Answer key · Unit 15

Modern physics.

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 7

Lesson 1 of 7 · About 25 minutes

Atoms and nuclei

What an atom is made of: protons and neutrons in a tiny nucleus with electrons around it, the atomic number and the mass number, isotopes, electron shells, and why an electron is not a planet.

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The 6 steps and their answers

Step 1 of 6
Open this step
What is inside an atom
No practice question in this step.
The Scene at this step
Protons
6
Electron view
Shells
Step 2 of 6
Open this step
Counting neutrons
Practice · Calculate
Question
A sodium nucleus is $^{23}_{11}\mathrm{Na}$ ⁠. How many neutrons does it have?

Correct answer
12 ± 0.01
Setup, choices and feedback
The Scene at this step
Protons
11fixed
Electron view
Shells
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Neutrons in the most common isotope/nodes/atom/outputs/neutrons
Wrong answers it recognises
- 23 ± 0.01
  Misconception: Takes the mass number A to be the number of neutrons.
  Feedback from the catalogue
  The mass number $A$ ⁠ counts every particle in the nucleus, protons and neutrons together: $A = Z + N$ ⁠. To get the number of neutrons, subtract the atomic number from it: $N = A - Z$ ⁠.
- 11 ± 0.01
  Misconception: Assumes every nucleus has as many neutrons as protons.
  Feedback from the catalogue
  Only in light nuclei is the number of neutrons close to the number of protons. The heavier the nucleus, the more neutrons it has than protons: sodium-23 has $11$ ⁠ protons and $12$ ⁠ neutrons, and uranium-238 has $92$ ⁠ protons and $146$ ⁠ neutrons. Work the number of neutrons out: $N = A - Z$ ⁠.
Feedback for a right answer
$N = A - Z = 23 - 11 = 12$ ⁠. Sodium has one more neutron than protons, and in heavier nuclei the gap grows.

Hint
The mass number counts every particle in the nucleus.

Worked solution
1. Given: the mass number $A = 23$ ⁠, the atomic number $Z = 11$ ⁠.
2. $A = Z + N$ ⁠, so $N = A - Z$ ⁠.
3. $N = 23 - 11 = 12$ ⁠ neutrons.
If you answered $23$ ⁠, you gave the mass number: it counts the protons too.
Step 3 of 6
Open this step
Isotopes
No practice question in this step.
The Scene at this step
Protons
6
Electron view
Shells
Step 4 of 6
Open this step
Shells
Practice · Predict, then look
Question
Neon has two full shells. How many electrons will the outer shell of sodium hold, with one more proton?

Correct answer
1 ± 0.1
A number prediction counts as right within ± 0.1
Setup, choices and feedback
The Scene at this step
Protons
10
Electron view
Shells
Once the prediction is locked in, the Scene changes
Protons
11
The Scene then shows: 1 · Electrons in the outermost occupied shell/nodes/atom/outputs/outerElectrons
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The $11$ ⁠th electron finds no room in the two full shells, so it starts a third shell on its own. A lone electron like that leaves easily, which is why sodium reacts so eagerly.

Worked solution
1. Sodium has $Z = 11$ ⁠, so the neutral atom has $11$ ⁠ electrons.
2. The first shell takes $2$ ⁠ and the second $8$ ⁠: $2 + 8 = 10$ ⁠.
3. That leaves $11 - 10 = 1$ ⁠ electron, in the third shell.
If you predicted $9$ ⁠, you added the electron to the second shell, but it is already full with eight.
Step 5 of 6
Open this step
A full outer shell
Practice · Reach the target
Question
Use Protons to find an atom with three shells, the outer one full, and press Check.

Goal
- 3 ± 0.1
  At the start of the step: 2
- 8 ± 0.1
  At the start of the step: 4
Setup, choices and feedback
The Scene at this step
Protons
6
Electron view
Shells
Read from the Scene at
Shells holding at least one electron/nodes/atom/outputs/shells
Read from the Scene at
Electrons in the outermost occupied shell/nodes/atom/outputs/outerElectrons
One setting that reaches it
Protons
18
Feedback for a right answer
Argon, $Z = 18$ ⁠: $2 + 8 + 8$ ⁠. Like helium and neon it is a noble gas, and about one per cent of the air you breathe is argon.

Feedback for a wrong answer
Not yet. Count how many electrons it takes to fill three shells.

Hint
The third shell fills with eight electrons, after $2 + 8$ ⁠ in the first two.

Worked solution
1. Three full shells take $2 + 8 + 8 = 18$ ⁠ electrons.
2. In a neutral atom the number of electrons equals the number of protons, so $Z = 18$ ⁠: argon.
3. At $Z = 19$ ⁠ the $19$ ⁠th electron already starts a fourth shell.
If you stopped at neon, $Z = 10$ ⁠, the outer shell is full, but there are only two shells.
Step 6 of 6
Open this step
Clouds, not orbits
Practice · Choose
Question
What best describes an electron in an atom?

Correct answer
- Correct You cannot say where it is or what path it follows; you can only say how likely you are to find it at each place round the nucleus
Setup, choices and feedback
The Scene at this step
Protons
6
Electron view
Cloud
Choices
Shown in a random order.
- Wrong It circles the nucleus in a fixed circular orbit, like a planet round the Sun
  Misconception: Pictures electrons as moving round the nucleus in fixed circular orbits, like planets round the Sun.
  Feedback from the catalogue
  The planets picture is handy for counting, but it is not right. An electron in an atom does not follow a path you could draw: quantum mechanics gives only the chance of finding it in each place, and that chance is spread as a cloud round the nucleus. A shell is a region where the chance is high, not an orbit.
- Correct You cannot say where it is or what path it follows; you can only say how likely you are to find it at each place round the nucleus
- Wrong It sits still at one spot on its ring
  Feedback
  The rings are a drawing. An electron does not sit still, and it does not move along the ring either.
- Wrong It is inside the nucleus, next to the protons
  Feedback
  The nucleus holds only protons and neutrons. The electrons are round it, in a space about $100{,}000$ ⁠ times wider.
Feedback for a right answer
The cloud in the Scene shows that chance: it is bright where the electron is likely to be found. The shells are regions of high chance, not orbits.

Worked solution
1. In quantum mechanics an electron in an atom has no path you could draw.
2. What can be worked out is the chance of finding it at each place, and that is the cloud in the Scene.
3. The shells are layers of the cloud, one inside another.
If you chose the planet, that is Rutherford’s and Bohr’s picture: an important step in history, and still handy for counting electrons, but not what happens in an atom.

Lesson 2 of 7 · About 30 minutes

Radioactive decay

What comes out of an unstable nucleus and what is left of it: alpha decay, beta decay and gamma rays, nuclear equations, a nucleus with no memory, and the activity of a radioactive source.

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The 8 steps and their answers

Step 1 of 8
Open this step
Unstable nuclei
No practice question in this step.
The Scene at this step
Half-life
5 s
Nuclei in the sample
200
Radiation
Alpha
Nuclei / Decay curve
Nuclei
Step 2 of 8
Open this step
Alpha decay
Practice · Calculate
Question
Uranium-238, $^{238}_{92}\mathrm{U}$ ⁠, gives off an alpha particle. What is the mass number of the daughter nucleus?

Correct answer
234 ± 0.01
Setup, choices and feedback
The Scene at this step
Half-life
5 s
Nuclei in the sample
200
Radiation
Alpha
Nuclei / Decay curve
Nuclei
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 236 ± 0.01
  Misconception: Subtracts 2 from the mass number in alpha decay, using the atomic number of the alpha particle instead of its mass number.
  Feedback from the catalogue
  An alpha particle is $^{4}_{2}\mathrm{He}$ ⁠: two protons and two neutrons. The nucleus loses four particles, so the mass number falls by $4$ ⁠. It is the atomic number, the number of protons, that falls by $2$ ⁠.
- 238 ± 0.01
  Misconception: Keeps the mass number unchanged in alpha decay, as in beta decay.
  Feedback from the catalogue
  The mass number stays the same in beta decay, because there a neutron only turns into a proton. In alpha decay four particles leave the nucleus, so the mass number falls by $4$ ⁠: the mass numbers add up to the same total on both sides of the equation.
Feedback for a right answer
$238 - 4 = 234$ ⁠, and $92 - 2 = 90$ ⁠: the daughter is thorium-234, $^{234}_{90}\mathrm{Th}$ ⁠.

Hint
An alpha particle is $^{4}_{2}\mathrm{He}$ ⁠.

