Key Idea: This topic builds the modern picture of the atom and shows that the small-scale world comes in fixed lumps rather than smooth amounts. An atom is a tiny, dense, positive nucleus (protons + neutrons) with electrons around it; the electrons sit only in discrete energy levels; energy is exchanged in single photons; and charge itself only ever comes in whole multiples of one smallest amount. It is examined on both papers. Paper 1A is quick multiple-choice ā count the particles in a nuclide or ion, read an emission spectrum off an energy-level diagram, pick the longest-wavelength transition, or spot which charge is impossible. Paper 2 is longer structured work ā describe and interpret the alpha-scattering experiment, use E = hf or E = hc/Ī» on a level gap, convert between eV, joules and MeV, or deduce a charge using the fact that it must be a whole-number multiple of e.
āļø The nuclear atom ā counting particles
A nuclide is written $AZ\mathrm{X}$. The top number A is the nucleon number (protons + neutrons); the bottom number Z is the proton number (it fixes the element). From those two you can count every particle. None of these counting rules is printed in the data booklet ā you reason them out.
- nucleon (mass) number ā the top number = protons + neutrons
- proton (atomic) number ā the bottom number; it defines the element
- number of neutrons in the nucleus
- proton number (number of protons)
- the ion's charge in units of e (0 for a neutral atom, +2 for a 2+ ion, ā1 for a 1ā ion)
Alpha-scattering: observation ā conclusion
| What was observed | What it tells us |
|---|---|
| Almost all alpha particles passed straight through, barely deflected | The atom is mostly empty space |
| A small number were deflected through large angles | The positive charge and nearly all the mass sit in a tiny core |
| A very few bounced almost straight back | That core ā the nucleus ā is small, dense and positively charged (it repels the positive alpha) |
Most pass through ā the atom is mostly empty. A few bounce back hard ā all the positive charge and almost all the mass is squeezed into a tiny central nucleus. This replaced the old 'spread-out positive pudding' picture.
š” Energy levels, photons and spectra
An atom holds only a few allowed energies (the levels are quantised). An electron dropping between levels emits a single photon whose energy equals the gap; an electron absorbing a matching photon jumps up. The data booklet links that photon energy to the light two ways:
- energy of the photon ā equals the energy lost (or gained) in the jump (J)
- Planck constant, 6.63 Ć 10ā»Ā³ā“ J s (given)
- frequency of the emitted (or absorbed) light (Hz)
- energy of the photon ā equals the gap between the two levels (J)
- Planck constant, 6.63 Ć 10ā»Ā³ā“ J s (given)
- speed of light, 3.00 Ć 10āø m sā»Ā¹ (given)
- wavelength of the light (m)
- the level the electron starts from (counting the ground state as level 1)
Recap: an atom holds only a few allowed (discrete) energies. An electron DROPS to a lower level and the lost energy leaves as one photon, E = gap = hf = hc/Ī». A bigger gap ā a shorter wavelength; each distinct gap = one bright line in the emission spectrum.
š Interactive diagram
Explore the labelled diagram, charts and maps for this topic in study mode.
Emission vs absorption line spectra
| Emission spectrum | Absorption spectrum | |
|---|---|---|
| What the electron does | Drops to a lower level and gives OUT a photon | Absorbs a photon and jumps UP to a higher level |
| What you see | Bright coloured lines on a dark background | Dark lines missing from a continuous rainbow |
| Where the lines are | One line per allowed energy gap | The SAME gaps are missing ā same atom, same energies |
| Why it's a fingerprint | Each element has its own set of levels ā its own line pattern | Match the pattern to known spectra to identify the element |
š The electronvolt and the quantisation of charge
Atomic and nuclear energies are tiny fractions of a joule, so we measure them in electronvolts. And charge itself comes in fixed lumps ā the same elementary charge e turns up in both ideas.
- the energy expressed in joules (J)
- the same energy expressed in electronvolts (eV)
- the joules in one electronvolt ā the elementary charge e in coulombs
- the total charge on the object (C, coulombs)
- a whole number ā how many elementary charges (extra or missing electrons)
- the elementary charge, e = 1.60 Ć 10ā»Ā¹ā¹ C (given in the data booklet)
The value 1.60 Ć 10ā»Ā¹ā¹ appears twice in this topic ā and it is the same physics. It is the elementary charge e in coulombs, and it is also the joules in one electronvolt, because energy = charge Ć voltage, so one electron (charge e) crossing 1 volt gains e Ć 1 = 1.60 Ć 10ā»Ā¹ā¹ J.
āļø Worked exam-style questions
An ion has 20 protons, 24 neutrons and 18 electrons. (a) State its proton number Z and nucleon number A. (b) Determine its overall charge. (c) Write its nuclide symbol (the element with Z = 20 is calcium, Ca).
