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Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles
40 free, high-quality past-paper-style questions on Atomic structure and particle accelerators for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
A nucleus of copper is represented by 6529Cu.
How many neutrons does the nucleus contain?
| Answer | Mark | Notes |
|---|---|---|
| The nucleon number 65 counts protons and neutrons; the proton number is 29. Neutrons = 65 − 29 = 36. | ✓ 1 | Answer B |
Answer: B
Common error: The top number is the total number of nucleons, not the number of neutrons. Subtract the proton number.
A beam of alpha particles is directed at a very thin gold foil in a vacuum.
Which observation is evidence that most of the volume of an atom is empty space?
| Answer | Mark | Notes |
|---|---|---|
| If most alpha particles meet nothing that deflects them, most of each atom must contain nothing massive or strongly charged: it is mostly empty space. The rare large deflections are the evidence for a small, massive, positive nucleus. | ✓ 1 | Answer D |
Answer: D
Exam tip: Match each observation to its own conclusion. “Most pass straight through” goes with “mostly empty space”; “a few come back” goes with “small, massive, positive nucleus”.
An electron is accelerated from rest through a potential difference of 500 V.
What is the kinetic energy gained by the electron?
| Answer | Mark | Notes |
|---|---|---|
| Kinetic energy gained = eV = 1.60 × 10−19 × 500 = 8.0 × 10−17 J (500 eV). | ✓ 1 | Answer C |
Answer: C
Exam tip: The energy gained by a charge Q moving through a p.d. V is QV. For an electron, the number of volts is the energy in eV.
In a bubble chamber with a uniform magnetic field, the track of a charged particle is a spiral of decreasing radius.
What is the reason for the decreasing radius?
| Answer | Mark | Notes |
|---|---|---|
| The particle loses kinetic energy each time it ionises an atom, so its momentum falls. Since r = p/BQ and B and Q are constant, the radius decreases as the momentum decreases. | ✓ 1 | Answer C |
Answer: C
Exam tip: A spiralling track also tells you the direction of travel: the particle moves from the large-radius end towards the small-radius end.
Naturally occurring chlorine is a mixture of two stable isotopes, 3517Cl and 3717Cl.
State the number of protons and the number of neutrons in a nucleus of 3717Cl.
Explain why these two nuclei are isotopes of the same element.
Calculate the charge on a chlorine nucleus.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 17 protons | ✓ 1 | |
| 37 − 17 = 20 neutrons | ✓ 1 | |
| Part (b) | ||
| same proton number (17) but different numbers of neutrons / different nucleon numbers | ✓ 1 | |
| Part (c) | ||
| 17 × 1.60 × 10−19 = 2.72 × 10−18 C | ✓ 1 | |
Common error: Including the electrons when asked for the charge on the nucleus. Only the protons count.
In an X-ray tube, electrons are released from a hot cathode and accelerated from rest towards a metal target through a p.d. of 1.2 kV.
Define the electronvolt.
Calculate the kinetic energy of each electron when it reaches the target, in joules.
Calculate the speed of the electrons when they reach the target.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the energy gained (or work done) when an electron is accelerated through a p.d. of 1 volt | ✓ 1 | |
| Part (b) | ||
| 1.60 × 10−19 × 1200 = 1.92 × 10−16 J | ✓ 1 | |
| Part (c) | ||
| ½ × 9.11 × 10−31 × v2 = 1.92 × 10−16 | ✓ 1 | |
| v = 2.1 × 107 m s−1 (2.05 × 107) | ✓ 1 | |
Common error: Forgetting the ½ in the kinetic energy, which gives a speed √2 times too small. Write the full equation before rearranging.
Figure 1 shows the protons and neutrons in the nucleus of an atom of element X.
Which symbol represents this nucleus?
| Answer | Mark | Notes |
|---|---|---|
| Count from the figure: 5 protons and 6 neutrons. The proton number (bottom) is 5 and the nucleon number (top) is 5 + 6 = 11: 115X. | ✓ 1 | Answer A |
Answer: A
Exam tip: The top number counts every nucleon, protons and neutrons together; the bottom number counts protons only, so the top number is never the smaller one.
Figure 1 shows three models of the atom, X, Y and Z, that scientists have used at different times.
The results of the alpha-particle scattering experiment showed that one of these models had to be replaced by another. Which change did the results lead to?
| Answer | Mark | Notes |
|---|---|---|
| When the experiment was done, the accepted model was Y: positive charge spread through the atom with electrons embedded in it. Such a diffuse charge cannot turn an alpha particle back, so the few large-angle deflections required a tiny, massive, positive nucleus: model Y was replaced by model Z. | ✓ 1 | Answer D |
Answer: D
Exam tip: Know the order of the models and the evidence for each change: the discovery of the electron ended the indivisible sphere; alpha scattering ended the spread-out positive charge.
In a detector, a singly charged positive ion passes between two parallel plates, as shown in Figure 1.