Worked solution
1. The equation: $^{238}_{92}\mathrm{U} \to {}^{A}_{Z}\mathrm{X} + {}^{4}_{2}\mathrm{He}$ ⁠.
2. The mass numbers add up the same: $238 = A + 4$ ⁠, so $A = 234$ ⁠.
3. The atomic numbers add up the same: $92 = Z + 2$ ⁠, so $Z = 90$ ⁠, thorium.
If you answered $236$ ⁠, you took off $2$ ⁠, the alpha particle’s atomic number, instead of its mass number, $4$ ⁠.
Step 3 of 8
Open this step
Beta decay
Practice · Calculate
Question
Carbon-14, $^{14}_{6}\mathrm{C}$ ⁠, decays by beta decay. What is the atomic number of the daughter nucleus?

Correct answer
7 ± 0.01
Setup, choices and feedback
The Scene at this step
Half-life
5 s
Nuclei in the sample
200
Radiation
Beta
Nuclei / Decay curve
Nuclei
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 5 ± 0.01
  Misconception: Thinks beta decay lowers the atomic number, as losing a negative particle seemed to.
  Feedback from the catalogue
  In beta decay a neutron, which has no charge, turns into a proton and an electron. The electron leaves and the proton stays in the nucleus. So the nucleus has one more proton, and the atomic number rises by $1$ ⁠. Check it by the charge: $6 = 7 + (-1)$ ⁠.
- 6 ± 0.01
  Misconception: Thinks the beta electron comes from the electron shells, so the nucleus and the element stay the same.
  Feedback from the catalogue
  The electron of beta radiation does not come from the shells. It is made inside the nucleus, when a neutron turns into a proton. The number of protons in the nucleus goes up by one, so it is now another element: carbon-14 becomes nitrogen-14.
Feedback for a right answer
A neutron turned into a proton, so $Z$ ⁠ went up from $6$ ⁠ to $7$ ⁠: nitrogen. The mass number stays $14$ ⁠: $^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\mathrm{e}$ ⁠.

Hint
Inside the nucleus a neutron turns into a proton.

Worked solution
1. The equation: $^{14}_{6}\mathrm{C} \to {}^{A}_{Z}\mathrm{X} + {}^{0}_{-1}\mathrm{e}$ ⁠.
2. Mass numbers: $14 = A + 0$ ⁠, so $A = 14$ ⁠.
3. Atomic numbers, that is the charge: $6 = Z + (-1)$ ⁠, so $Z = 7$ ⁠: nitrogen-14.
If you answered $5$ ⁠, you subtracted the electron’s charge instead of balancing it: the electron takes away negative charge, so the nucleus is left with more positive charge.
Step 4 of 8
Open this step
Gamma rays
No practice question in this step.
The Scene at this step
Half-life
5 s
Nuclei in the sample
200
Radiation
Gamma
Nuclei / Decay curve
Nuclei
Step 5 of 8
Open this step
No memory
Practice · Choose
Question
The half-life is 5 s. One nucleus in the sample has not decayed after 15 s, three half-lives. What is the chance that it decays in the next 5 s?

Correct answer
- Correct $\frac{1}{2}$ ⁠
Setup, choices and feedback
The Scene at this step
Half-life
5 s
Nuclei in the sample
20
Radiation
Alpha
Nuclei / Decay curve
Decay curve
Choices
Shown in a random order.
- Correct $\frac{1}{2}$ ⁠
- Wrong More than $\frac{1}{2}$ ⁠, because its time has come
  Misconception: Thinks a nucleus that has survived a long time is more likely to decay soon, as if it were old.
  Feedback from the catalogue
  A nucleus has no age. Its chance of decaying in each second is the same for as long as it has not decayed, just as a die always has the same chance of landing on six, even after ten throws without one. So a nucleus that has survived three half-lives has a chance of one half of decaying in the next half-life, exactly like a new one.
- Wrong $\frac{1}{16}$ ⁠
  Feedback
  $\frac{1}{16}$ ⁠ is the chance that a new nucleus survives four half-lives. Here we already know this nucleus has survived three, and the only question is what happens in the next one.
- Wrong Less than $\frac{1}{2}$ ⁠, because it is probably an extra-strong nucleus
  Feedback
  All nuclei of one isotope are the same. There are no strong or weak nuclei, and no sign of a nucleus about to decay.
Feedback for a right answer
A nucleus has no memory. For as long as it has not decayed, its chance of decaying in the next half-life is one half, just like a new one’s.

Worked solution
1. A half-life is the time in which half of any group of not-yet-decayed nuclei decays.
2. The nuclei that have survived 15 s are such a group, and half of them will decay in the next 5 s.
3. So the chance for each of them is $\frac{1}{2}$ ⁠.
If you chose “more than half”, you thought of the nucleus as something that ages. But a nucleus has no age, and that is exactly what makes decay exponential.
Step 6 of 8
Open this step
Activity
Practice · Calculate
Question
The sample has $400$ ⁠ nuclei, and the half-life is 5 s. What is the sample’s activity as it starts, in becquerels?

Correct answer
55.45 Bq ± 2%
Setup, choices and feedback
The Scene at this step
Half-life
5 sfixed
Nuclei in the sample
400fixed
Radiation
Alpha
Nuclei / Decay curve
Nuclei
Accepted within
± 2%
Computed from the Scene at this step’s setting
Expected decays per second of the fresh sample, A = λ·N/nodes/sample/outputs/activity
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 80 Bq ± 1 Bq
  Misconception: Takes the activity as N divided by the half-life, leaving out ln 2.
  Feedback from the catalogue
  The decay constant is not $\dfrac{1}{T_{1/2}}$ ⁠ but $\lambda = \dfrac{\ln 2}{T_{1/2}} \approx \dfrac{0.693}{T_{1/2}}$ ⁠. The factor $\ln 2$ ⁠ comes from the link between $e^{-\lambda t}$ ⁠ and $2^{-t/T_{1/2}}$ ⁠. So $A = \lambda N$ ⁠ is about $30\%$ ⁠ less than $\dfrac{N}{T_{1/2}}$ ⁠.
Feedback for a right answer
$\lambda \approx 0.139\ \mathrm{1/s}$ ⁠: each nucleus has a chance of about $14\%$ ⁠ of decaying in each second. Out of $400$ ⁠ nuclei, about $55$ ⁠ decay each second on average.

Hint
First work out the decay constant, $\lambda = \dfrac{\ln 2}{T_{1/2}}$ ⁠.

Worked solution
1. Given: $N = 400$ ⁠, $T_{1/2} = 5\ \mathrm{s}$ ⁠.
2. The decay constant: $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5} \approx 0.139\ \mathrm{1/s}$ ⁠.
3. The activity: $A = \lambda N = 0.139 \cdot 400 \approx 55.5\ \mathrm{Bq}$ ⁠.
If you answered 80 Bq, you divided $N$ ⁠ by the half-life without $\ln 2$ ⁠.
Step 7 of 8
Open this step
The activity falls
Practice · Calculate
Question
In the same sample, what will the activity be after 10 s?

Correct answer
13.86 Bq ± 2%
Setup, choices and feedback
The Scene at this step
Half-life
5 sfixed
Nuclei in the sample
400fixed
Radiation
Alpha
Nuclei / Decay curve
Decay curve
Accepted within
± 2%
Computed from the Scene at this step’s setting
Expected decays per second of the fresh sample, A = λ·N/nodes/sample/outputs/activity
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 Bq ± 0.5 Bq
  Misconception: Thinks two half-lives make a whole: nothing is left, or the activity is zero, after two half-lives.
  Feedback from the catalogue
  A half-life does not take away half of the starting amount, but half of what is left. After one half-life half is left, and after two, half of the half: a quarter. The amount, and the activity with it, never reach zero after a set number of half-lives.
- 55.45 Bq ± 2%
  Misconception: Thinks the activity is a fixed property of the isotope and does not fall as the sample decays.
  Feedback from the catalogue
  The half-life is a property of the isotope, but the activity also depends on how many nuclei are left: $A = \lambda N$ ⁠. As the nuclei decay, $N$ ⁠ falls, so the activity falls too, by half every half-life.
Feedback for a right answer
After two half-lives a quarter of the nuclei are left, so the activity is a quarter too: about 13.9 Bq.

Hint
10 s is two half-lives, and the activity is proportional to the number of nuclei left.

