IB Physics Topic 5.1: Structure of the atom | Notes & Flashcards | Aimnova

IB Physics — Structure of the atom

Topic 5.1 of IB Physics covers Structure of the atom, which is part of Unit 5: Nuclear and quantum physics. Students explore key concepts including Nuclear model and atomic structure, Energy levels and atomic spectra, The electronvolt, Quantisation of charge. A strong understanding of structure of the atom is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Structure of the atom

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 — 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 — describe and interpret the alpha-scattering experiment, use E = hf or E = hc/λ 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.

⚛️ The nuclear atom — 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 — you reason them out.

N = A - Z

Number of neutrons = nucleon number − proton number (top minus bottom). NOT a booklet equation — you derive it.

$A$: nucleon (mass) number — the top number = protons + neutrons
$Z$: proton (atomic) number — the bottom number; it defines the element
$N$: number of neutrons in the nucleus

\text{electrons} = Z - q

Electrons = proton number − charge. A neutral atom has q = 0 (electrons = Z). A positive ion has LOST electrons (fewer than Z); a negative ion has gained them. Only the electron count changes when an atom becomes an ion — the protons and neutrons are untouched.

$Z$: proton number (number of protons)
$q$: the ion's charge in units of e (0 for a neutral atom, +2 for a 2+ ion, −1 for a 1− ion)

Most pass through ⇒ the atom is mostly empty. A few bounce back hard ⇒ 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.

💡 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:

E = hf

Photon energy from frequency (given). Use this when a frequency is involved. E comes out in joules.

$E$: energy of the photon — equals the energy lost (or gained) in the jump (J)
$h$: Planck constant, 6.63 × 10⁻³⁴ J s (given)
$f$: frequency of the emitted (or absorbed) light (Hz)

E = \frac{hc}{\lambda}

Photon energy from wavelength (given). Bigger gap → bigger E → SHORTER wavelength. The smallest drop makes the longest-wavelength line.

$E$: energy of the photon — equals the gap between the two levels (J)
$h$: Planck constant, 6.63 × 10⁻³⁴ J s (given)
$c$: speed of light, 3.00 × 10⁸ m s⁻¹ (given)
$\lambda$: wavelength of the light (m)

\text{lines} = \frac{n(n-1)}{2}

Number of distinct emission wavelengths when an electron falls from level n to the ground state by every route — count the distinct GAPS, not the levels. NOT in the booklet; e.g. n = 3 → 3 lines, n = 4 → 6 lines. (You can also just list the jumps.)

$n$: the level the electron starts from (counting the ground state as level 1)

[Diagram: phys-energy-levels]

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/λ. A bigger gap → a shorter wavelength; each distinct gap = one bright line in the emission spectrum.

🔋 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 — the same elementary charge e turns up in both ideas.

E_{\text{(J)}} = E_{\text{(eV)}} \times 1.60\times10^{-19}

The eV ↔ J conversion (given on the data booklet's unit-conversions page). One eV is the energy an electron gains across 1 V. eV → J MULTIPLY by 1.60 × 10⁻¹⁹; J → eV DIVIDE. (1 keV = 10³ eV, 1 MeV = 10⁶ eV.)

$E_{\text{(J)}}$: the energy expressed in joules (J)
$E_{\text{(eV)}}$: the same energy expressed in electronvolts (eV)
$1.60\times10^{-19}$: the joules in one electronvolt — the elementary charge e in coulombs

Q = N e

Charge is quantised: every charge is a whole number N of elementary charges. NOT a separate booklet line, but e = 1.60 × 10⁻¹⁹ C IS given. To count electrons, rearrange to N = Q ÷ e — and the answer must be a WHOLE number.

$Q$: the total charge on the object (C, coulombs)
$N$: a whole number — how many elementary charges (extra or missing electrons)
$e$: the elementary charge, e = 1.60 × 10⁻¹⁹ C (given in the data booklet)

The value 1.60 × 10⁻¹⁹ appears twice in this topic — 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 × voltage, so one electron (charge e) crossing 1 volt gains e × 1 = 1.60 × 10⁻¹⁹ J.

✏️ Worked exam-style questions

IB-style question — count the particles, then write the symbol

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).

