $+\tfrac{2}{3}$ ⁠, and down and strange quarks $-\tfrac{1}{3}$ ⁠ each. The omega is made of three strange quarks. Predict: what is its charge, in units of the proton’s charge?

Correct answer
-1 ± 0.05
A number prediction counts as right within ± 0.05
Setup, choices and feedback
The Scene at this step
Particle
Proton
Pull a quark out
0 fmfixed
Zoom out to the Standard Model
Inside the particle
Once the prediction is locked in, the Scene changes
Particle
Omega Ω⁻
The Scene then shows: -1 · The hadron's electric charge, in units of the proton's/nodes/particle/outputs/charge
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$-\tfrac{1}{3} - \tfrac{1}{3} - \tfrac{1}{3} = -1$ ⁠. In 1964 Gell-Mann and Ne’eman predicted a particle like this, and it was found with almost exactly the predicted mass.

Worked solution
1. Each strange quark has a charge of $-\tfrac{1}{3}$ ⁠.
2. $3 \times \left(-\tfrac{1}{3}\right) = -1$ ⁠.
3. As in the proton, $+\tfrac{2}{3} + \tfrac{2}{3} - \tfrac{1}{3} = +1$ ⁠, the quarks’ fractional charges always add up to a whole number.
Step 2 of 6
Open this step
Where the mass comes from
Practice · Choose
Question
A proton weighs 938 MeV, and its three quarks, each on its own, weigh about 9 MeV together. Where does the rest of the mass come from?

Correct answer
- Correct From the energy of the moving quarks and of the gluon field that holds them
Setup, choices and feedback
The Scene at this step
Particle
Proton
Pull a quark out
0 fm
Zoom out to the Standard Model
Inside the particle
Choices
Shown in a random order.
- Correct From the energy of the moving quarks and of the gluon field that holds them
- Wrong From the Higgs field, which gives all matter its mass
  Misconception: Thinks the Higgs field gives all matter, the proton included, all its mass.
  Feedback from the catalogue
  The Higgs field gives mass to the elementary particles themselves: the quarks, the electron and the W and Z bosons. But the quarks’ mass is only about one per cent of the proton’s. The other 99 per cent comes from the energy of the moving quarks and the gluon field, not from the Higgs.
- Wrong There are more heavy quarks hidden in the proton
  Misconception: Thinks a proton's mass is the sum of its three quarks' masses.
  Feedback from the catalogue
  The proton’s three quarks together weigh about 9 MeV, about one per cent of 938 MeV. All the rest is energy: the quarks move close to the speed of light, and the gluon field that holds them stores a great deal of energy. By $E = mc^2$ ⁠ energy has mass, and that is where almost all the proton’s mass comes from.
Feedback for a right answer
By $E = mc^2$ ⁠ energy has mass. About 99% of the proton’s mass, and of the mass of almost everything you can weigh, is the energy of motion and of the strong force’s field.

Worked solution
1. $\dfrac{9}{938} \approx 0.01$ ⁠: the quarks themselves are about one per cent of the mass.
2. The quarks move inside the proton close to the speed of light, and the gluon field stores a great deal of energy.
3. That energy, by $E = mc^2$ ⁠, is about 99% of the mass.
Step 3 of 6
Open this step
Heavy quarks
Practice · Predict, then look
Question
The J/psi is made of a charm quark and an anticharm quark, each hundreds of times heavier than a down quark. Predict: what share of the J/psi’s mass is the quarks’ own mass?

Correct answer
- Correct Most of it
Setup, choices and feedback
The Scene at this step
Particle
Proton
Pull a quark out
0 fmfixed
Zoom out to the Standard Model
Inside the particle
Choices
- Correct Most of it
- Wrong About one per cent, as in the proton
  Feedback
  The field’s energy is indeed similar in the two particles, but here the quarks themselves are heavy, so they are most of the mass.
- Wrong Exactly all of it
  Misconception: Thinks a proton's mass is the sum of its three quarks' masses.
  Feedback from the catalogue
  The proton’s three quarks together weigh about 9 MeV, about one per cent of 938 MeV. All the rest is energy: the quarks move close to the speed of light, and the gluon field that holds them stores a great deal of energy. By $E = mc^2$ ⁠ energy has mass, and that is where almost all the proton’s mass comes from.
Once the prediction is locked in, the Scene changes
Particle
J/psi
The Scene then shows: 0.8221 · Share of the mass that is the quarks' own mass, 0..1/nodes/particle/outputs/quarkShare
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 82%: the two charm quarks weigh about 2546 MeV out of 3097. With heavy quarks, their own mass is most of the story.

Worked solution
1. A charm quark weighs about 1273 MeV, so two weigh about 2546 MeV.
2. The J/psi weighs 3097 MeV.
3. $\dfrac{2546}{3097} \approx 0.82$ ⁠.
Step 4 of 6
Open this step
The string
Practice · Calculate
Question
The quark is pulled one femtometre out, and the gluon field’s tube stores 0.9 GeV. How hard does it pull the quark back? 1 GeV = 1.602 × 10⁻¹⁰ J, and 1 fm = 10⁻¹⁵ m.