Worked solution
1. $t = 10\ \mathrm{s}$ ⁠ and $T_{1/2} = 5\ \mathrm{s}$ ⁠, so $n = \dfrac{10}{5} = 2$ ⁠ half-lives have passed.
2. $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ ⁠ of the nuclei are left: $100$ ⁠ out of $400$ ⁠.
3. $A = \lambda N = 0.139 \cdot 100 \approx 13.9\ \mathrm{Bq}$ ⁠, a quarter of 55.5 Bq.
If you answered $0$ ⁠, you thought two halves make a whole. But each half-life takes half of what is left, not half of the start.
Step 8 of 8
Open this step
Choosing a half-life
Practice · Reach the target
Question
The sample has $200$ ⁠ nuclei, locked. Change Half-life so that the starting activity is 20 Bq, and press Check.

Goal
- 20 Bq ± 0.5 Bq
  At the start of the step: 27.73 Bq
Setup, choices and feedback
The Scene at this step
Half-life
5 s
Nuclei in the sample
200fixed
Radiation
Alpha
Nuclei / Decay curve
Nuclei
Read from the Scene at
Expected decays per second of the fresh sample, A = λ·N/nodes/sample/outputs/activity
One setting that reaches it
Half-life
7 s
Radiation
Alpha
Feedback for a right answer
At 7 s the activity is about 19.8 Bq. A longer half-life gives a smaller activity from the same number of nuclei.

Feedback for a wrong answer
Not yet. If the activity is too high, make the half-life longer.

Hint
From $A = \lambda N$ ⁠ find $\lambda$ ⁠, then $T_{1/2} = \dfrac{\ln 2}{\lambda}$ ⁠.

Worked solution
1. Given: $A = 20\ \mathrm{Bq}$ ⁠, $N = 200$ ⁠.
2. $\lambda = \dfrac{A}{N} = \dfrac{20}{200} = 0.1\ \mathrm{1/s}$ ⁠.
3. $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.1} \approx 6.9\ \mathrm{s}$ ⁠. The slider moves in half seconds, and 7 s gives 19.8 Bq.
If you set 10 s, you used $A = \dfrac{N}{T_{1/2}}$ ⁠ without $\ln 2$ ⁠: there the activity is only about 13.9 Bq.

Lesson 3 of 7 · About 30 minutes

Photons

The photoelectric effect: a threshold frequency that brightness cannot get past, the energy of a photon, the work function, the stopping voltage, and the speed of the fastest electron.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
A puzzle
Practice · Predict, then look
Question
The red light, at 460 THz, frees no electrons from the sodium. What will happen when Light power is turned up to 100 µW?

Correct answer
- Correct Still no electron will leave
Setup, choices and feedback
The Scene at this step
Light frequency
460 THz
Light power
50 µW
Metal plate
Sodiumfixed
Show the photons
Light beam
Choices
- Correct Still no electron will leave
- Wrong Electrons will start to leave, because there is more energy
  Misconception: Expects bright enough light of any frequency to free electrons from a metal.
  Feedback from the catalogue
  Each photon gives its energy to one electron, and an electron cannot build up energy from several photons. If $hf$ ⁠ is less than the work function, no photon is enough, and billions more photons will not help. Brighter light brings more photons, not stronger ones.
- Wrong Electrons will leave after a while, once enough energy has built up
  Misconception: Thinks that below the threshold, electrons will leave once they have soaked up enough light for long enough.
  Feedback from the catalogue
  That is what the wave picture predicted, and it does not happen. Below the threshold frequency no electron leaves, even after a long time. Each photon is absorbed on its own and warms the metal a little; the electron does not save up its energy until there is enough.
Once the prediction is locked in, the Scene changes
Light power
100 µW
The Scene then shows: 0 A · Photocurrent/nodes/cell/outputs/current
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The needle stays at zero. A photon of red light carries 1.90 eV, and freeing an electron from sodium takes 2.3 eV. More photons like that do not help.

Worked solution
1. The photon energy: $E = hf = 6.63 \times 10^{-34} \cdot 4.6 \times 10^{14} \approx 3.05 \times 10^{-19}\ \mathrm{J}$ ⁠, that is 1.90 eV.
2. Sodium’s work function is 2.3 eV, more than the photon’s energy.
3. Each photon gives its energy to one electron, so no photon is enough, and neither are twice as many photons.
If you expected brighter light to free electrons, you thought of light as a wave whose energy builds up. That is exactly the puzzle photons solved.
Step 2 of 7
Open this step
Light arrives as photons
No practice question in this step.
The Scene at this step
Light frequency
650 THz
Light power
50 µW
Metal plate
Sodium
Show the photons
0.3 (Light beam → Photons and energy)
Step 3 of 7
Open this step
The threshold frequency
Practice · Calculate
Question
Calcium’s work function is 2.9 eV. What is its threshold frequency, in terahertz?

Correct answer
701.2 THz ± 1%
Setup, choices and feedback
The Scene at this step
Light frequency
650 THz
Light power
50 µW
Metal plate
Calciumfixed
Show the photons
0.62 (Light beam → Photons and energy)
Accepted within
± 1%
Computed from the Scene at this step’s setting
W / h/nodes/cell/outputs/threshold
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4.38 × 10²¹ THz ± 5%
  Misconception: Divides a work function in electronvolts by Planck's constant in joule-seconds.
  Feedback from the catalogue
  In $f_0 = \dfrac{W}{h}$ ⁠ both quantities must be in the same system of units. $h \approx 6.63 \times 10^{-34}\ \mathrm{J \cdot s}$ ⁠ is in joules, so convert $W$ ⁠ to joules first: 1 eV = 1.6 × 10⁻¹⁹ J. Visible light has a frequency of a few hundred terahertz; a result of $10^{33}$ ⁠ hertz is a sign of mixed units.
Feedback for a right answer
About 701 THz, violet light. Blue light at 650 THz frees no electrons from calcium, though it frees them from sodium.

Hint
Convert $W$ ⁠ to joules: 1 eV = 1.6 × 10⁻¹⁹ J.

Worked solution
1. $W = 2.9 \cdot 1.6 \times 10^{-19} \approx 4.64 \times 10^{-19}\ \mathrm{J}$ ⁠.
2. $f_0 = \dfrac{W}{h} = \dfrac{4.64 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 7.0 \times 10^{14}\ \mathrm{Hz}$ ⁠.
3. $7.0 \times 10^{14}$ ⁠ hertz is 700 THz, since 1 THz = 10¹² Hz.
If you got $4.4 \times 10^{21}$ ⁠, you divided electronvolts by joule-seconds.
Step 4 of 7
Open this step
The stopping voltage
Practice · Predict, then look
Question
The plate is sodium, W = 2.3 eV. Predict the stopping voltage when the light is switched to Ultraviolet, at 1200 THz.

Correct answer
2.663 V ± 0.05 V
A number prediction counts as right within ± 0.05 V
Setup, choices and feedback
The Scene at this step
Light frequency
650 THz
Light power
50 µW
Metal plate
Sodiumfixed
Show the photons
0.62 (Light beam → Photons and energy)
Once the prediction is locked in, the Scene changes
Light frequency
1200 THz
The Scene then shows: 2.663 V · Reverse voltage that stops the fastest electron/nodes/cell/outputs/stoppingVoltage
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$hf \approx 4.96\ \mathrm{eV}$ ⁠, so $K_{\max} = 4.96 - 2.3 = 2.66\ \mathrm{eV}$ ⁠, and the stopping voltage is 2.66 V.

Worked solution
1. $E = hf = 6.63 \times 10^{-34} \cdot 1.2 \times 10^{15} \approx 7.96 \times 10^{-19}\ \mathrm{J}$ ⁠, that is 4.96 eV.
2. $K_{\max} = hf - W = 4.96 - 2.3 = 2.66\ \mathrm{eV}$ ⁠.
3. $eV_s = K_{\max}$ ⁠, so $V_s = 2.66\ \mathrm{V}$ ⁠.
If you predicted 4.96 V, you forgot the work function: part of the energy goes on getting out of the metal.
Step 5 of 7
Open this step
Brightness
Practice · Predict, then look
Question
Blue light on sodium. What will happen to the energy of the fastest electron when Light power is doubled?

