Collisions of electrons with atoms — AQA A-Level Physics
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Collisions of electrons with atoms is the process where an incident electron transfers energy to a bound atomic electron, causing excitation or ionisation. This underpins the discrete line spectra of atoms and provided key evidence for quantised energy levels in the Bohr model of the atom.
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Collisions of electrons with atoms explained
Collisions of electrons with atoms covers 3 specification points, including Ionisation and excitation; understanding of… and The electron volt..
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Open a point below for its explanation and revision guidance.
Ionisation and excitation; understanding of ionisation and excitation in the fluorescent tube.
This statement requires you to understand ionisation and excitation, and how these processes occur in a fluorescent tube. Ionisation is when an electron gains enough energy to escape from an atom, leaving a positive ion. Excitation is when an electron gains a specific amount of energy and moves to a higher energy level within the atom. In a fluorescent tube, electrons are accelerated by a potential difference and collide with mercury atoms. These collisions can cause excitation of mercury atoms, which then emit ultraviolet photons as they de-excite. The ultraviolet photons are absorbed by the fluorescent coating on the tube, which then emits visible light. Ionisation can also occur, producing free electrons that sustain the discharge. You should be able to describe these processes and explain how they lead to the emission of visible light.
The electron volt.
The electron volt (eV) is a unit of energy convenient at atomic and nuclear scales. It is defined as the work done when an electron, carrying elementary charge e = 1.60 × 10⁻¹⁹ C, is accelerated through a potential difference of 1 V. Since work done W = QV, one electron volt equals 1.60 × 10⁻¹⁹ J. This unit appears whenever single particles gain or lose energy in electric fields, such as electrons colliding with atoms in a discharge tube. To use it, identify the charge in coulombs and the potential difference in volts, multiply, and express the result in eV or J as required. For example, an electron accelerated through 50 V gains 50 eV, equivalent to 50 × 1.60 × 10⁻¹⁹ J = 8.0 × 10⁻¹⁸ J.
Students will be expected to be able to convert eV into J and vice versa.
Converting between electron volts and joules uses the definition 1 eV = 1.60 × 10⁻¹⁹ J. To change eV into J, multiply the number of electron volts by 1.60 × 10⁻¹⁹. To change J into eV, divide the energy in joules by 1.60 × 10⁻¹⁹. For example, 25 eV becomes 25 × 1.60 × 10⁻¹⁹ J = 4.0 × 10⁻¹⁸ J, while 3.2 × 10⁻¹⁷ J becomes 3.2 × 10⁻¹⁷ ÷ 1.60 × 10⁻¹⁹ = 200 eV. This skill is needed whenever collision energies, ionisation energies or photon energies are quoted in different units, so always check which unit the final answer requires and keep powers of ten consistent.
Your focus
- Distinguish between ionisation and excitation in terms of energy and atomic structure.
- Describe the sequence of events in a fluorescent tube that leads to the emission of visible light.
- Explain how collisions between accelerated electrons and mercury atoms result in excitation and ionisation.
Show all 9 objectives
- Define the electron volt in terms of the work done on an electron accelerated through 1 V.
- Derive and recall the equivalence 1 eV = 1.60 × 10⁻¹⁹ J.
- Convert between electron volts and joules accurately in calculations involving atomic-scale energies.
- Convert an energy given in electron volts into joules using the factor 1.60 × 10⁻¹⁹.
- Convert an energy given in joules into electron volts by dividing by 1.60 × 10⁻¹⁹.
- Apply conversions correctly to energies expressed in keV or MeV and present answers with suitable units.
Which command word is this?
A brief, factual answer with no explanation required. For example, 'State what is meant by excitation.' Answer: 'The promotion of an electron to a higher energy level.'
Collisions of electrons with atoms exam tips
Topic Overview
This topic explores what happens when free electrons collide with atoms, focusing on energy transfer and the quantised nature of atomic energy levels. You will learn how inelastic collisions can excite or ionise atoms, and how subsequent de-excitation leads to the emission of photons with discrete energies, explaining atomic line spectra.
Understanding these collisions is crucial because it provides experimental evidence for the Bohr model of the atom and underpins many areas of physics, including spectroscopy, lasers, and quantum mechanics. It also connects to the wider A-Level curriculum, such as the photoelectric effect and energy level diagrams.
Key Concepts
- →Electrons in atoms can only exist at specific discrete energy levels; energy is quantised.