Which row gives the electric force on the ion while it is between the plates?
| Answer | Mark | Notes |
|---|---|---|
| E = V/d = 450 ÷ 0.030 = 1.5 × 104 V m−1, so F = EQ = 1.5 × 104 × 1.60 × 10−19 = 2.4 × 10−15 N. The field points from the higher potential (lower plate) to the lower potential (upper plate), and the force on a positive ion is along the field: 2.4 × 10−15 N towards the upper plate. | ✓ 1 | Answer D |
Answer: D
Common error: Leaving the plate separation in centimetres. Convert 3.0 cm to 0.030 m before using E = V/d.
Figure 1 shows the proton number Z and neutron number N of four nuclides, P, Q, R and S.
Which two nuclides have the same nucleon number?
| Answer | Mark | Notes |
|---|---|---|
| Nucleon number = Z + N. P: 6 + 8 = 14; Q: 7 + 7 = 14; R: 6 + 6 = 12; S: 8 + 7 = 15. P and Q both have 14 nucleons. | ✓ 1 | Answer C |
Answer: C
Exam tip: Same Z means isotopes (same element); same Z + N means the same nucleon number. Read both values for each point before deciding.
Figure 1 is a sketch of tracks in a bubble chamber. A uniform magnetic field of flux density 0.20 T acts into the page. A γ-ray photon enters from the left and, at P, produces an electron and a positron (an antielectron, charge +1.60 × 10−19 C). The initial radius of track X is 4.0 cm.
Explain why the γ-ray leaves no track.
Explain how conservation of charge accounts for the two tracks curving in opposite directions.
Deduce which track, X or Y, is made by the positron.
Calculate the initial momentum of the particle that made track X.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a γ-ray photon is uncharged, so it does not ionise the liquid along its path | ✓ 1 | |
| Part (b) | ||
| the photon has zero charge, so the two particles must have equal and opposite charges | ✓ 1 | |
| the magnetic forces on them are therefore in opposite directions, so they curve opposite ways | ✓ 1 | |
| Part (c) | ||
| a positive particle moving to the right in a field into the page has a force towards the top of the page (Fleming's left-hand rule) | ✓ 1 | |
| track Y curves upwards, so Y is the positron | ✓ 1 | |
| Part (d) | ||
| p = BQr = 0.20 × 1.60 × 10−19 × 0.040 = 1.3 × 10−21 kg m s−1 (1.28 × 10−21) | ✓ 1 | |
Common error: Applying the left-hand rule to the electron's direction of motion as if it were a current. For an electron, the conventional current is opposite to its motion.
Figure 1 shows the circuit of an electron gun. The electrons leave the filament with negligible speed.
Explain the purpose of the 6.3 V supply.
Explain why the tube must be evacuated.
Calculate the speed of the electrons as they pass through the hole in the anode.
Use the reading on the milliammeter to calculate the number of electrons reaching the anode each second.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the current from it heats the filament | ✓ 1 | |
| some electrons in the hot metal gain enough energy to escape from its surface (thermionic emission) | ✓ 1 | |
| Part (b) | ||
| so that the electrons do not collide with gas molecules and lose energy / get scattered | ✓ 1 | |
| Part (c) | ||
| eV = ½mv2: 1.60 × 10−19 × 3600 = ½ × 9.11 × 10−31 × v2 | ✓ 1 | |
| v = 3.6 × 107 m s−1 (3.56 × 107) | ✓ 1 | |
| Part (d) | ||
| N = I/e = 0.40 × 10−3 ÷ 1.60 × 10−19 = 2.5 × 1015 | ✓ 1 | |
Exam tip: Only the accelerating p.d. between filament and anode sets the electrons' energy. The heater p.d. releases the electrons but does not accelerate them.
An alpha particle travels directly towards a nucleus and is turned back.
Which change would decrease its distance of closest approach to the nucleus?
| Answer | Mark | Notes |
|---|---|---|
| At closest approach the initial kinetic energy has all become electric potential energy, Ek = Q1Q2/4πε0r. A larger Ek gives a smaller r; a larger proton number or a lower speed gives a larger r. | ✓ 1 | Answer B |
Answer: B
Exam tip: Write the energy equation first, then read off how r depends on each quantity: r ∝ Z and r ∝ 1/Ek.
In an electron gun, electrons are released from a hot filament and accelerated from rest through a potential difference of 1.8 kV.
What is the speed of the electrons as they leave the gun?
| Answer | Mark | Notes |
|---|---|---|
| eV = ½mv2, so v = √(2eV/m) = √(2 × 1.60 × 10−19 × 1800 ÷ 9.11 × 10−31) = 2.5 × 107 m s−1. | ✓ 1 | Answer D |
Answer: D
Common error: Leaving the p.d. in kilovolts. Convert 1.8 kV to 1800 V before substituting.
An alpha particle has momentum 9.6 × 10−20 kg m s−1. It moves at right angles to a uniform magnetic field of flux density 0.60 T.
What is the radius of its circular path?
| Answer | Mark | Notes |
|---|---|---|
| The charge of an alpha particle is 2e = 3.20 × 10−19 C. r = p/BQ = 9.6 × 10−20 ÷ (0.60 × 3.20 × 10−19) = 0.50 m. | ✓ 1 | Answer B |
Answer: B
Common error: Using e for the charge of every particle. An alpha particle carries two proton charges.