Correct answer
- Correct It will stay at 0.39 eV
Setup, choices and feedback
The Scene at this step
Light frequency
650 THzfixed
Light power
50 µW
Metal plate
Sodiumfixed
Show the photons
0.3 (Light beam → Photons and energy)
Choices
- Correct It will stay at 0.39 eV
- Wrong It will double, to 0.78 eV
  Misconception: Thinks brighter light makes the photoelectrons faster.
  Feedback from the catalogue
  The speed of the fastest electron is set by $K_{\max} = hf - W$ ⁠, and the power is not in the formula. Brighter light at the same frequency brings more photons each second, each the same as before, so more electrons leave and the current grows, but none of them is faster.
- Wrong It will fall, because the electrons will share the energy
  Feedback
  Each photon gives its energy to one electron. The electrons do not share out the energy of the whole beam.
Once the prediction is locked in, the Scene changes
Light power
100 µW
The Scene then shows: 0.3882 eV · Fastest electron, h·f − W, or 0/nodes/cell/outputs/maxEnergy
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The greatest energy stayed the same, and the stream of electrons thickened: twice as many photons each second, each with the same energy $hf$ ⁠.

Worked solution
1. $K_{\max} = hf - W = 2.69 - 2.3 = 0.39\ \mathrm{eV}$ ⁠, and the power is not in the formula.
2. Doubling the power doubles the number of photons each second, $\dfrac{P}{hf}$ ⁠, not the energy of each one.
3. So twice as many electrons leave and the current doubles, but the fastest of them still has 0.39 eV.
If you chose that the energy would double, you gave each photon the brightness of the whole beam.
Step 6 of 7
Open this step
The electron's speed
Practice · Calculate
Question
Ultraviolet light at 1000 THz on sodium. The energy of the fastest electron is in the readings beside the Scene. What is its speed?

Correct answer
804000 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Light frequency
1000 THzfixed
Light power
50 µW
Metal plate
Sodiumfixed
Show the photons
0.62 (Light beam → Photons and energy)
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1210000 m/s ± 2%
  Misconception: Gives the photoelectron all of the photon's energy, forgetting the work function.
  Feedback from the catalogue
  Part of the photon’s energy goes on getting out of the metal: the work function $W$ ⁠. The fastest electron is left with only $K_{\max} = hf - W$ ⁠, and that is what you use to find the speed.
- 568000 m/s ± 2%
  Misconception: Uses K = mv² instead of K = ½mv² when finding a speed from a kinetic energy.
  Feedback from the catalogue
  Kinetic energy is $K = \tfrac12 m v^2$ ⁠, so $v = \sqrt{\dfrac{2K}{m}}$ ⁠. Without the $2$ ⁠ the speed comes out $\sqrt{2}$ ⁠ times too small, about $0.71$ ⁠ of the right one.
Feedback for a right answer
About 8.0 × 10⁵ m/s, eight hundred kilometres a second: very fast, and still less than one per cent of the speed of light.

Hint
Convert $K_{\max}$ ⁠ to joules, then $v = \sqrt{\dfrac{2K}{m}}$ ⁠.

Worked solution
1. $hf = 4.14\ \mathrm{eV}$ ⁠ and $W = 2.3\ \mathrm{eV}$ ⁠, so $K_{\max} = 1.84\ \mathrm{eV}$ ⁠.
2. In joules: $K = 1.84 \cdot 1.6 \times 10^{-19} \approx 2.94 \times 10^{-19}\ \mathrm{J}$ ⁠.
3. $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \cdot 2.94 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 8.0 \times 10^{5}\ \mathrm{m/s}$ ⁠.
If you got $1.2 \times 10^6\ \mathrm{m/s}$ ⁠, you gave the electron all the photon’s energy and forgot the work function.
Step 7 of 7
Open this step
Measuring Planck's constant
No practice question in this step.
The Scene at this step
Light frequency
1000 THz
Light power
50 µW
Metal plate
Sodium
Show the photons
Photons and energy

Lesson 4 of 7 · About 25 minutes

Matter waves

Single electrons that build an interference pattern: the de Broglie wavelength, how the gun voltage sets it, what happens when you check which slit each electron went through, and what this wave really is.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
One electron at a time
Practice · Predict, then look
Question
The electrons are sent one at a time. What will the screen show after six thousand electrons?

Correct answer
- Correct Bright and dark stripes, side by side
Setup, choices and feedback
The Scene at this step
Electrons sent
10
Gun voltage
250 V
Watch the slits
offfixed
Build up the pattern
Dots
Choices
- Wrong Two bright bands, one opposite each slit
  Misconception: Expects electrons, as particles, to make two bright bands behind the two slits and no stripes.
  Feedback from the catalogue
  That is how tiny balls would behave. But electrons avoid some stripes of the screen and crowd into others: an interference pattern, which only waves make. Each electron arrives as one dot, and yet the dots together build the stripes, even when the electrons are sent one at a time.
- Correct Bright and dark stripes, side by side
- Wrong An even patch over the whole screen
  Feedback
  Electrons do not land everywhere with the same chance: there are places they hardly ever reach.
Once the prediction is locked in, the Scene changes
Electrons sent
6000
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
An interference pattern, dot by dot. Each electron arrives as a particle, and yet something about it went through both slits and interfered.

Worked solution
1. Each electron lands as one dot, at a place no one can predict.
2. But the places are not wholly random: on some stripes of the screen the chance of landing is high, and on others it is almost zero.
3. After thousands of electrons the dots build bright and dark stripes, like the interference of waves through two slits.
If you expected two bands, you thought of the electron as a ball that goes through one slit. That is indeed what happens when you check which slit it went through, as we will see later.
Step 2 of 5
Open this step
The electron's speed
Practice · Calculate
Question
Gun voltage is 100 V. How fast do the electrons leave the gun?

Correct answer
5930000 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Electrons sent
1500
Gun voltage
100 Vfixed
Watch the slits
off
Build up the pattern
Dots
Accepted within
± 2%
Computed from the Scene at this step’s setting
Speed of the electrons after the gun/nodes/bench/outputs/speed
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4190000 m/s ± 2%
  Misconception: Uses K = mv² instead of K = ½mv² when finding a speed from a kinetic energy.
  Feedback from the catalogue
  Kinetic energy is $K = \tfrac12 m v^2$ ⁠, so $v = \sqrt{\dfrac{2K}{m}}$ ⁠. Without the $2$ ⁠ the speed comes out $\sqrt{2}$ ⁠ times too small, about $0.71$ ⁠ of the right one.
Feedback for a right answer
About 5.9 × 10⁶ m/s, some six thousand kilometres a second: about $2\%$ ⁠ of the speed of light.

Hint
$eV = \tfrac12 m v^2$ ⁠.

Worked solution
1. Given: $V = 100\ \mathrm{V}$ ⁠, $e = 1.6 \times 10^{-19}\ \mathrm{C}$ ⁠, $m = 9.11 \times 10^{-31}\ \mathrm{kg}$ ⁠.
2. $eV = \tfrac12 m v^2$ ⁠, so $v = \sqrt{\dfrac{2eV}{m}}$ ⁠.
3. $v = \sqrt{\dfrac{2 \cdot 1.6 \times 10^{-19} \cdot 100}{9.11 \times 10^{-31}}} \approx 5.9 \times 10^{6}\ \mathrm{m/s}$ ⁠.
If you got $4.2 \times 10^6\ \mathrm{m/s}$ ⁠, you used $K = mv^2$ ⁠ without the half.
Step 3 of 5
Open this step
The de Broglie wavelength
Practice · Calculate
Question
How many times shorter is the electrons’ wavelength at 400 V than at 100 V? Answer, then check on the label of the gun’s power supply.

Correct answer
2 ± 0.05
Setup, choices and feedback
The Scene at this step
Electrons sent
1500
Gun voltage
100 V
Watch the slits
off
Build up the pattern
Dots
Accepted within
± 0.05
A fixed value in the question
Wrong answers it recognises
- 4 ± 0.05
  Misconception: Takes the de Broglie wavelength to be inversely proportional to the voltage, not to its square root.
  Feedback from the catalogue
  The wavelength is inverse to the momentum, $\lambda = \dfrac{h}{p}$ ⁠, but the momentum is not proportional to the voltage. From $eV = \tfrac12 m v^2$ ⁠ comes $p = \sqrt{2meV}$ ⁠, so $\lambda$ ⁠ is inverse to $\sqrt{V}$ ⁠. Four times the voltage is twice the momentum, and so half the wavelength.
Feedback for a right answer
Twice: from 0.123 nm to 0.061 nm. A shorter wave, and so closer stripes.

Hint
$\lambda = \dfrac{h}{\sqrt{2meV}}$ ⁠: how does $\lambda$ ⁠ depend on $V$ ⁠?