- →An incident electron can excite a bound electron to a higher level only if it transfers an amount of energy exactly equal to the energy level difference.
- →Ionisation occurs when the incident electron transfers enough energy to remove the bound electron from the atom completely.
- →In an inelastic collision, kinetic energy is conserved overall but some is transferred to the atom; in an elastic collision, no energy is transferred to the atom's internal energy.
- →De-excitation of an electron from a higher to a lower energy level results in the emission of a photon with energy E = hf = E_initial - E_final.
Marking Points
- Define ionisation as the removal of an electron from an atom, requiring energy greater than or equal to the ionisation energy.
- Define excitation as the promotion of an electron to a higher energy level within an atom, requiring a specific amount of energy.
- Explain that in a fluorescent tube, accelerated electrons collide with mercury atoms, causing excitation and ionisation.
- Describe how excited mercury atoms emit ultraviolet photons when they de-excite.
- Explain that the fluorescent coating absorbs ultraviolet photons and re-emits visible light.
- States that the electron volt is defined using the work done on an electron accelerated through a potential difference of 1 V.
- Uses W = QV with Q equal to the elementary charge 1.60 × 10⁻¹⁹ C and V equal to 1 V to derive 1 eV = 1.60 × 10⁻¹⁹ J.
- Applies the conversion factor correctly in either direction, multiplying by 1.60 × 10⁻¹⁹ to change eV to J and dividing to change J to eV.
- Recognises that the eV is an energy unit suited to atomic, nuclear and particle contexts, not a unit of potential difference or charge.
- Calculates the energy gained by a charged particle accelerated through a stated potential difference using E = QV and expresses it in eV or J.
- Recalls the conversion factor 1 eV = 1.60 × 10⁻¹⁹ J and applies it in the correct direction.
- Multiplies an energy in eV by 1.60 × 10⁻¹⁹ to obtain the energy in joules.
- Divides an energy in joules by 1.60 × 10⁻¹⁹ to obtain the energy in electron volts.
- Handles powers of ten correctly when converting values such as keV or MeV, for example 1 keV = 1.60 × 10⁻¹⁶ J.
- Presents the converted value with an appropriate unit and a sensible number of significant figures.
Examiner Tips
- 💡Use clear diagrams to show energy levels and transitions when explaining excitation and ionisation.
- 💡Link the processes step by step: acceleration of electrons, collisions, excitation/ionisation, emission of UV, absorption by coating, emission of visible light.
- 💡Remember that ionisation energy is the minimum energy needed to remove an electron from an atom, and it is specific to each element.
- 💡Write the conversion factor as 1 eV = 1.60 × 10⁻¹⁹ J and show the multiplication or division explicitly so the method is visible.
- 💡Check the magnitude of your answer: atomic-scale energies are usually a few eV, which is of order 10⁻¹⁹ J to 10⁻¹⁸ J.
- 💡When a question gives a potential difference, use E = QV with Q as the particle charge, then convert to eV by dividing by 1.60 × 10⁻¹⁹ J eV⁻¹ if needed.
- 💡Write the conversion as a clear equation, such as E(J) = E(eV) × 1.60 × 10⁻¹⁹, before substituting numbers.
- 💡Estimate the expected order of magnitude first so an incorrect power of ten is easy to spot.
- 💡Give the final answer in the unit requested by the question and include the unit symbol to avoid losing credit.
- 💡Always use the term 'exact energy level difference' when describing excitation, and specify that collisions are inelastic when energy is transferred to the atom.
- 💡When calculating photon wavelengths, show clearly that you use E = hf and c = fλ, and convert between eV and J correctly.
- 💡In explanations, structure your answer logically: describe the collision, energy transfer, excitation, de-excitation, and photon emission.
Common Mistakes
- Confusing ionisation with excitation. Correction: ionisation removes an electron completely, while excitation moves an electron to a higher energy level within the atom.
- Thinking that any collision energy can cause excitation. Correction: excitation requires a specific energy equal to the difference between energy levels.
- Believing that the mercury vapour emits visible light directly. Correction: mercury atoms emit ultraviolet photons, which are converted to visible light by the fluorescent coating.
- Treating the electron volt as a unit of potential difference; correct this by stating that the volt is the potential difference unit while the electron volt is an energy unit.
- Using the electron mass instead of its charge in the defining calculation; correct this by using the elementary charge 1.60 × 10⁻¹⁹ C.