š Model answer plan
See the mark-by-mark plan ā for / against / judgement, with marking guidance ā in study mode.
In a sample of atoms, electrons are excited to the third level (n = 3) and then fall back to the ground state by every possible route. (a) How many different wavelengths can appear in the emission spectrum? (b) One bright line comes from the smallest energy drop, 2.9 Ć 10ā»Ā¹ā¹ J. Find its wavelength. (h = 6.63 Ć 10ā»Ā³ā“ J s, c = 3.00 Ć 10āø m sā»Ā¹.)
š Model answer plan
See the mark-by-mark plan ā for / against / judgement, with marking guidance ā in study mode.
An electron in an atom drops from a level at ā3.4 eV to a level at ā13.6 eV, emitting a photon. (a) Find the photon energy in eV. (b) Convert it to joules. (c) Hence find the photon's frequency. (1 eV = 1.60 Ć 10ā»Ā¹ā¹ J, h = 6.63 Ć 10ā»Ā³ā“ J s.)
š Model answer plan
See the mark-by-mark plan ā for / against / judgement, with marking guidance ā in study mode.
In an oil-drop experiment a drop carries a charge of 8.0 Ć 10ā»Ā¹ā¹ C. (a) Deduce how many elementary charges this is. (b) The drop then splits so that one piece carries 4.8 Ć 10ā»Ā¹ā¹ C; deduce the charge on the other piece and check both are allowed. (e = 1.60 Ć 10ā»Ā¹ā¹ C.)
š Model answer plan
See the mark-by-mark plan ā for / against / judgement, with marking guidance ā in study mode.
š§ Quick self-check
Tap each card to reveal the answer.
From $AZ\mathrm{X}$, how do you count protons, neutrons and electrons? Protons = Z (bottom number). Neutrons = A ā Z (top minus bottom). Electrons = Z ā charge. For an ion, only the electron count changes ā the nucleus is untouched.
What did the alpha-scattering experiment show, and how? Most alphas pass straight through ā the atom is mostly empty space. A few bounce back ā a tiny, dense, positive nucleus holds the charge and the mass. This gave the nuclear model.
Which transition gives the LONGEST-wavelength photon? The smallest energy drop. Because E = hc/Ī», a small energy means a large wavelength. (The biggest drop gives the shortest wavelength.)
How many emission lines come from level n down to the ground state? Count the distinct GAPS: n(n ā 1) Ć· 2. So n = 3 ā 3 lines, n = 4 ā 6 lines. Each distinct gap is one bright line.
How do you convert between eV, joules and MeV? eV ā J: MULTIPLY by 1.60 Ć 10ā»Ā¹ā¹. J ā eV: DIVIDE by it. For MeV, convert Ć10ā¶ to plain eV first. E = hf needs joules ā convert before substituting.
Why must N in Q = N e be a whole number? An object charges only by gaining or losing WHOLE electrons, each carrying e. So charge changes only in steps of e ā N = Q Ć· e is always a whole number. Millikan's oil drops proved it.
šÆ Exam tips
Exam Tips
- Counting from a nuclide: protons = Z (bottom), neutrons = A ā Z (top ā bottom), electrons = Z ā charge. When an atom becomes an ion, ONLY the electron count changes ā never adjust the protons or neutrons.
- Paper 2 alpha-scattering: pair every observation with its conclusion ā 'most pass through ā mostly empty space' and 'a few bounce back ā a tiny, dense, positive nucleus'. A bare list of observations loses the interpretation marks.
- Photon energy = the GAP between two levels. Then E = hf links it to frequency and E = hc/Ī» links it to wavelength. A bigger drop gives a bigger E, so a SHORTER wavelength; the smallest drop gives the longest wavelength.
- Count emission lines by counting distinct gaps, not levels: from level n to the ground state there are n(n ā 1) Ć· 2 different wavelengths. Emission = bright lines on dark; absorption = dark lines in a rainbow at the same wavelengths.
- eV ā J: MULTIPLY by 1.60 Ć 10ā»Ā¹ā¹. J ā eV: DIVIDE. Convert keV/MeV to plain eV first (Ć10³ or Ć10ā¶). E = hf and E = hc/Ī» give joules ā convert to eV only at the very end if asked.
- Charge is quantised: every charge is a whole-number multiple of e = 1.60 Ć 10ā»Ā¹ā¹ C, so N = Q Ć· e must come out whole. A fractional N (like 1.5 or 2.5) means the charge is impossible ā that is exactly what Millikan's oil-drop results demonstrated.
- The value 1.60 Ć 10ā»Ā¹ā¹ does double duty: it is the elementary charge e (in coulombs) AND the joules in one electronvolt. Both are given in the data booklet, so you never have to memorise the number.