Worked solution
1. $\lambda = \dfrac{h}{p}$ ⁠ and $p = \sqrt{2meV}$ ⁠, so $\lambda$ ⁠ is proportional to $\dfrac{1}{\sqrt{V}}$ ⁠.
2. The ratio: $\dfrac{\lambda_{100}}{\lambda_{400}} = \sqrt{\dfrac{400}{100}} = \sqrt{4} = 2$ ⁠.
3. The wavelength at 400 V is $2$ ⁠ times shorter.
If you answered $4$ ⁠, you assumed the wavelength is inverse to the voltage itself. But the voltage sets the energy, and the momentum is its square root.
Step 4 of 5
Open this step
Watching the slits
Practice · Predict, then look
Question
We will switch on Watch the slits: the lamp will show which slit each electron went through. What will happen to the stripes?

Correct answer
- Correct The stripes will vanish, leaving one smooth band
Setup, choices and feedback
The Scene at this step
Electrons sent
1500
Gun voltage
250 V
Watch the slits
off
Build up the pattern
Dots
Choices
- Correct The stripes will vanish, leaving one smooth band
- Wrong The stripes will stay as they are; the lamp only looks
  Misconception: Thinks finding out which slit each electron goes through leaves the interference pattern as it was.
  Feedback from the catalogue
  The measurement changes the result. When the lamp finds out which slit each electron went through, there are no longer two parts of a wave to interfere, and the stripes vanish: one smooth band is left, the sum of two single-slit patterns. You can know which slit the electron went through, or get stripes, but not both.
- Wrong The stripes will be sharper
  Feedback
  The lamp does not improve the pattern. It finds out which slit each electron went through, and that does away with the interference.
Once the prediction is locked in, the Scene changes
Watch the slits
on
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Once it is known which slit each electron went through, the stripes vanish: there are no longer two parts of a wave to interfere.

Worked solution
1. Without the lamp, each electron’s wave goes through both slits, and the two parts interfere on the screen.
2. The lamp catches each electron in one slit. From then on the electron came from one particular slit, and there is no second part to interfere with.
3. What is left is the sum of two single-slit patterns: one smooth band, with no dark stripes.
If you chose that the stripes would stay, you thought a measurement only reveals what was already there. In the quantum world the measurement changes the result.
Step 5 of 5
Open this step
A wave of chance
Practice · Choose
Question
What does the wave in the Scene describe?

Correct answer
- Correct The chance of finding the electron at each place
Setup, choices and feedback
The Scene at this step
Electrons sent
1500
Gun voltage
250 V
Watch the slits
off
Build up the pattern
Wave
Choices
Shown in a random order.
- Correct The chance of finding the electron at each place
- Wrong The electron itself, smeared across the whole screen
  Misconception: Thinks the electron's wave means the electron itself is spread out across the screen.
  Feedback from the catalogue
  The electron never arrives smeared out: it always lands at one point, with all its charge and mass. The wave does not describe smeared-out matter but the chance of finding the electron in each place.
- Wrong The winding path along which the electron moves
  Misconception: Pictures the electron's wave as the electron moving along a wavy path.
  Feedback from the catalogue
  The electron’s wave is not a winding path. Nobody sees the electron move in a wavy line; what interferes is the wave of chance, which goes through both slits together and sets where the electron is likely to land.
- Wrong Vibrations of the foil
  Feedback
  The foil does not vibrate. The wave belongs to the electrons, and its wavelength is set by their momentum.
Feedback for a right answer
The wave sets where the electron is likely to land: many dots where the wave is strong, and almost none where the two parts cancel.

Worked solution
1. Each electron lands at one point, with all its charge and mass, so it is not smeared out.
2. Still, the dots spread out by the wave’s pattern: many where the blue curve is high, and none where it is zero.
3. So the wave describes the chance of finding the electron, not its matter and not its path.
If you chose that the electron is smeared, look at the screen again: each electron is a dot.

Lesson 5 of 7 · About 25 minutes

Quantum tunnelling

An electron can be found past a wall higher than its energy: the wave fading inside the wall, the depth at which the chance halves, a slightly thicker or higher wall letting far less through, and energy kept on the far side.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A wall too high
Practice · Predict, then look
Question
The electron has an energy of 1 eV. Right now the wall is lower than that, 0.5 eV. We will raise it to 2 eV, twice the electron’s energy, 0.3 nm thick. What will happen to the wave packet?

Correct answer
- Correct Most of it will bounce back, but some will get through to the other side
Setup, choices and feedback
The Scene at this step
Wall height
0.5 eV
Wall thickness
0.3 nmfixed
Look into the wall
Wave packet
Choices
- Correct Most of it will bounce back, but some will get through to the other side
- Wrong All of it will bounce back: the electron does not have enough energy to get over
  Misconception: Thinks an electron can never be found past a wall higher than its energy, as a ball could not roll over a hill too high for it.
  Feedback from the catalogue
  That is true of a ball, but not of an electron. The electron is described by a wave, and inside a wall higher than its energy the wave is not cut off but fades. If the wall is thin, some of it is left on the far side, and there is a chance of finding the electron there: quantum tunnelling.
- Wrong All of it will get through, only more slowly
  Feedback
  An electron cannot climb a wall higher than its energy. Most of the wave bounces back, and only a small part ends up on the far side.
Once the prediction is locked in, the Scene changes
Wall height
2 eV
The Scene then shows: 0.1718 · Share of the packet that ends up past the wall, 0..1/nodes/wall/outputs/transmitted
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About $17\%$ ⁠ of the packet gets through: about one time in six, the electron will be found past the wall. That is quantum tunnelling. A ball rolling at a hill too high for it would never reach the other side.

Worked solution
1. In classical physics, a body with energy $E$ ⁠ cannot pass a wall of height $V > E$ ⁠: halfway across its kinetic energy would be negative.
2. But the electron is described by a wavefunction. Inside the wall the wave is not cut off: it fades.
3. If the wall is thin, the wave has not faded to zero by the far side, and carries on from there. Here about $17\%$ ⁠ gets through.
If you chose that all of it bounces back, that is the classical answer. It is nearly right for a thick wall, but not for a wall a few atoms thick.
Step 2 of 5
Open this step
Inside the wall
Practice · Calculate
Question
Inside the wall the chance of finding the electron fades as $e^{-2\kappa x}$ ⁠, where $\kappa = \dfrac{\sqrt{2m(V-E)}}{\hbar}$ ⁠. Here V − E = 1 eV, and for the electron $\kappa \approx 5.12$ ⁠ per nanometre. At what depth into the wall has the chance halved?

Correct answer
0.06765 nm ± 0.003 nm
Setup, choices and feedback
The Scene at this step
Wall height
2 eVfixed
Wall thickness
0.3 nmfixed
Look into the wall
Wave packetfixed
Accepted within
± 0.003 nm
Computed from the Scene at this step’s setting
Depth in the wall over which the chance of finding the electron halves (0 over the top)/nodes/wall/outputs/halving
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.1353 nm ± 0.005 nm
  Misconception: Uses the fading of the wave's amplitude, e^(−κx), for the chance of finding the electron, instead of its square, e^(−2κx).
  Feedback from the catalogue
  The chance of finding the electron is the square of the wave’s height. The wave fades as $e^{-\kappa x}$ ⁠, so the chance fades as $e^{-2\kappa x}$ ⁠: twice as fast.
Feedback for a right answer
$x = \dfrac{\ln 2}{2\kappa} \approx 0.068$ ⁠ nanometres, less than the width of an atom. A 0.3 nm wall is about $4.4$ ⁠ such depths, and $2^{-4.4} \approx 5\%$ ⁠. What actually gets through, about $17\%$ ⁠, is more than that, because what happens to the wave at the wall’s two faces adds a factor, here about $4$ ⁠.

Hint
Solve $e^{-2\kappa x} = \dfrac{1}{2}$ ⁠.

Worked solution
1. $e^{-2\kappa x} = \dfrac{1}{2}$ ⁠, so $2\kappa x = \ln 2$ ⁠.
2. $x = \dfrac{\ln 2}{2\kappa} = \dfrac{0.693}{2 \times 5.12} \approx 0.068$ ⁠ nanometres.
3. Every further $0.068$ ⁠ nanometres into the wall halves the chance.
If you got about 0.135 nm, you used the fading of the wave itself, $e^{-\kappa x}$ ⁠. The chance is the square of the wave, so it fades twice as fast.
Step 3 of 5
Open this step
Thicker walls
Practice · Reach the target
Question
Wall height is fixed at 2 eV, and at 0.3 nm thick about $17\%$ ⁠ gets through. Change Wall thickness until only about one in forty, about $2.5\%$ ⁠, gets through, and press Check.