In a cyclotron, charged particles are accelerated each time they cross the gap between the dees by an alternating p.d. of fixed frequency.
Why can the frequency stay fixed while the particles speed up?
| Answer | Mark | Notes |
|---|---|---|
| From r = mv/BQ, the radius grows in proportion to the speed, so the time for a semicircle, πr/v = πm/BQ, does not depend on the speed. Each crossing of the gap therefore comes at the same interval. | ✓ 1 | Answer C |
Answer: C
Exam tip: In a cyclotron the electric field (in the gap) changes the speed; the magnetic field changes only the direction.
In the early twentieth century, a beam of alpha particles was fired at a very thin metal foil. The results led to the nuclear model of the atom.
Describe the experiment and explain how its results provide evidence for the nuclear model of the atom. (6 marks: indicative content, one mark per point)
| Answer | Mark | Notes |
|---|---|---|
| a narrow beam of alpha particles is directed at a very thin (gold) foil, in a vacuum | ✓ 1 | |
| a detector is moved round the foil to count the alpha particles scattered at different angles | ✓ 1 | |
| most alpha particles pass straight through / are deflected through small angles, so the atom is mostly empty space | ✓ 1 | |
| some are deflected through large angles, so there is a concentrated charge that repels them, so it is positive | ✓ 1 | |
| a very small number are deflected by more than 90° / come back, so the nucleus contains most of the mass of the atom | ✓ 1 | |
| the nucleus is very small compared with the atom (because so few alpha particles come close to it) | ✓ 1 |
Model sketch
Exam tip: In a six-mark answer, keep the method short and make every result lead to a conclusion: observation, then “so”, then what it shows about the atom.
A student uses a demonstration tube to study an electron beam. Electrons released from a heated filament are accelerated from rest through a p.d. of 250 V. The tube contains a gas at very low pressure that glows faintly where the electrons pass. A pair of coils produces a uniform magnetic field of flux density 1.0 mT at right angles to the beam, and the beam forms a complete circle.
Explain how electrons are released from the filament.
Show that the speed of the electrons is about 9.4 × 106 m s−1.
Calculate the diameter of the circle.
The accelerating p.d. is increased to 1000 V with the same magnetic field. Determine the new diameter of the circle.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the filament is heated by an electric current | ✓ 1 | |
| some electrons gain enough energy to escape from the surface of the metal (thermionic emission) | ✓ 1 | |
| Part (b) | ||
| eV = ½mv2: 1.60 × 10−19 × 250 = ½ × 9.11 × 10−31 × v2 | ✓ 1 | |
| v = 9.37 × 106 m s−1 | ✓ 1 | |
| Part (c) | ||
| r = mv/Be = 9.11 × 10−31 × 9.37 × 106 ÷ (1.0 × 10−3 × 1.60 × 10−19) = 0.0534 m | ✓ 1 | |
| diameter = 0.11 m (0.107 m) | ✓ 1 | |
| Part (d) | ||
| four times the kinetic energy, so twice the speed and twice the radius: diameter = 0.21 m | ✓ 1 | |
Exam tip: For a scaling question, write how the quantity depends on the one that changes: r ∝ v ∝ √V, so ×4 on the p.d. gives ×2 on the radius.
A linear accelerator for protons consists of a line of hollow, cylindrical metal drift tubes in an evacuated pipe. The tubes are connected alternately to the terminals of an alternating supply of frequency 30 MHz. In one of the tubes the protons each have a kinetic energy of 0.90 MeV.
Explain why the protons are not accelerated while they are inside a drift tube.
Explain why the p.d. supplied to the tubes must alternate.
Calculate the length of the tube in which the protons have a kinetic energy of 0.90 MeV.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| there is no electric field inside a hollow metal tube (the whole tube is at one potential), so there is no force on the protons | ✓ 1 | |
| Part (b) | ||
| the protons are accelerated only in the gaps, so at each gap the field must point the right way (forwards) as the protons cross | ✓ 1 | |
| the potential of each tube must reverse while the protons are inside it, ready for the next gap | ✓ 1 | |
| Part (c) | ||
| v = √(2 × 0.90 × 106 × 1.60 × 10−19 ÷ 1.67 × 10−27) = 1.31 × 107 m s−1 | ✓ 1 | |
| time in the tube = half a period = 1 ÷ (2 × 30 × 106) = 1.67 × 10−8 s | ✓ 1 | |
| length = 1.31 × 107 × 1.67 × 10−8 = 0.22 m (0.219 m) | ✓ 1 | |
Common error: Using a whole period for the time in a tube. The p.d. reverses every half period, so each tube is crossed in half a period.
A cloud chamber contains air supersaturated with alcohol vapour. A uniform magnetic field acts out of the plane of the page. A charged particle enters the chamber from the left, moving to the right, and its track curves towards the bottom of the page.
Explain how the particle produces a visible track.
Deduce the sign of the charge on the particle.
The radius of the track decreases gradually along its length. Explain why.
