Atomic, nuclear and quantum physics for IB Physics HL

Reading cue. The arrow length means nothing here, the diagram is not to scale: take every energy from the printed labels, and note every level is negative because 0 eV is the just-free electron.

1. aState State the energy that must be supplied to a hydrogen atom to remove its electron completely when that electron occupies the level marked n = 2, and state why every level below n = ∞ carries a negative value.

   recall2 marks

   Check answer · 2 marks

   1. 3.40 eV, the difference between the −3.40 eV level and the 0 eV level marked n = ∞
   2. Zero is defined as the electron free of the atom and at rest, so a bound electron has less energy than this and its value is negative

2. bCalculate Calculate the wavelength of the photon emitted when the electron falls from the level marked n = 4 to the level marked n = 2. Use hc = 1240 eV nm.

   routine3 marks

   Check answer · 3 marks

   1. ΔE = −0.85 − (−3.40) = 2.55 eV
   2. λ = 1240/2.55
   3. λ = 486 nm

3. cDetermine An electron begins in the level marked n = 4. Using only the levels drawn, determine how many different photon wavelengths could be emitted as it returns to the ground state, and determine the shortest of those wavelengths.

   demanding4 marks

   Check answer · 4 marks

   1. Every downward transition between the four bound levels is available: 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1
   2. Six different wavelengths
   3. Shortest wavelength comes from the largest energy drop, 4→1: ΔE = −0.85 − (−13.6) = 12.75 eV
   4. λ = 1240/12.75 = 97.3 nm

4. dExplain Hydrogen atoms in the ground state are bombarded first with free electrons of kinetic energy 12.5 eV, and then with photons of energy 12.5 eV. Explain, using the levels drawn, why the electrons excite the atoms but the photons do not.

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   Check answer · 4 marks

   1. Excitation from the ground state requires exactly 10.20 eV, 12.09 eV or 12.75 eV to reach n = 2, n = 3 or n = 4
   2. A photon is absorbed whole or not at all, and 12.5 eV matches none of these gaps, so the photons pass through unabsorbed
   3. A free electron transfers energy by collision and need not give up all of it, so it can supply exactly 12.09 eV and raise the atom to n = 3
   4. The bombarding electron then moves on with the remaining 12.5 − 12.09 = 0.41 eV of kinetic energy

Reading cue. Follow the supply plus terminal: it reaches the collector, so electrons are being pulled across. As drawn the meter reads a collected current, not anything about a stopping potential.

1. aState State the direction of the conventional current in the wire joining the plate to the microammeter, and state why the tube must be evacuated.

   recall2 marks

   Check answer · 2 marks

   1. Conventional current flows from the plate through the microammeter towards the negative terminal of the supply, opposite to the electron flow in that wire
   2. Evacuated so that photoelectrons are not scattered or absorbed by gas molecules before reaching the collector

2. bDetermine The incident light has a wavelength of 400 nm and the plate has a work function of 2.30 eV. Determine the reverse potential difference that would have to be applied to the collector to bring the microammeter reading to zero, and determine the reading in μA when 1.2 × 10¹² electrons leave the plate each second and all are collected. Use hc = 1240 eV nm.

   routine4 marks

   Check answer · 4 marks

   1. Photon energy = 1240/400 = 3.10 eV
   2. Maximum kinetic energy = 3.10 − 2.30 = 0.80 eV
   3. Stopping potential difference = 0.80 V, with the collector made negative
   4. Current = 1.2 × 10¹² × 1.60 × 10⁻¹⁹ = 1.9 × 10⁻⁷ A, that is 0.19 μA

3. cSketch Sketch the variation of the microammeter reading with the potential difference applied to the collector, from a reverse value, through zero, to a forward value large enough for the reading to stop rising. On the same axes sketch the result of replacing the light with light of shorter wavelength delivering the same number of photons per second.

   demanding4 marks

   Check answer · 4 marks

   1. First curve rises from zero at a negative potential difference and levels off at a constant saturation current once the collector is positive
   2. The arrangement drawn, with the collector held positive, lies on the flat saturated part of that curve
   3. Second curve saturates at the same current, because the number of photons arriving each second is unchanged
   4. Second curve meets the potential-difference axis at a more negative value, because the photoelectrons now leave with greater maximum kinetic energy

4. dDiscuss The collecting electrode is drawn small and clear of the light beam. Discuss what the microammeter would record if it were enlarged so that it faced the whole plate and was itself illuminated.