Goal
- 0.025 ± 0.004
  At the start of the step: 0.1718
Setup, choices and feedback
The Scene at this step
Wall height
2 eVfixed
Wall thickness
0.3 nm
Look into the wall
Wave packet
Read from the Scene at
Share of the packet that ends up past the wall, 0..1/nodes/wall/outputs/transmitted
One setting that reaches it
Wall thickness
0.5 nm
Feedback for a right answer
At 0.5 nm about $2.5\%$ ⁠ gets through. Two more tenths of a nanometre, less than an atom’s width, cut what gets through sevenfold: each 0.1 nm divides it by about $2.8$ ⁠.

Feedback for a wrong answer
Not yet. A thicker wall lets less through, and the change is fast: a little thickness changes a lot.

Hint
In a thick wall, each extra 0.1 nm cuts what gets through almost threefold.

Worked solution
1. In a thick wall, each 0.1 nm divides what gets through by a fixed factor: $e^{2\kappa \times 0.1} = e^{1.02} \approx 2.8$ ⁠.
2. From $17\%$ ⁠ it must fall about $7$ ⁠ times, and $2.8^2 \approx 7.8$ ⁠: two more tenths of a nanometre.
3. Set Wall thickness to 0.5 nm.
Step 4 of 5
Open this step
Higher walls
Practice · Predict, then look
Question
Through a 2 eV, 0.3 nm wall about $17\%$ ⁠ gets through. The wall is 1 eV above the electron. We will raise it to 4 eV, so 3 eV above the electron. What will happen to the share that gets through?

Correct answer
- Correct It will fall about ten times
Setup, choices and feedback
The Scene at this step
Wall height
2 eV
Wall thickness
0.3 nmfixed
Look into the wall
Wave packet
Choices
- Correct It will fall about ten times
- Wrong It will fall three times, like the rise in the gap
  Misconception: Thinks the share of electrons that tunnel falls in proportion to how far the wall rises above their energy, or to its thickness.
  Feedback from the catalogue
  What gets through a thick wall goes as $e^{-2\kappa d}$ ⁠, where $\kappa$ ⁠ grows with $\sqrt{V-E}$ ⁠. Both sit in an exponent, so a wall a little higher or a little thicker cuts what gets through many times over, far more than in proportion.
- Wrong Nothing will get through: the wall is too high
  Misconception: Thinks an electron can never be found past a wall higher than its energy, as a ball could not roll over a hill too high for it.
  Feedback from the catalogue
  That is true of a ball, but not of an electron. The electron is described by a wave, and inside a wall higher than its energy the wave is not cut off but fades. If the wall is thin, some of it is left on the far side, and there is a chance of finding the electron there: quantum tunnelling.
Once the prediction is locked in, the Scene changes
Wall height
4 eV
The Scene then shows: 0.01491 · Share of the packet that ends up past the wall, 0..1/nodes/wall/outputs/transmitted
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About $1.5\%$ ⁠ gets through, $11$ ⁠ times less. $\kappa$ ⁠ grows with $\sqrt{V - E}$ ⁠, by $\sqrt{3} \approx 1.7$ ⁠, and it sits in an exponent: so what gets through plunges far more than threefold.

Worked solution
1. $\kappa = \dfrac{\sqrt{2m(V-E)}}{\hbar}$ ⁠: up by $\sqrt{3}$ ⁠ when the gap grows threefold, from $5.1$ ⁠ to about $8.9$ ⁠ per nanometre.
2. Through a $0.3$ ⁠ nanometre wall, $e^{-2\kappa d}$ ⁠ falls from $e^{-3.1}$ ⁠ to $e^{-5.3}$ ⁠: by $e^{2.2} \approx 9$ ⁠.
3. With the factor the wall’s two faces add, what gets through falls from $17\%$ ⁠ to about $1.5\%$ ⁠.
Tunnelling depends on the height and thickness through an exponential, so small changes in them change it a lot.
Step 5 of 5
Open this step
The far side
Practice · Choose
Question
The part of the packet that got past the wall travels on. What is the electron’s energy on the far side?

Correct answer
- Correct The same energy, 1 eV, as before the wall
Setup, choices and feedback
The Scene at this step
Wall height
2 eV
Wall thickness
0.3 nm
Look into the wall
Inside the wall
Choices
Shown in a random order.
- Correct The same energy, 1 eV, as before the wall
- Wrong Less: it spent energy getting through the wall
  Misconception: Thinks an electron loses energy tunnelling through a wall, so it comes out the far side slower.
  Feedback from the catalogue
  The wall does not change in time, so the electron’s energy is conserved. Past the wall the wave carries on with the same wavelength, and the electron with the same energy. What is smaller is the chance of finding it there, not its energy.
- Wrong More: the wall pushed it on
  Feedback
  A wall that does not change in time neither gives energy nor takes it. The electron comes out the far side with the energy it had.
Feedback for a right answer
Energy is conserved: past the wall the wave carries on with the same wavelength, and the electron has the same energy, only with a smaller chance. What tunnelling changes is where it is found, not how much energy it has.

Worked solution
1. The wall does not change in time, so the electron’s total energy is conserved.
2. Outside the wall all the energy is kinetic, so the de Broglie wavelength, which depends on it, is the same on both sides.
3. What is smaller on the far side is the height of the wave, that is, the chance of finding the electron there.
This is how an alpha particle leaves a nucleus in alpha decay: it tunnels through the barrier and comes out with all the energy it had inside.

Lesson 6 of 7 · About 25 minutes

Special relativity

A light clock on a fast train: the speed of light is the same for every observer, the Lorentz factor from Pythagoras, time dilation, length contraction, and muons that reach the ground.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Two light clocks
No practice question in this step.
The Scene at this step
Train speed
0 c
Show the Lorentz factor
Train
Step 2 of 6
Open this step
The slanted path
Practice · Calculate
Question
The train runs at 0.6 c. What is the Lorentz factor $\gamma$ ⁠?

Correct answer
1.25 ± 0.005
Setup, choices and feedback
The Scene at this step
Train speed
0.6 cfixed
Show the Lorentz factor
Light paths
Accepted within
± 0.005
Computed from the Scene at this step’s setting
Lorentz factor γ = 1/√(1 − v²/c²)/nodes/station/outputs/gamma
Wrong answers it recognises
- 1.563 ± 0.005
  Misconception: Leaves out the square root in the Lorentz factor: γ = 1/(1 − v²/c²).
  Feedback from the catalogue
  The Lorentz factor has a square root: $\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$ ⁠. The root comes from Pythagoras, when you solve $(ct)^2 = L^2 + (vt)^2$ ⁠ for $t$ ⁠. At 0.6 c: $\sqrt{1 - 0.36} = 0.8$ ⁠, and $\gamma = 1.25$ ⁠.
- 0.8 ± 0.005
  Misconception: Gives √(1 − v²/c²) as γ, the reciprocal of the Lorentz factor.
  Feedback from the catalogue
  $\sqrt{1 - v^2/c^2}$ ⁠ is the denominator, not the factor. $\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$ ⁠ is always $1$ ⁠ or more: a tick of a moving clock takes more time, not less.
Feedback for a right answer
$\gamma = 1.25$ ⁠: each tick of the train clock lasts a quarter longer than a tick of the platform clock.

Hint
$\dfrac{v}{c} = 0.6$ ⁠, so $\dfrac{v^2}{c^2} = 0.36$ ⁠.

Worked solution
1. $\dfrac{v^2}{c^2} = 0.6^2 = 0.36$ ⁠.
2. $\sqrt{1 - 0.36} = \sqrt{0.64} = 0.8$ ⁠.
3. $\gamma = \dfrac{1}{0.8} = 1.25$ ⁠.
If you answered $1.5625$ ⁠, you left out the square root: $\dfrac{1}{0.64} = 1.5625$ ⁠.
Step 3 of 6
Open this step
Time dilation
Practice · Calculate
Question
The train runs at 0.8 c. While the platform clock counts $30$ ⁠ ticks, how many ticks does the train clock count?

Correct answer
18 ± 0.1
Setup, choices and feedback
The Scene at this step
Train speed
0.8 cfixed
Show the Lorentz factor
Train
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Ticks of the train's clock per tick of the platform's, 1/γ/nodes/station/outputs/clockRate
Wrong answers it recognises
- 50 ± 0.1
  Misconception: Multiplies by γ instead of dividing: thinks the moving clock counts more ticks.
  Feedback from the catalogue
  The moving clock is the slow one. Each of its ticks lasts $\gamma$ ⁠ ticks of the platform clock, so while the platform clock counts $n$ ⁠ ticks, the train clock counts $\dfrac{n}{\gamma}$ ⁠: fewer, not more.
Feedback for a right answer
$\gamma = \dfrac{5}{3}$ ⁠, so the train clock counts $\dfrac{30}{5/3} = 18$ ⁠ ticks.