A second particle has the same momentum as the first but twice its charge. State how the radius of its track compares with that of the first.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the particle ionises molecules along its path | ✓ 1 | |
| the vapour condenses on the ions, forming a line of droplets | ✓ 1 | |
| Part (b) | ||
| the force is towards the centre of curvature, i.e. towards the bottom of the page; with the field out of the page and the motion to the right, Fleming's left-hand rule gives a downward force for a positive charge | ✓ 1 | |
| so the particle is positive | ✓ 1 | |
| Part (c) | ||
| the particle loses kinetic energy as it ionises the gas | ✓ 1 | |
| so its momentum decreases and, since r = p/BQ, the radius decreases | ✓ 1 | |
| Part (d) | ||
| half the radius | ✓ 1 | |
Exam tip: For a sign-of-charge question, state the field direction, the direction of motion and the direction of the force (towards the centre of the curve) before naming the rule.
Physicists study the internal structure of protons and neutrons by scattering very high-energy electrons from them. A nucleon has a diameter of about 1.6 × 10−15 m.
Explain why the de Broglie wavelength of the electrons must be no larger than the size of the structure being studied.
Calculate the momentum of an electron with a de Broglie wavelength of 0.80 × 10−15 m.
A school electron diffraction tube accelerates electrons from rest through 5.0 kV. Show that their de Broglie wavelength is about 1.7 × 10−11 m and explain why these electrons cannot be used to study the structure of nucleons.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the electrons are diffracted / scattered significantly by structure only when their wavelength is comparable to or smaller than its size | ✓ 1 | |
| detail smaller than the wavelength cannot be resolved | ✓ 1 | |
| Part (b) | ||
| p = h/λ = 6.63 × 10−34 ÷ 0.80 × 10−15 = 8.3 × 10−19 kg m s−1 | ✓ 1 | |
| Part (c) | ||
| p = √(2meV) = √(2 × 9.11 × 10−31 × 1.60 × 10−19 × 5000) = 3.82 × 10−23 kg m s−1, so λ = h/p = 1.74 × 10−11 m | ✓ 1 | |
| this is about 104 times the size of a nucleon, so the electrons would not be scattered by its internal structure / much higher energies are needed | ✓ 1 | |
Exam tip: Link the three ideas in order: smaller detail → shorter wavelength → larger momentum (λ = h/p) → higher energy.
Figure 1 shows how the electric potential energy of an alpha particle and a tin nucleus depends on the separation r of their centres.
An alpha particle with an initial kinetic energy of 6.0 MeV travels directly towards a tin nucleus, which does not move. What is the distance of closest approach?
| Answer | Mark | Notes |
|---|---|---|
| Energy is conserved: at closest approach the alpha particle is momentarily at rest, so all 6.0 MeV of kinetic energy has become electric potential energy. Read across from 6.0 MeV to the curve: r = 24 × 10−15 m = 2.4 × 10−14 m. | ✓ 1 | Answer C |
Answer: C
Exam tip: Closest approach is where kinetic energy has all become potential energy, so read the graph at the starting kinetic energy. Then apply the axis power of ten carefully.
Figure 1 shows the path of a proton in a cyclotron, seen from above. A uniform magnetic field acts at right angles to the plane of the page, and the arrows show the direction in which the proton travels.
What is the direction of the magnetic field?
| Answer | Mark | Notes |
|---|---|---|
| The magnetic force on the proton points towards the centre of each semicircle. On the right-hand dee the proton moves up the page and the force is to the left. Fleming's left-hand rule (current up the page, force to the left) gives a field into the page. A field in the plane of the page could not make the proton circle in that plane. | ✓ 1 | Answer A |
Answer: A
Exam tip: For a circular path, the force is always towards the centre. Pick one point on the path, mark the velocity and the force there, then apply the left-hand rule.
Protons move at right angles to a uniform magnetic field. Figure 1 shows how the radius r of their circular path depends on their momentum p.
What is the magnetic flux density?
| Answer | Mark | Notes |
|---|---|---|
| Use the marked point: r = 0.50 m when p = 4.0 × 10−20 kg m s−1. r = p/BQ, so B = p/rQ = 4.0 × 10−20 ÷ (0.50 × 1.60 × 10−19) = 0.50 T. | ✓ 1 | Answer B |
Answer: B
Common error: Missing the power of ten on the axis label. The momentum axis is in units of 10−20 kg m s−1, so the marked value is 4.0 × 10−20, not 4.0.
Four charged particles enter a uniform magnetic field at the same point, all moving to the right with the same momentum. The field is directed into the page. Figure 1 shows their tracks, A to D.
Which track was made by an alpha particle?
| Answer | Mark | Notes |
|---|---|---|
| An alpha particle is positive. For a positive charge moving to the right in a field into the page, Fleming's left-hand rule gives a force towards the top of the page, so it curves upwards (B or C). With the same momentum, r = p/BQ ∝ 1/Q, and its charge 2e gives half the radius of a singly charged particle: track C. | ✓ 1 | Answer C |
Answer: C
Exam tip: Use the direction of curvature for the sign of the charge and the size of the radius for the momentum or charge, one at a time.
Figure 1 shows the sizes of three structures on a logarithmic scale. A beam of electrons, each with momentum 2.2 × 10−19 kg m s−1, is used to study them. Detail can only be resolved if the de Broglie wavelength of the electrons is smaller than the size of the structure.