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   Check answer · 5 marks

   1. Light reaching the enlarged collector ejects photoelectrons from the collector as well as from the plate, provided its work function is small enough for the wavelength used
   2. With the collector held positive, as drawn, the field between the electrodes returns those electrons to the collector, so the saturation reading is barely altered by them
   3. An electrode large enough to face the whole plate does stand in the path of the beam, so less light reaches the plate and the saturation current falls for that reason instead
   4. Once the potential difference is reversed to look for the stopping potential, the field then drives the collector's own photoelectrons across to the plate, giving a current in the opposite sense
   5. The reading therefore does not fall to zero at the true stopping potential difference, so any maximum kinetic energy obtained from it would be wrong

Reading cue. The vertical axis is ln (A / Bq) and starts at 5.2, not zero. Where the line meets t = 0 you read ln A0, so it must be exponentiated before it is an activity in Bq.

1. aDetermine The plotted crosses run from ln (A / Bq) = 6.4 at t = 0 down to ln (A / Bq) = 5.2 at t = 40 minutes. Determine the activity of the source at the moment the first reading was taken.

   recall2 marks

   Check answer · 2 marks

   1. Recognition that the intercept is ln A₀, so A₀ = e⁶.4 rather than 6.4
   2. A₀ = 6.0 × 10² Bq

2. bDetermine Determine the activity of the source, in Bq, 25 minutes after the first reading.

   routine3 marks

   Check answer · 3 marks

   1. Gradient = (5.2 − 6.4)/40 = −0.030 min⁻¹
   2. ln (A / Bq) at t = 25 min is 6.4 − 0.030 × 25 = 5.65
   3. A = e⁵.65 = 2.8 × 10² Bq

3. cDetermine Determine the number of undecayed nuclei of this nuclide present in the source at the moment the first reading was taken.

   demanding4 marks

   Check answer · 4 marks

   1. Use of A = λN
   2. Decay constant converted from the graph's minutes to seconds: 0.030/60 = 5.0 × 10⁻⁴ s⁻¹
   3. N₀ = 602/(5.0 × 10⁻⁴)
   4. N₀ = 1.2 × 10⁶ nuclei

4. dEvaluate The background in this laboratory corresponds to an activity of 0.50 Bq. A student extends the straight line through these five crosses in order to predict the time at which the corrected activity falls to that value. Evaluate the prediction.

   top of the paper4 marks

   Check answer · 4 marks

   1. Extending the line gives ln 0.50 = −0.69, so t = (6.4 + 0.69)/0.030 = 2.4 × 10² minutes, about four hours
   2. This is roughly six times beyond the last plotted point, so the prediction rests entirely on a single decay constant continuing to hold
   3. That assumption fails if the source contains a second nuclide or if the daughter is itself active, and nothing in the plotted range would reveal either
   4. As the corrected activity approaches the background, it becomes the small difference between two comparable and randomly fluctuating counts, so the prediction could not be tested with any precision anyway

Reading cue. Count what crosses the drawing: one neutron in, two fragments out, nothing else. The free neutrons that carry the chain on are not drawn, so the nucleon numbers as pictured do not balance.

1. aState State the form in which almost all of the 200 MeV appears immediately after the split, and state what the two arrows drawn on the fragments show about the momentum of the system.

   recall2 marks

   Check answer · 2 marks

   1. Almost all appears as kinetic energy of the two fragments, driven apart by the electrostatic repulsion between their positive charges
   2. The arrows point in opposite directions, so the fragments carry equal and opposite momenta and the total momentum stays essentially that of the slow incoming neutron, close to zero

2. bDetermine Determine the energy released in the single event drawn, in joules, and hence determine the number of such events needed each second to sustain a thermal output of 3.2 GW.