- Forgetting to convert eV to J before substituting into an equation that requires SI units; correct this by multiplying the eV value by 1.60 × 10⁻¹⁹ J eV⁻¹ first.
- Multiplying when converting from J to eV; correct this by dividing the joule value by 1.60 × 10⁻¹⁹.
- Using 1.60 × 10⁻¹⁹ as a power of ten adjustment rather than as the charge factor; correct this by treating it as the conversion factor between the two energy units.
- Mixing up keV and eV by a factor of 1000; correct this by converting keV to eV first, then applying the joule conversion.
- Students often think that any amount of energy can be transferred to an atom, leading to continuous spectra. Correction: Only exact energy differences between levels are allowed, so spectra are discrete.
- Confusing excitation with ionisation. Correction: Excitation moves an electron to a higher bound level; ionisation removes it completely from the atom.
- Believing that the incident electron must have exactly the excitation energy to cause excitation. Correction: It must have at least that energy; any excess remains as kinetic energy of the incident electron.
Revision Plan
- 1Day 1-2: Revise the Bohr model and energy level diagrams. Learn the definitions of excitation, ionisation, and de-excitation.
- 2Day 3-4: Study the conditions for excitation and ionisation in electron collisions. Practice calculations involving energy transfer and kinetic energy.
- 3Day 5-6: Learn how line spectra are produced from electron transitions. Practice explaining the origin of emission and absorption spectra.
- 4Day 7-8: Work through past paper questions on this topic, focusing on 4-6 mark explanations and calculations.
- 5Day 9-10: Review common misconceptions and examiner tips. Create flashcards for key definitions and formulas.
Exam Question Types
- 📋Calculation of energy required for excitation or ionisation, often combined with kinetic energy of the incident electron. Advice: Convert units carefully and use conservation of energy.
- 📋Explanation of why a line spectrum is produced rather than a continuous spectrum. Advice: Link discrete energy levels to specific photon energies.
- 📋Determination of whether a collision is elastic or inelastic based on energy transfer. Advice: Compare incident electron energy to energy level differences.
- 📋Calculation of photon wavelength or frequency from an energy level transition. Advice: Use E = hf and c = fλ, and ensure correct unit conversions.
Command Word Expectations (AQA)
A brief, factual answer with no explanation required. For example, 'State what is meant by excitation.' Answer: 'The promotion of an electron to a higher energy level.'
A detailed answer linking cause and effect, using correct terminology. For example, 'Explain why only certain wavelengths are emitted.' Must include reference to discrete energy levels and photon energy.
Show all working, include the formula, substitute values with units, and give the final answer with appropriate significant figures and units.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: An electron is accelerated through a potential difference of 12.0 V and collides with a hydrogen atom in its ground state. The ground state energy of hydrogen is -13.6 eV. Determine whether the electron can excite the atom to the first excited state at -3.40 eV. If so, calculate the speed of the electron after the collision.
- 1.Step 1: Calculate the kinetic energy gained by the incident electron: E = eV = 1.60e-19 C x 12.0 V = 1.92e-18 J. Convert to eV: 12.0 eV.
- 2.Step 2: Calculate the energy required for excitation: ΔE = (-3.40 eV) - (-13.6 eV) = 10.2 eV. Since 12.0 eV > 10.2 eV, excitation is possible.
- 3.Step 3: After excitation, the electron retains kinetic energy: 12.0 eV - 10.2 eV = 1.8 eV. Convert to joules: 1.8 eV x 1.60e-19 J/eV = 2.88e-19 J.
- 4.Step 4: Use KE = 1/2 mv^2 to find speed: v = sqrt(2KE/m) = sqrt(2 x 2.88e-19 / 9.11e-31) = 7.95e5 m/s.
Question: Explain, in terms of electron collisions, why a gas discharge tube containing hydrogen emits light only at specific wavelengths.
- 1.Step 1: In the tube, free electrons are accelerated by the electric field and collide with hydrogen atoms.
- 2.Step 2: These collisions can transfer energy to bound electrons, but only if the energy transferred equals the exact difference between two atomic energy levels. This excites the bound electron to a higher level.
- 3.Step 3: The excited electron is unstable and de-excites, emitting a photon. The photon energy equals the energy level difference: E = hf = hc/λ.
- 4.Step 4: Since energy levels are discrete, only specific energy differences are possible, so only specific photon energies (and hence wavelengths) are emitted, producing a line spectrum.