Which structures in Figure 1 can these electrons resolve?
| Answer | Mark | Notes |
|---|---|---|
| λ = h/p = 6.63 × 10−34 ÷ 2.2 × 10−19 = 3.0 × 10−15 m. This is smaller than an atom (10−10 m) and a large nucleus (10−14 m) but larger than a proton (10−15 m): the atom and the large nucleus only. | ✓ 1 | Answer B |
Answer: B
Exam tip: To see finer structure you need a shorter wavelength, so a larger momentum and therefore a higher energy. That is why nucleon structure needs very high-energy beams.
A beam of neon ions contains two isotopes, 2010Ne+ and 2210Ne+. Each ion has a charge of +1.60 × 10−19 C and all the ions have the same speed. They enter a uniform magnetic field of flux density 0.25 T through a slit S, as shown in Figure 1, and land on a detector plate at marks M1 and M2. Mass of a 2010Ne+ ion = 20 × 1.66 × 10−27 kg.
State the number of protons and the number of neutrons in the nucleus of each isotope.
The 2010Ne+ ions land at M1. Use Figure 1 to calculate the momentum of these ions.
Calculate the speed of the ions.
Explain why the 2210Ne+ ions land at M2, and show that the position of M2 in Figure 1 agrees with this.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| both have 10 protons | ✓ 1 | |
| 20Ne has 10 neutrons and 22Ne has 12 neutrons | ✓ 1 | |
| Part (b) | ||
| distance S to M1 = 10.0 cm (allow 9.9–10.1 cm), so radius = 0.050 m | ✓ 1 | |
| p = BQr = 0.25 × 1.60 × 10−19 × 0.050 = 2.0 × 10−21 kg m s−1 | ✓ 1 | |
| Part (c) | ||
| v = p/m = 2.0 × 10−21 ÷ (20 × 1.66 × 10−27) = 6.0 × 104 m s−1 | ✓ 1 | |
| Part (d) | ||
| same speed and charge, so r = mv/BQ ∝ m: the heavier ions move in a larger semicircle and land further from S | ✓ 1 | |
| predicted distance = 10.0 × 22/20 = 11.0 cm, which matches M2 at 11.0 cm on the scale | ✓ 1 | |
Common error: Using the distance from S to the mark as the radius. The ions travel a semicircle, so that distance is the diameter: halve it first.
A narrow beam of alpha particles is directed at a very thin metal foil in a vacuum. A detector is moved around the foil and the count rate is measured at different scattering angles θ. Figure 1 shows the results, with the background count rate already subtracted.
Suggest why the count rate is plotted on a logarithmic scale.
Use Figure 1 to determine the count rate at 30° and at 150°, and calculate the ratio (count rate at 30°) ÷ (count rate at 150°).
Explain what the detection of alpha particles at 150° shows about the structure of the atom.
Before this experiment, the positive charge of an atom was thought to be spread evenly through the whole atom. Explain why that model predicts that almost no alpha particles would be detected at 150°.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the count rates range over several orders of magnitude / from about 1 to several hundred, so a linear scale would hide the small values | ✓ 1 | |
| Part (b) | ||
| about 450 min−1 at 30° (allow 420–470) and 2.3 min−1 at 150° (allow 2.1–2.5) | ✓ 1 | |
| ratio ≈ 190 (allow 165–225) | ✓ 1 | |
| Part (c) | ||
| the alpha particles are turned back, so they must meet a very strong repulsive force: the positive charge is concentrated in a very small region | ✓ 1 | |
| to turn back an alpha particle this region must also contain most of the mass of the atom (a small, massive, positive nucleus) | ✓ 1 | |
| Part (d) | ||
| spread-out charge produces only a weak electric field, so each atom deflects an alpha particle through a very small angle | ✓ 1 | |
| many small random deflections cannot add up to a deflection of 150° | ✓ 1 | |
Exam tip: On a logarithmic axis, the gridlines between 100 and 1000 are 200, 300, 400…, not equally spaced values of 110, 120… Read the decade first, then the line within it.
Figure 1 shows the tracks in a liquid-hydrogen bubble chamber with no magnetic field. A proton with kinetic energy 3.0 MeV strikes a stationary proton (a hydrogen nucleus) at V. The two protons move off in the directions shown.
Calculate the momentum of the incoming proton.
Use conservation of momentum to calculate the momentum of each proton after the collision.
Show that kinetic energy is conserved in this collision.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Ek = 3.0 × 106 × 1.60 × 10−19 = 4.8 × 10−13 J | ✓ 1 | |
| p = √(2mEk) = √(2 × 1.67 × 10−27 × 4.8 × 10−13) = 4.0 × 10−20 kg m s−1 | ✓ 1 | |
| Part (b) | ||
| at right angles to the original direction: p1 sin 35° = p2 sin 55° | ✓ 1 | |
| along the original direction: p1 cos 35° + p2 cos 55° = 4.0 × 10−20 | ✓ 1 | |
| p1 = 3.3 × 10−20 kg m s−1 and p2 = 2.3 × 10−20 kg m s−1 | ✓ 1 | |
| Part (c) | ||
| Ek = p2/2m: proton 1 has 3.22 × 10−13 J (2.0 MeV) and proton 2 has 1.58 × 10−13 J (0.99 MeV) | ✓ 1 | |
| total = 4.8 × 10−13 J (3.0 MeV), equal to the initial kinetic energy | ✓ 1 | |
Exam tip: With no magnetic field the tracks are straight, so the angles in the photograph are the directions of the momenta. Resolve along and at right angles to the incoming track.