Solution:

(a) The proton number is the number of protons; the nucleon number is protons + neutrons:
$Z = 20, \qquad A = 20 + 24 = 44$
(b) Charge (in units of e) = protons − electrons. Fewer electrons than protons ⇒ positive:
$q = 20 - 18 = +2$
(c) Build the symbol $AZ\mathrm{X}$ with the charge as a superscript:
${}^{44}_{20}\mathrm{Ca}^{2+}$

Final answer:

(a) Z = 20, A = 44. (b) charge +2 (a 2+ ion). (c) ⁴⁴₂₀Ca²⁺. Only the electron count differs from a neutral atom — the 2 missing electrons make it a 2+ ion; the nucleus (protons and neutrons) is untouched.

IB-style question — count the lines, then the longest-wavelength photon

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 × 10⁻¹⁹ J. Find its wavelength. (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹.)

Solution:

(a) Each distinct gap gives one line. Use the counting rule (or list 3→2, 3→1, 2→1):
$\text{lines} = \frac{n(n-1)}{2} = \frac{3(2)}{2} = 3$
(b) The longest wavelength is the smallest drop. Start from the given formula and rearrange for λ:
$\lambda = \frac{hc}{E}$
(b) Substitute (h = 6.63 × 10⁻³⁴, c = 3.00 × 10⁸, E = 2.9 × 10⁻¹⁹):
$\lambda = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{2.9\times10^{-19}}$
(b) Evaluate — keep the unit:
$\lambda = 6.9\times10^{-7}\ \text{m}$

Final answer:

(a) 3 different wavelengths (3→2, 3→1, 2→1). (b) λ ≈ 6.9 × 10⁻⁷ m (about 690 nm, red light). The longest-wavelength line always comes from the SMALLEST energy gap, because E = hc/λ.

IB-style question — a transition energy in electronvolts

An electron in an atom drops from a level at −3.4 eV to a level at −13.6 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 × 10⁻¹⁹ J, h = 6.63 × 10⁻³⁴ J s.)

Solution:

(a) The photon energy is the SIZE of the gap between the two levels:
$E = (-3.4) - (-13.6) = 10.2\ \text{eV}$
(b) Going eV → J, MULTIPLY by the given conversion:
$E = 10.2 \times 1.60\times10^{-19} = 1.63\times10^{-18}\ \text{J}$
(c) E = hf works in joules. Rearrange and substitute:
$f = \frac{E}{h} = \frac{1.63\times10^{-18}}{6.63\times10^{-34}}$
(c) Evaluate — keep the unit:
$f = 2.5\times10^{15}\ \text{Hz}$

Final answer:

(a) E = 10.2 eV. (b) E ≈ 1.63 × 10⁻¹⁸ J. (c) f ≈ 2.5 × 10¹⁵ Hz (ultraviolet). Convert the eV to joules BEFORE using E = hf — that equation only works in joules. eV → J means multiply by 1.60 × 10⁻¹⁹.

IB-style question — quantisation of charge (Millikan)

In an oil-drop experiment a drop carries a charge of 8.0 × 10⁻¹⁹ C. (a) Deduce how many elementary charges this is. (b) The drop then splits so that one piece carries 4.8 × 10⁻¹⁹ C; deduce the charge on the other piece and check both are allowed. (e = 1.60 × 10⁻¹⁹ C.)

Solution:

(a) A real charge is a whole number of elementary charges. Use N = Q ÷ e:
$N = \frac{Q}{e} = \frac{8.0\times10^{-19}}{1.60\times10^{-19}} = 5$
(b) Charge is conserved, so the other piece carries the rest:
$Q_{2} = 8.0\times10^{-19} - 4.8\times10^{-19} = 3.2\times10^{-19}\ \text{C}$
(b) Check each piece is a whole number of elementary charges:
$\frac{4.8\times10^{-19}}{1.60\times10^{-19}} = 3, \qquad \frac{3.2\times10^{-19}}{1.60\times10^{-19}} = 2$
(b) Both N are whole numbers, so both charges are allowed (and 3 + 2 = 5, conserving the original 5e).
$Q_{2} = 2e \;\checkmark$

Final answer:

(a) N = 5 (a charge of 5e). (b) the other piece carries 3.2 × 10⁻¹⁹ C = 2e; the first is 3e. Both are whole multiples of e, so both are allowed, and 3e + 2e = 5e conserves the charge. A non-whole N would mean an impossible charge — that is exactly what Millikan's whole-number results proved about charge.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Counting from a nuclide: protons = Z (bottom), neutrons = A − Z (top − bottom), electrons = Z − charge. When an atom becomes an ion, ONLY the electron count changes — never adjust the protons or neutrons.
Paper 2 alpha-scattering: pair every observation with its conclusion — 'most pass through ⇒ mostly empty space' and 'a few bounce back ⇒ 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/λ 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 − 1) ÷ 2 different wavelengths. Emission = bright lines on dark; absorption = dark lines in a rainbow at the same wavelengths.
eV → J: MULTIPLY by 1.60 × 10⁻¹⁹. J → eV: DIVIDE. Convert keV/MeV to plain eV first (×10³ or ×10⁶). E = hf and E = hc/λ give joules — convert to eV only at the very end if asked.
Charge is quantised: every charge is a whole-number multiple of e = 1.60 × 10⁻¹⁹ C, so N = Q ÷ e must come out whole. A fractional N (like 1.5 or 2.5) means the charge is impossible — that is exactly what Millikan's oil-drop results demonstrated.
The value 1.60 × 10⁻¹⁹ 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.

IB Physics — Structure of the atom

Key concepts in Structure of the atom

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 — 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 — describe and interpret the alpha-scattering experiment, use E = hf or E = hc/λ 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.

⚛️ The nuclear atom — counting particles

N = A - Z

Number of neutrons = nucleon number − proton number (top minus bottom). NOT a booklet equation — you derive it.

$A$: nucleon (mass) number — the top number = protons + neutrons
$Z$: proton (atomic) number — the bottom number; it defines the element
$N$: number of neutrons in the nucleus

\text{electrons} = Z - q

$Z$: proton number (number of protons)
$q$: the ion's charge in units of e (0 for a neutral atom, +2 for a 2+ ion, −1 for a 1− ion)

Most pass through ⇒ the atom is mostly empty. A few bounce back hard ⇒ 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.

💡 Energy levels, photons and spectra

E = hf

Photon energy from frequency (given). Use this when a frequency is involved. E comes out in joules.

$E$: energy of the photon — equals the energy lost (or gained) in the jump (J)
$h$: Planck constant, 6.63 × 10⁻³⁴ J s (given)
$f$: frequency of the emitted (or absorbed) light (Hz)

E = \frac{hc}{\lambda}

Photon energy from wavelength (given). Bigger gap → bigger E → SHORTER wavelength. The smallest drop makes the longest-wavelength line.

$E$: energy of the photon — equals the gap between the two levels (J)
$h$: Planck constant, 6.63 × 10⁻³⁴ J s (given)
$c$: speed of light, 3.00 × 10⁸ m s⁻¹ (given)
$\lambda$: wavelength of the light (m)

\text{lines} = \frac{n(n-1)}{2}

$n$: the level the electron starts from (counting the ground state as level 1)

[Diagram: phys-energy-levels]

🔋 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 — the same elementary charge e turns up in both ideas.

E_{\text{(J)}} = E_{\text{(eV)}} \times 1.60\times10^{-19}

$E_{\text{(J)}}$: the energy expressed in joules (J)
$E_{\text{(eV)}}$: the same energy expressed in electronvolts (eV)
$1.60\times10^{-19}$: the joules in one electronvolt — the elementary charge e in coulombs

Q = N e

$Q$: the total charge on the object (C, coulombs)
$N$: a whole number — how many elementary charges (extra or missing electrons)
$e$: the elementary charge, e = 1.60 × 10⁻¹⁹ C (given in the data booklet)

The value 1.60 × 10⁻¹⁹ appears twice in this topic — 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 × voltage, so one electron (charge e) crossing 1 volt gains e × 1 = 1.60 × 10⁻¹⁹ J.

✏️ Worked exam-style questions

IB-style question — count the particles, then write the symbol

Solution:

(a) The proton number is the number of protons; the nucleon number is protons + neutrons:
$Z = 20, \qquad A = 20 + 24 = 44$
(b) Charge (in units of e) = protons − electrons. Fewer electrons than protons ⇒ positive:
$q = 20 - 18 = +2$
(c) Build the symbol $AZ\mathrm{X}$ with the charge as a superscript:
${}^{44}_{20}\mathrm{Ca}^{2+}$

Final answer:

IB-style question — count the lines, then the longest-wavelength photon

Solution:

(a) Each distinct gap gives one line. Use the counting rule (or list 3→2, 3→1, 2→1):
$\text{lines} = \frac{n(n-1)}{2} = \frac{3(2)}{2} = 3$
(b) The longest wavelength is the smallest drop. Start from the given formula and rearrange for λ:
$\lambda = \frac{hc}{E}$
(b) Substitute (h = 6.63 × 10⁻³⁴, c = 3.00 × 10⁸, E = 2.9 × 10⁻¹⁹):
$\lambda = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{2.9\times10^{-19}}$
(b) Evaluate — keep the unit:
$\lambda = 6.9\times10^{-7}\ \text{m}$

Final answer:

(a) 3 different wavelengths (3→2, 3→1, 2→1). (b) λ ≈ 6.9 × 10⁻⁷ m (about 690 nm, red light). The longest-wavelength line always comes from the SMALLEST energy gap, because E = hc/λ.

IB-style question — a transition energy in electronvolts

Solution:

(a) The photon energy is the SIZE of the gap between the two levels:
$E = (-3.4) - (-13.6) = 10.2\ \text{eV}$
(b) Going eV → J, MULTIPLY by the given conversion:
$E = 10.2 \times 1.60\times10^{-19} = 1.63\times10^{-18}\ \text{J}$
(c) E = hf works in joules. Rearrange and substitute:
$f = \frac{E}{h} = \frac{1.63\times10^{-18}}{6.63\times10^{-34}}$
(c) Evaluate — keep the unit:
$f = 2.5\times10^{15}\ \text{Hz}$

Final answer:

IB-style question — quantisation of charge (Millikan)

Solution:

(a) A real charge is a whole number of elementary charges. Use N = Q ÷ e:
$N = \frac{Q}{e} = \frac{8.0\times10^{-19}}{1.60\times10^{-19}} = 5$
(b) Charge is conserved, so the other piece carries the rest:
$Q_{2} = 8.0\times10^{-19} - 4.8\times10^{-19} = 3.2\times10^{-19}\ \text{C}$
(b) Check each piece is a whole number of elementary charges:
$\frac{4.8\times10^{-19}}{1.60\times10^{-19}} = 3, \qquad \frac{3.2\times10^{-19}}{1.60\times10^{-19}} = 2$
(b) Both N are whole numbers, so both charges are allowed (and 3 + 2 = 5, conserving the original 5e).
$Q_{2} = 2e \;\checkmark$

Final answer:

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Counting from a nuclide: protons = Z (bottom), neutrons = A − Z (top − bottom), electrons = Z − charge. When an atom becomes an ion, ONLY the electron count changes — never adjust the protons or neutrons.
Paper 2 alpha-scattering: pair every observation with its conclusion — 'most pass through ⇒ mostly empty space' and 'a few bounce back ⇒ 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/λ 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 − 1) ÷ 2 different wavelengths. Emission = bright lines on dark; absorption = dark lines in a rainbow at the same wavelengths.
eV → J: MULTIPLY by 1.60 × 10⁻¹⁹. J → eV: DIVIDE. Convert keV/MeV to plain eV first (×10³ or ×10⁶). E = hf and E = hc/λ give joules — convert to eV only at the very end if asked.
Charge is quantised: every charge is a whole-number multiple of e = 1.60 × 10⁻¹⁹ C, so N = Q ÷ e must come out whole. A fractional N (like 1.5 or 2.5) means the charge is impossible — that is exactly what Millikan's oil-drop results demonstrated.
The value 1.60 × 10⁻¹⁹ 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.

Key concepts in Structure of the atom

⚛️ The nuclear atom — counting particles

💡 Energy levels, photons and spectra

🔋 The electronvolt and the quantisation of charge

✏️ Worked exam-style questions

IB-style question — count the particles, then write the symbol

IB-style question — count the lines, then the longest-wavelength photon

IB-style question — a transition energy in electronvolts

IB-style question — quantisation of charge (Millikan)

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 5.1

Study resources — 5.1 Structure of the atom

Nuclear model and atomic structure

Energy levels and atomic spectra

The electronvolt

Quantisation of charge

Ready to study Structure of the atom?

Ready to practice?

Key concepts in Structure of the atom

⚛️ The nuclear atom — counting particles

💡 Energy levels, photons and spectra

🔋 The electronvolt and the quantisation of charge

✏️ Worked exam-style questions

IB-style question — count the particles, then write the symbol

IB-style question — count the lines, then the longest-wavelength photon

IB-style question — a transition energy in electronvolts

IB-style question — quantisation of charge (Millikan)

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 5.1

Study resources — 5.1 Structure of the atom

Nuclear model and atomic structure

Energy levels and atomic spectra

The electronvolt

Quantisation of charge

Ready to study Structure of the atom?

Ready to practice?