Correct answer
144200 N ± 3%
Setup, choices and feedback
The Scene at this step
Particle
Protonfixed
Pull a quark out
1 fmfixed
Zoom out to the Standard Model
Inside the particle
Accepted within
± 3%
Computed from the Scene at this step’s setting
The force the stretched gluon string pulls back with, 0.9 GeV per fm; 0 at rest and once it has snapped/nodes/particle/outputs/stringForce
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$F = \dfrac{0.9 \times 1.602 \times 10^{-10}}{10^{-15}} \approx 1.4 \times 10^5$ ⁠ newtons: the weight of about 15 tonnes, on a single quark.

Hint
Energy that grows at a steady rate with distance is a steady force: $F = \dfrac{E}{d}$ ⁠.

Worked solution
1. The energy in joules: $0.9 \times 1.602 \times 10^{-10} \approx 1.44 \times 10^{-10}$ ⁠.
2. The distance in metres: $10^{-15}$ ⁠.
3. Every femtometre adds the same energy, so the force is steady: $F = \dfrac{E}{d} = \dfrac{1.44 \times 10^{-10}}{10^{-15}} \approx 1.44 \times 10^5$ ⁠ newtons.
4. $\dfrac{1.44 \times 10^5}{9.8} \approx 14\,700$ ⁠ kilograms: the weight of about 15 tonnes.
Step 5 of 6
Open this step
Keep pulling
Practice · Predict, then look
Question
The quark is pulled further, to 1.5 fm. Predict what will happen.

Correct answer
- Correct The tube will snap into a new quark and antiquark, and the quark will leave inside a new particle
Setup, choices and feedback
The Scene at this step
Particle
Protonfixed
Pull a quark out
1 fm
Zoom out to the Standard Model
Inside the particle
Choices
- Correct The tube will snap into a new quark and antiquark, and the quark will leave inside a new particle
- Wrong The quark will come out free
  Misconception: Thinks pulling hard enough on a quark will free it from the particle.
  Feedback from the catalogue
  The strong force between quarks does not weaken as they move apart. The gluon field gathers into a narrow tube, and every extra femtometre of it adds the same energy. When it holds enough energy to make matter, the tube snaps into a new quark and antiquark. Instead of a free quark you get a new particle. This is confinement: no one has ever seen a quark on its own.
- Wrong The tube will snap the quark back in, like a spring
  Feedback
  If you let go, yes. But keep pulling and the energy in the tube grows until it is enough to make new matter.
Once the prediction is locked in, the Scene changes
Pull a quark out
1.5 fm
The Scene then shows: 1 · 1 once the string has snapped into a new quark pair, else 0/nodes/particle/outputs/snapped
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At about 1.2 fm the tube holds enough energy, and it snaps into a new up quark and anti-up quark. The pulled quark leaves with the anti-up as a neutral pion, and the new one takes its place in the proton. This is confinement.

Worked solution
1. The strong force doesn’t weaken with distance: each femtometre adds about 0.9 GeV.
2. Once the tube’s energy is enough to make a pair of quarks, it is easier for nature to make a new pair than to stretch further.
3. The new pair forms at the break, and each end of the tube gets a quark: two particles come out, and no quark leaves alone.
Step 6 of 6
Open this step
The Standard Model
Practice · Choose
Question
Zoom out to the Standard Model is at Standard Model. Which matter particles is almost everything around you made of?

Correct answer
- Correct The first generation alone: up and down quarks and electrons
Setup, choices and feedback
The Scene at this step
Particle
Proton
Pull a quark out
0 fm
Zoom out to the Standard Model
Standard Model
Choices
Shown in a random order.
- Correct The first generation alone: up and down quarks and electrons
- Wrong All twelve matter particles, about equally
  Feedback
  The second and third generations are made only in colliders and cosmic-ray showers, and decay into lighter ones in a fraction of a second.
- Wrong Photons and gluons
  Feedback
  Photons and gluons carry forces. They hold matter together, but are not the matter itself.
Feedback for a right answer
Protons and neutrons are made of up and down quarks, with electrons around them. The heavier generations behave the same way in every force, only heavier, and no one yet knows why there are exactly three.

Worked solution
1. A proton: two up and one down. A neutron: one up and two down.
2. Electrons move around the nuclei.
3. Those three, the first generation, build almost all ordinary matter.

Across the unit · 11 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Nuclear fission

A nucleus that tears

Question

Correct answer

The Scene at this step

Choices

Worked solution

Where the energy comes from

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The multiplication factor

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A steady chain

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Pulling the rods out

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Ten generations

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Nuclear fusion

Repulsion

Question

Correct answer

The Scene at this step

Choices

Worked solution

Heat is speed

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A thousandfold

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Fuel density

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The neutron that escapes

Question

Correct answer

The Scene at this step

Worked solution

Why the Sun gets away with less

Question

Correct answer

The Scene at this step

Choices