A student accelerates electrons from rest through a p.d. V in an electron diffraction tube and finds their de Broglie wavelength λ from the diffraction rings, for five values of V between 2.0 kV and 6.0 kV. Figure 1 shows a graph of λ against 1/√V with a line of best fit.
Show that λ = h/√(2meeV).
Explain why the graph is expected to be a straight line through the origin.
Use Figure 1 to determine a value for the Planck constant.
A nucleon has a diameter of about 10−15 m. Use the graph to explain why the electrons in this tube cannot be used to study the internal structure of a nucleon, and what must be done instead.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| eV = ½mev2 = p2/2me, so p = √(2meeV) | ✓ 1 | |
| λ = h/p = h/√(2meeV) | ✓ 1 | |
| Part (b) | ||
| λ = [h/√(2mee)] × 1/√V, and the bracket is a constant, so λ ∝ 1/√V | ✓ 1 | |
| Part (c) | ||
| gradient = 2.95 × 10−11 ÷ 2.40 × 10−2 = 1.23 × 10−9 | ✓ 1 | |
| gradient = h/√(2mee), so h = 1.23 × 10−9 × √(2 × 9.11 × 10−31 × 1.60 × 10−19) | ✓ 1 | |
| h = 6.6 × 10−34 J s (allow 6.5–6.8 × 10−34) | ✓ 1 | |
| Part (d) | ||
| the wavelengths are about 10−11 m, roughly 104 times larger than a nucleon, so no detail inside a nucleon can be resolved | ✓ 1 | |
| a wavelength of 10−15 m or less is needed, so the electrons must have a much larger momentum: they must be accelerated to much higher energies | ✓ 1 | |
Exam tip: When using a gradient, take values from the line, not from a data point, and use points far apart. Keep the powers of ten from both axis labels.
A proton has a kinetic energy of 2.0 MeV.
What is the de Broglie wavelength of the proton?
| Answer | Mark | Notes |
|---|---|---|
| Ek = 2.0 × 106 × 1.60 × 10−19 = 3.2 × 10−13 J. p = √(2mEk) = √(2 × 1.67 × 10−27 × 3.2 × 10−13) = 3.27 × 10−20 kg m s−1. λ = h/p = 2.0 × 10−14 m. | ✓ 1 | Answer A |
Answer: A
Exam tip: Get from kinetic energy to momentum with Ek = p2/2m, then use λ = h/p. The photon relation λ = hc/E does not apply to a particle with mass.
A particle X with charge +e is at rest in a uniform magnetic field. It decays into two charged particles only. One of them, Y, has charge +2e; the other is Z. Both move at right angles to the field.
What is (radius of the path of Y) ÷ (radius of the path of Z)?
| Answer | Mark | Notes |
|---|---|---|
| Charge conservation: +e = +2e + QZ, so QZ = −e. Momentum conservation: X was at rest, so Y and Z have equal and opposite momenta. With the same p and B, r ∝ 1/Q, so rY/rZ = e/2e = 0.50. | ✓ 1 | Answer A |
Answer: A
Exam tip: Two conservation laws are needed here: charge gives the unknown charge, momentum shows the two momenta are the same size.
An alpha particle with an initial kinetic energy of 5.3 MeV travels directly towards a platinum nucleus. The proton number of platinum is 78. Assume the platinum nucleus does not move.
Show that the initial kinetic energy is about 8.5 × 10−13 J.
Calculate the distance of closest approach of the alpha particle to the centre of the nucleus.
Explain why your answer is an upper limit for the radius of a platinum nucleus.
An alpha particle with the same kinetic energy is aimed directly at an aluminium nucleus (proton number 13). State and explain how the distance of closest approach changes.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 5.3 × 106 × 1.60 × 10−19 = 8.48 × 10−13 J | ✓ 1 | |
| Part (b) | ||
| at closest approach all the kinetic energy is electric potential energy: Ek = Q1Q2/4πε0r | ✓ 1 | |
| 8.48 × 10−13 = 8.99 × 109 × (2 × 1.60 × 10−19) × (78 × 1.60 × 10−19) ÷ r | ✓ 1 | |
| r = 4.2 × 10−14 m (4.23 × 10−14 m) | ✓ 1 | |
| Part (c) | ||
| the alpha particle does not reach / touch the nucleus, so the nucleus must be smaller than the closest-approach distance | ✓ 1 | |
| Part (d) | ||
| smaller, by a factor of 78/13 = 6, because the potential energy at a given distance is proportional to the charge of the nucleus | ✓ 1 | |
Common error: Using e for both charges. The alpha particle carries 2e and the nucleus carries Ze.
A cyclotron accelerates deuterons (mass 3.34 × 10−27 kg, charge +1.60 × 10−19 C) in a uniform magnetic field of flux density 1.4 T. The deuterons leave the cyclotron when the radius of their path reaches 0.60 m. Their speed stays well below the speed of light.