Hint
First work out $\gamma$ ⁠ at 0.8 c.

Worked solution
1. $\sqrt{1 - 0.8^2} = \sqrt{0.36} = 0.6$ ⁠, so $\gamma = \dfrac{1}{0.6} \approx 1.67$ ⁠.
2. Each tick of the train clock lasts $\gamma$ ⁠ platform ticks, so the train clock counts $\dfrac{n}{\gamma}$ ⁠.
3. $\dfrac{30}{\gamma} = 30 \cdot 0.6 = 18$ ⁠ ticks.
If you answered $50$ ⁠, you multiplied by $\gamma$ ⁠ instead of dividing: the moving clock is the slow one, so it counts fewer.
Step 4 of 6
Open this step
Length contraction
Practice · Reach the target
Question
Change Train speed so that the carriage, 20 m long at rest, measures 16 m from the platform, and press Check.

Goal
- 16 m ± 0.05 m
  At the start of the step: 19.08 m
Setup, choices and feedback
The Scene at this step
Train speed
0.3 c
Show the Lorentz factor
Train
Read from the Scene at
Length of the 20 m carriage as measured from the platform/nodes/station/outputs/length
One setting that reaches it
Train speed
0.597 c
Feedback for a right answer
At 0.6 c, $\gamma = 1.25$ ⁠, and the carriage is $1.25$ ⁠ times shorter: 16 m.

Feedback for a wrong answer
Not yet. To shorten the carriage more, raise the speed.

Hint
You need $\gamma = \dfrac{20}{16} = 1.25$ ⁠. At what speed did that already happen in this Lesson?

Worked solution
1. $L = \dfrac{L_0}{\gamma}$ ⁠, so $\gamma = \dfrac{L_0}{L} = \dfrac{20}{16} = 1.25$ ⁠.
2. $\sqrt{1 - v^2/c^2} = \dfrac{1}{1.25} = 0.8$ ⁠, so $\dfrac{v^2}{c^2} = 1 - 0.64 = 0.36$ ⁠.
3. $\dfrac{v}{c} = 0.6$ ⁠: set Train speed to 0.6 c.
If you set 0.8 c, you thought the carriage shrinks by $20\%$ ⁠ when the speed is $80\%$ ⁠ of $c$ ⁠. But there it is already 12 m.
Step 5 of 6
Open this step
Whose clock is slow?
Practice · Choose
Question
A passenger on the train, at 0.8 c, compares her clock with the platform clock. What does she measure?

Correct answer
- Correct The platform clock ticks slower than hers, by the same $\gamma$ ⁠
Setup, choices and feedback
The Scene at this step
Train speed
0.8 c
Show the Lorentz factor
Train
Choices
Shown in a random order.
- Correct The platform clock ticks slower than hers, by the same $\gamma$ ⁠
- Wrong Her clock is slow, and the platform clock ticks faster
  Misconception: Thinks the passenger sees her own clock slow, or the platform clock fast.
  Feedback from the catalogue
  The passenger may say she is standing still. For her, her own clock ticks as usual and the platform is moving, so the platform clock is the slow one, by the same $\gamma$ ⁠. Every observer measures the clock moving relative to them to run slow. This is no contradiction: the two observers do not agree on which events happen at the same moment.
- Wrong Both clocks tick at the same rate; the slowing only looks that way from the platform
  Misconception: Thinks moving clocks only look slow, an illusion from the time light takes to reach the observer, and do not really run slow.
  Feedback from the catalogue
  Time dilation is not an illusion of light’s delay. It remains after the time the light took to arrive is worked out, and it leaves traces no one can argue with: muons reach the ground, and atomic clocks flown round the world came back out of step with clocks that stayed on the ground.
Feedback for a right answer
For her, she is standing still and the platform is moving, so the platform clock is the slow one. Every observer measures the clock moving relative to them to run slow.

Worked solution
1. By the first principle, no experiment shows who is standing still: the passenger may say she is at rest.
2. For her, her flash goes straight up and down, and it is the platform clock that moves at 0.8 c.
3. So she measures the platform clock to be slow, by $\gamma \approx 1.67$ ⁠, exactly as the people on the platform measure the train clock.
If you chose that the platform clock is faster, you assumed the platform is “really” standing still. In relativity there is no such thing.
Step 6 of 6
Open this step
Muons
Practice · Choose
Question
At rest a muon lives 2.2 µs on average, and in that time it covers only about 660 m. Why do many of the muons made 15 km up reach the ground?

Correct answer
- Correct At 0.995 c their clock runs about $10$ ⁠ times slow, as measured from the ground, and they live about 22 µs
Setup, choices and feedback
The Scene at this step
Train speed
0.995 c
Show the Lorentz factor
Lorentz factor
Choices
Shown in a random order.
- Correct At 0.995 c their clock runs about $10$ ⁠ times slow, as measured from the ground, and they live about 22 µs
- Wrong It is only an illusion of light’s delay; really they decay after 660 m
  Misconception: Thinks moving clocks only look slow, an illusion from the time light takes to reach the observer, and do not really run slow.
  Feedback from the catalogue
  Time dilation is not an illusion of light’s delay. It remains after the time the light took to arrive is worked out, and it leaves traces no one can argue with: muons reach the ground, and atomic clocks flown round the world came back out of step with clocks that stayed on the ground.
- Wrong They move faster than light
  Misconception: Explains the muons' journey by their moving faster than light.
  Feedback from the catalogue
  Nothing with mass reaches the speed of light: as the speed approaches $c$ ⁠, $\gamma$ ⁠ grows without limit. The muons move slower than that, at about 0.995 c. They get far because their clock runs slow, not because they are faster than light.
Feedback for a right answer
In 22 µs a muon covers about 6.6 km on average, and many go further. From the muon’s point of view, it is the atmosphere that is $10$ ⁠ times thinner.

Worked solution
1. At 0.995 c, $\gamma = \dfrac{1}{\sqrt{1 - 0.995^2}} \approx 10$ ⁠.
2. As measured from the ground, the muon lives $\gamma \cdot 2.2\ \mu\mathrm{s} \approx 22\ \mu\mathrm{s}$ ⁠.
3. In that time it covers $0.995 \cdot 3 \times 10^8 \cdot 22 \times 10^{-6} \approx 6.6\ \mathrm{km}$ ⁠ on average, $10$ ⁠ times 660 m, so far more muons reach the ground.
If you chose an illusion, remember that detectors on the ground count the muons that arrive. An illusion leaves no particles in a detector.

Lesson 7 of 7 · About 20 minutes

Superconductivity

Below its critical temperature a material's resistance vanishes completely, and it keeps magnetic fields out: the Meissner effect, a floating magnet, why cold copper does not do it, and why YBCO and liquid nitrogen changed the picture.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Colder, and then zero
Practice · Predict, then look
Question
The disc is at 100 K, and the meter shows a small resistance that falls in a straight line as the disc cools. What will happen to the resistance when we cool it to 77 K, the temperature of liquid nitrogen?

Correct answer
- Correct At 92 K it will drop at once to zero
Setup, choices and feedback
The Scene at this step
Temperature
100 K
Show the field lines
Off
Choices
- Correct At 92 K it will drop at once to zero
- Wrong It will keep falling in a straight line, very small but not zero
  Misconception: Thinks a superconductor is just a very good conductor, with a small resistance that keeps falling as it cools.
  Feedback from the catalogue
  In an ordinary conductor, like copper, the resistance falls as it cools but never reaches zero. In a superconductor it drops to exactly zero at the critical temperature. Only with no resistance does a current flow on forever, and only then can surface currents keep a magnetic field out.
- Wrong It will rise, because electrons find it harder to move in the cold
  Feedback
  In metals and in this ceramic, cooling calms the atoms’ shaking, so the resistance falls, not rises.
Once the prediction is locked in, the Scene changes
Temperature
77 K
The Scene then shows: 0 Ω · The disc’s resistance as the meter reads it: 0 below Tc, proportional to T above/nodes/disc/outputs/resistance
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At 92 K, YBCO’s critical temperature, the resistance vanishes. That is superconductivity: not a very small resistance but zero, and a current started round a superconducting loop flows on for years with no battery.