   routine4 marks

   Check answer · 4 marks

   1. 200 MeV = 200 × 10⁶ × 1.60 × 10⁻¹⁹ J
   2. = 3.2 × 10⁻¹¹ J
   3. Number per second = 3.2 × 10⁹/(3.2 × 10⁻¹¹)
   4. = 1.0 × 10²⁰ events per second

3. cDetermine The two fragments drawn have nucleon numbers 141 and 92, and about 170 MeV of the energy released appears as their kinetic energy. Determine the kinetic energy carried by each fragment.

   demanding4 marks

   Check answer · 4 marks

   1. Momentum conservation gives the two fragments equal and opposite momenta
   2. Since kinetic energy equals p²/2m, the energy is shared in inverse proportion to mass, so the shares are in the ratio 141 : 92 in favour of the lighter fragment
   3. Lighter fragment, nucleon number 92: 170 × 141/233 = 103 MeV
   4. Heavier fragment, nucleon number 141: 170 × 92/233 = 67 MeV

4. dDiscuss Nothing is drawn leaving the reaction except the two fragments. Discuss what else must leave the nucleus, and what decides whether the single event drawn grows into a self-sustaining chain.

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   Check answer · 4 marks

   1. Nucleon numbers must balance: 235 + 1 = 236, while 141 + 92 = 233, so three free neutrons must also be released
   2. A chain is sustained only if, on average, exactly one neutron from each fission goes on to cause a further fission
   3. The remaining neutrons are lost by escaping through the surface or by being absorbed without causing fission, so the mass and shape of the sample decide the outcome
   4. The released neutrons are fast and are captured by ²³⁵U far less readily until repeated collisions in a moderator have slowed them

Reading cue. Only the 420 nm label fixes the photon energy, the slant of the rays does not. The single drawn arrow is one electron among many, not necessarily the fastest one to leave.

1. aState State the minimum energy that must be given to a single electron for it to leave the surface drawn, and state how that energy reaches the electron.

   recall2 marks

   Check answer · 2 marks

   1. 2.1 eV, the work function labelled on the block
   2. From one photon of the incident light, absorbed whole in a single one-to-one interaction

2. bDetermine The rays drawn represent a beam delivering 1.5 mW to the surface. Determine the number of photons striking the surface each second. Use hc = 1240 eV nm.

   routine4 marks

   Check answer · 4 marks

   1. Photon energy = 1240/420 = 2.95 eV
   2. = 2.95 × 1.60 × 10⁻¹⁹ = 4.72 × 10⁻¹⁹ J
   3. Rate = 1.5 × 10⁻³/(4.72 × 10⁻¹⁹)
   4. = 3.2 × 10¹⁵ photons per second

3. cDetermine The photoelectron drawn leaves at 60° to the surface, carrying the maximum possible kinetic energy. A uniform retarding field of 500 V m⁻¹ is now applied at right angles to the surface. Determine how far from the surface this electron travels before it stops moving away from it.

   demanding4 marks

   Show a hint

   Resolve the velocity into components along and normal to the surface; the field acts on only one of them.

   Check answer · 4 marks

   1. Maximum kinetic energy = 2.95 − 2.1 = 0.85 eV
   2. Only the velocity component normal to the surface is retarded, so the energy to be removed is 0.85 sin²60° = 0.64 eV
   3. Distance = energy in eV divided by the field in V m⁻¹, d = 0.64/500
   4. d = 1.3 × 10⁻³ m (1.3 mm), the electron still moving parallel to the surface at that moment

4. dDiscuss The single electron is drawn leaving the surface at a point some distance along from where the rays land. Discuss how faithful this drawing is to the photon model of the effect.

   top of the paper4 marks

   Check answer · 4 marks

   1. In the photon model one photon is absorbed by one electron and emission follows with no measurable delay, so electrons leave from the illuminated region itself
   2. The drawn separation implies the energy travels along the surface before emission, which the model does not allow and which would introduce a delay that is not observed
   3. Only one arrow is drawn, whereas photoelectrons leave in all directions above the surface
   4. The drawn electron need not carry the maximum kinetic energy either: electrons freed below the surface lose energy on the way out, giving a spread of energies from zero up to 0.85 eV