Show that the time taken by a deuteron to travel one semicircle inside a dee does not depend on its speed.
Calculate the frequency of the alternating p.d. applied to the dees.
Calculate the maximum kinetic energy of the deuterons. Give your answer in MeV.
Suggest why a cyclotron cannot be used to accelerate particles to speeds close to the speed of light.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| BQv = mv2/r, so r = mv/BQ | ✓ 1 | |
| time = πr/v = πm/BQ, which does not contain v | ✓ 1 | |
| Part (b) | ||
| one cycle of the p.d. per revolution: f = BQ/2πm | ✓ 1 | |
| = 1.4 × 1.60 × 10−19 ÷ (2π × 3.34 × 10−27) = 1.07 × 107 Hz | ✓ 1 | |
| Part (c) | ||
| p = BQr = 1.4 × 1.60 × 10−19 × 0.60 = 1.34 × 10−19 kg m s−1 | ✓ 1 | |
| Ek = p2/2m = 2.70 × 10−12 J | ✓ 1 | |
| ÷ 1.60 × 10−13 J per MeV = 17 MeV (16.9 MeV) | ✓ 1 | |
| Part (d) | ||
| close to c the time per revolution no longer stays constant (it increases), so the particles fall out of step with the fixed-frequency p.d. | ✓ 1 | |
Exam tip: In a “show that” derivation, start from the physics statement — the magnetic force provides the centripetal force — before writing any algebra.
A photograph from a bubble chamber with a uniform magnetic field of 1.8 T shows a neutral particle decaying in flight into two particles, each with a charge of magnitude 1.60 × 10−19 C. Measurements on the photograph, corrected to real size, give:
track 1: radius 0.50 m, leaving at 25° on one side of the line of the neutral particle;
track 2: radius 0.33 m, leaving at 40° on the other side of that line.
State what conservation of charge tells you about the charges of the two particles.
Calculate the momentum of each particle just after the decay.
Show that these momenta are consistent with conservation of momentum at right angles to the line of the neutral particle.
Calculate the momentum of the neutral particle.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| they are equal and opposite (+1.60 × 10−19 C and −1.60 × 10−19 C), because the total charge is zero before and after | ✓ 1 | |
| Part (b) | ||
| p1 = BQr = 1.8 × 1.60 × 10−19 × 0.50 = 1.44 × 10−19 kg m s−1 | ✓ 1 | |
| p2 = 1.8 × 1.60 × 10−19 × 0.33 = 9.50 × 10−20 kg m s−1 | ✓ 1 | |
| Part (c) | ||
| perpendicular components: 1.44 × 10−19 sin 25° = 6.09 × 10−20 and 9.50 × 10−20 sin 40° = 6.11 × 10−20 | ✓ 1 | |
| these are equal (within 1%) and in opposite directions, so the total perpendicular momentum is zero, as it was before the decay | ✓ 1 | |
| Part (d) | ||
| sum of components along the line: 1.44 × 10−19 cos 25° + 9.50 × 10−20 cos 40° = 1.31 × 10−19 + 7.28 × 10−20 | ✓ 1 | |
| = 2.0 × 10−19 kg m s−1 (2.03 × 10−19) | ✓ 1 | |
Exam tip: Momentum is a vector. Resolve each momentum along and at right angles to a chosen line, then apply conservation to each direction separately.
Positive ions start from rest and are accelerated through a potential difference V. Figure 1 shows how the square of their final speed, v2, varies with V.
What is the charge-to-mass ratio Q/m of the ions?
| Answer | Mark | Notes |
|---|---|---|
| QV = ½mv2, so v2 = (2Q/m)V and the gradient is 2Q/m. Gradient = 7.66 × 1011 ÷ 4.0 × 103 = 1.92 × 108 m2 s−2 V−1, so Q/m = 1.92 × 108 ÷ 2 = 9.6 × 107 C kg−1. | ✓ 1 | Answer A |
Answer: A
Exam tip: Before using a gradient, write the equation in the form y = (gradient) × x with the plotted quantities: here v2 = (2Q/m) × V.
Figure 1 shows the first five drift tubes of a linear accelerator for protons, with the length of each tube. The alternating supply has a frequency of 8.0 MHz. Protons leave the source with negligible kinetic energy and gain the same energy at every gap, including the gap between the source and tube 1. Each proton spends half a period of the supply inside each tube.
Explain why each drift tube must be longer than the one before it.
Use the length of tube 5 to show that the speed of the protons in tube 5 is about 6.9 × 106 m s−1.
Calculate the kinetic energy of the protons in tube 5. Give your answer in keV.
Deduce the p.d. across each gap as the protons cross it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the protons gain energy and speed at every gap | ✓ 1 | |
| each tube must be crossed in the same time (half a period), so at a higher speed the tube must be longer | ✓ 1 | |
| Part (b) | ||
| time in the tube = 1 ÷ (2 × 8.0 × 106) = 6.25 × 10−8 s | ✓ 1 | |
| v = 0.433 ÷ 6.25 × 10−8 = 6.93 × 106 m s−1 | ✓ 1 | |
| Part (c) | ||
| ½ × 1.67 × 10−27 × (6.93 × 106)2 = 4.01 × 10−14 J | ✓ 1 | |
| ÷ 1.60 × 10−16 J per keV = 250 keV (251 keV) | ✓ 1 | |
| Part (d) | ||
| five gaps have been crossed before tube 5, so energy per gap = 250 ÷ 5 = 50 keV and the p.d. = 50 kV (5.0 × 104 V) | ✓ 1 | |
Exam tip: Count the gaps carefully: protons in tube n have crossed n gaps when the first gap is between the source and tube 1.
Protons are released at the centre of a cyclotron with negligible speed. The magnetic flux density is 0.80 T. Each time a proton crosses the gap between the dees it is accelerated through the same p.d. V. A probe measures the radius r of the semicircle a proton travels after its nth crossing. Figure 1 shows a graph of r2 against n.
Explain why the radius increases each time the proton crosses the gap.
Show that r2 should be proportional to n.
Use the gradient of the line in Figure 1 to determine V.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the electric field in the gap does work on the proton, increasing its speed / momentum | ✓ 1 | |
| r = p/BQ with B and Q constant, so a larger momentum gives a larger radius | ✓ 1 | |
| Part (b) | ||
| after n crossings the kinetic energy is neV, so p2 = 2m × neV | ✓ 1 | |
| r2 = p2/B2e2 = (2mV/B2e) × n, and everything in the bracket is constant | ✓ 1 | |
| Part (c) | ||
| gradient = 20.0 ÷ 10 = 2.0 cm2 = 2.0 × 10−4 m2 | ✓ 1 | |
| gradient = 2mV/B2e, so V = 2.0 × 10−4 × 0.802 × 1.60 × 10−19 ÷ (2 × 1.67 × 10−27) | ✓ 1 | |
| V = 6.1 × 103 V = 6.1 kV (6130 V) | ✓ 1 | |
Common error: Forgetting to convert cm2 to m2. 1 cm2 = 1 × 10−4 m2, not 10−2 m2.
Figure 1 shows the track of a charged particle in a cloud chamber. A uniform magnetic field of flux density 0.50 T acts into the page. The particle passes through a thin aluminium plate. The radius of the track is 0.60 m on one side of the plate and 0.40 m on the other.
Explain how Figure 1 shows the direction in which the particle was travelling.
Deduce the sign of the charge on the particle.
The particle is a proton. Calculate the kinetic energy it loses in the plate. Give your answer in MeV.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the particle loses kinetic energy, and so momentum, as it passes through the plate | ✓ 1 | |
| r = p/BQ, so the radius is smaller after the plate: the particle travelled from right to left | ✓ 1 | |
| Part (b) | ||
| moving to the left with the centre of curvature below the track, so the force is downwards (towards the centre) | ✓ 1 | |
| Fleming's left-hand rule with the field into the page gives a downward force for a positive charge, so the particle is positive | ✓ 1 | |
| Part (c) | ||
| p = BQr: before = 0.50 × 1.60 × 10−19 × 0.60 = 4.8 × 10−20 and after = 3.2 × 10−20 kg m s−1 | ✓ 1 | |
| Ek = p2/2m: before = 6.90 × 10−13 J and after = 3.07 × 10−13 J | ✓ 1 | |
| loss = 3.83 × 10−13 J ÷ 1.60 × 10−13 J per MeV = 2.4 MeV | ✓ 1 | |
Exam tip: Kinetic energy depends on p2, so the energy does not fall in the same ratio as the radius: here the radius falls by a third but the kinetic energy falls by more than half.
Figure 1 is a sketch of tracks in a liquid-hydrogen bubble chamber in a uniform magnetic field directed into the page. A negatively charged particle enters from the left and interacts with a proton (a hydrogen nucleus) at V1.
Discuss what the tracks in Figure 1 show about the particles produced at V1 and V2. Refer to conservation laws and to the shapes of the tracks. (6 marks: indicative content, one mark per point)
| Answer | Mark | Notes |
|---|---|---|
| tracks are trails of bubbles formed where charged particles ionise the liquid, so a neutral particle leaves no track | ✓ 1 | |
| the gap between V1 and V2 shows that a neutral particle was produced at V1 and travelled to V2, where it decayed into two charged particles, c and d | ✓ 1 | |
| with the field into the page, Fleming's left-hand rule shows that particles curving anticlockwise (a and c) are positive and those curving clockwise (b and d, like the incoming particle) are negative | ✓ 1 | |
| charge is conserved: at V1, −1 + 1 = 0 = (+1) + (−1) + 0 for a, b and the neutral particle; at V2, 0 = (+1) + (−1) | ✓ 1 | |
| r = p/BQ, so for the same charge a larger radius means a larger momentum: a has much more momentum than b | ✓ 1 | |
| the radius of b decreases steadily because it loses kinetic energy (and momentum) as it ionises the liquid; it spirals inwards | ✓ 1 |
Exam tip: In a track question, every feature is evidence: a gap means a neutral particle, the direction of curvature gives the sign, the radius gives the momentum, and a spiral shows energy loss.
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