Worked solution
1. Above 92 K the disc is an ordinary, poor conductor: its resistance is proportional to the temperature.
2. At 92 K it passes into a new state, and below it the resistance is exactly zero.
3. At 77 K the meter reads zero.
If you expected the straight line all the way, that is what copper does: its resistance falls as it cools but never reaches zero, because its flaws remain.
Step 2 of 6
Open this step
The Meissner effect
Practice · Choose
Question
The lines in the cut show the magnet’s field kept out of the disc, with glowing currents running round it. If we swap it for a copper disc of the same size, cooled to the same temperature, will the magnet float?

Correct answer
- Correct No: in copper the currents die away, and the field passes through it
Setup, choices and feedback
The Scene at this step
Temperature
77 K
Show the field lines
On
Choices
Shown in a random order.
- Correct No: in copper the currents die away, and the field passes through it
- Wrong Yes: cold copper conducts very well
  Misconception: Thinks a superconductor is just a very good conductor, with a small resistance that keeps falling as it cools.
  Feedback from the catalogue
  In an ordinary conductor, like copper, the resistance falls as it cools but never reaches zero. In a superconductor it drops to exactly zero at the critical temperature. Only with no resistance does a current flow on forever, and only then can surface currents keep a magnetic field out.
- Wrong No, but the magnet will stick to it
  Feedback
  Copper is not attracted to a magnet. The field simply passes through it as if it were not there, and the magnet rests on it.
Feedback for a right answer
The magnet floats because currents on the disc’s surface make a field that cancels its own inside the disc and pushes it up. Only without resistance do those currents not die away. In copper, even cold, resistance drains them within a moment. That is the Meissner effect: a superconductor keeps magnetic fields out.

Worked solution
1. When a magnet approaches a conductor, currents arise in it that oppose the change in the field.
2. In a superconductor there is no resistance, so the currents carry on, and their field repels the magnet for as long as it is there.
3. Copper has resistance, however small, and the currents die away fast. The field passes through it, and nothing pushes the magnet.
The difference between “very small resistance” and “zero” is the whole difference: only zero keeps the currents going.
Step 3 of 6
Open this step
How high it floats
Practice · Reach the target
Question
At 77 K the magnet floats about 20 mm above the disc, by the gauge behind it. Change the Temperature until the gap under the magnet is 15 mm, and press Check.

Goal
- 15 mm ± 0.5 mm
  At the start of the step: 19.93 mm
Setup, choices and feedback
The Scene at this step
Temperature
77 K
Show the field lines
Off
Read from the Scene at
The gap between the magnet’s lower face and the disc, as the gauge reads it; 0 when it rests on the disc/nodes/disc/outputs/float
One setting that reaches it
Temperature
88 K
Feedback for a right answer
At 88 K. The closer to 92 K, the smaller the current the disc can carry, the less of the field it keeps out, and the lower the magnet floats.

Feedback for a wrong answer
Not yet. Colder, the magnet rises. Warmer, but still below 92 K, it sinks.

Hint
Close to the critical temperature the disc keeps out less of the field, and the magnet sinks.

Worked solution
1. The magnet floats at the height where the currents’ push equals its weight.
2. The largest current the disc can carry shrinks near the critical temperature, so the push is weaker and the magnet sinks.
3. On the gauge, the gap is 15 mm at about 88 K.
Step 4 of 6
Open this step
Near the critical temperature
Practice · Predict, then look
Question
We warm the disc to 91 K, one degree below the critical temperature. What will the meter show, and what will the magnet do?

Correct answer
- Correct Zero resistance, but the magnet will float low: most of the field gets into the disc
Setup, choices and feedback
The Scene at this step
Temperature
77 K
Show the field lines
On
Choices
- Correct Zero resistance, but the magnet will float low: most of the field gets into the disc
- Wrong Zero resistance, and the magnet will float just as at 77 K
  Misconception: Thinks a superconductor below its critical temperature keeps out all of a magnetic field, the same at any temperature.
  Feedback from the catalogue
  The resistance is zero at any temperature below the critical one, but how much of the field stays out depends on the largest current the material can carry. Near the critical temperature that current is small, and most of the field gets in. The colder it is, the more of the field it keeps out.
- Wrong The resistance will come back, and the magnet will rest on the disc
  Feedback
  At 91 K the disc is still below the critical temperature, so it is still superconducting: the resistance is zero.
Once the prediction is locked in, the Scene changes
Temperature
91 K
The Scene then shows: 7.849 mm · The gap between the magnet’s lower face and the disc, as the gauge reads it; 0 when it rests on the disc/nodes/disc/outputs/float
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The resistance is still zero, but the disc keeps out only about $5\%$ ⁠ of the field, and the magnet floats about 8 mm above it, instead of 20 mm. The field lines show how many of them get into the disc.

Worked solution
1. Below 92 K the resistance is zero, at any temperature.
2. But how much of the field the disc keeps out depends on the largest current it can carry, and that is very small near the critical temperature.
3. At 91 K only a small part of the field is pushed out, the push is weak, and the magnet floats low.
Superconductivity is not an all-or-nothing switch: the colder the disc, the stronger it is.
Step 5 of 6
Open this step
Liquid nitrogen
Practice · Choose
Question
Kamerlingh Onnes discovered superconductivity in mercury in $1911$ ⁠, at 4.2 K, cooled in liquid helium. Why was the discovery of YBCO in $1987$ ⁠ such a big step?

Correct answer
- Correct Its critical temperature, 92 K, is above the boiling point of liquid nitrogen, 77 K, which is cheap and easy to handle
Setup, choices and feedback
The Scene at this step
Temperature
77 K
Show the field lines
Off
Choices
Shown in a random order.
- Correct Its critical temperature, 92 K, is above the boiling point of liquid nitrogen, 77 K, which is cheap and easy to handle
- Wrong It superconducts at room temperature
  Feedback
  92 K is about $-181$ ⁠ degrees Celsius, far from room temperature. A superconductor at room temperature and ordinary pressure has not yet been found.
- Wrong It was the first superconductor discovered
  Feedback
  The first was mercury, in $1911$ ⁠. What was new about YBCO was the temperature.
Feedback for a right answer
Liquid nitrogen boils at 77 K and cools YBCO below its critical temperature easily. Before, superconductivity needed liquid helium, costly and hard to handle. So it became an experiment a school lab could do.

Worked solution
1. To superconduct, a material must be cooled below its critical temperature.
2. For mercury: 4.2 K, which only liquid helium reaches.
3. For YBCO: 92 K, above the 77 K of liquid nitrogen, which is made from air and costs little.
Hospital MRI magnets still use niobium–titanium coils and liquid helium, but cables and magnets of ceramics like YBCO are coming into use.
Step 6 of 6
Open this step
Colder still
Practice · Predict, then look
Question
At 77 K the disc is already superconducting, and it keeps out about 75% of the magnet’s field. Predict: what will happen to that share when we cool it to 50 K?

Correct answer
- Correct It will grow
Setup, choices and feedback
The Scene at this step
Temperature
77 K
Show the field lines
On
Choices
- Wrong It will not change: a superconductor already keeps out all it can
  Misconception: Thinks a superconductor below its critical temperature keeps out all of a magnetic field, the same at any temperature.
  Feedback from the catalogue
  The resistance is zero at any temperature below the critical one, but how much of the field stays out depends on the largest current the material can carry. Near the critical temperature that current is small, and most of the field gets in. The colder it is, the more of the field it keeps out.
- Correct It will grow
- Wrong It will shrink, because in the cold the magnet is pulled in
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Temperature
50 K
The Scene then shows: 0.9336 · The share of the magnet’s field the disc keeps out, 0..1/nodes/disc/outputs/screening
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It grows to about 93%: the colder, the more electrons in pairs, and the higher the magnet floats.

Hint
Below the critical temperature, how many of the electrons join up into Cooper pairs? All at once, or more and more as it cools?

Worked solution
1. Just below 92 K only a small share of the electrons join into Cooper pairs, and the currents they make are weak.
2. As it cools, more electrons pair up, and the currents round the disc grow stronger.
3. At 50 K the disc keeps out about 93% of the field, and the magnet floats higher.
Superconductivity switches on at the critical temperature, but grows stronger gradually below it.

Across the unit · 34 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

Atoms and nuclei

What is inside an atom

Counting neutrons

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Isotopes

Shells

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

A full outer shell

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Clouds, not orbits

Question

Correct answer

The Scene at this step

Choices

Worked solution

Radioactive decay

Unstable nuclei

Alpha decay

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Beta decay

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Gamma rays

No memory

Question

Correct answer

The Scene at this step

Choices

Worked solution

Activity

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The activity falls

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Choosing a half-life

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Photons

A puzzle

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes