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Structure of the atom

Physics HL Chapter Notes

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Structure of the atomQuantum physicsRadioactive decayFissionFusion and stars
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1. The Nuclear Model of the Atom

Before 1911, the accepted model of the atom was J.J. Thomson's 'plum pudding' model, which proposed that an atom was a sphere of positive charge with negative electrons embedded within it. This was overturned by the Geiger-Marsden experiment, famously interpreted by Ernest Rutherford. In this experiment, a narrow beam of positively charged alpha particles was fired at a very thin sheet of gold foil inside a vacuum. A detector was moved around the foil to count the number of alpha particles scattered at different angles. The results were startling:

  1. Most alpha particles passed straight through the foil with no deflection. This implied that the atom is mostly empty space.
  2. A small number of particles were deflected by small angles. This suggested the presence of a concentration of positive charge that repelled the alpha particles.
  3. A very tiny fraction (about 1 in 8000) were deflected by large angles, some even 'bouncing back' (scattered by more than 90°). This could only happen if the atom's positive charge and mass were concentrated in a very small, dense region, which Rutherford called the nucleus.

These observations led to the Rutherford nuclear model: a dense, massive, positively charged nucleus at the centre, with tiny, negatively charged electrons orbiting it at a relatively large distance. The electrostatic force of attraction between the positive nucleus and negative electrons keeps the atom together.

Key term

Nucleus: The small, dense, positively charged region at the centre of an atom, containing almost all of the atom's mass in the form of protons and neutrons.

Examiner insight

Examiners expect you to clearly link each specific observation from the experiment to the conclusion it supports about atomic structure. For example, 'most passed through' -> 'atom is mostly empty space'.

Common pitfall

Simply stating that alpha particles 'hit' the nucleus and bounced back. It is the electrostatic repulsion between the positive alpha particle and the positive nucleus that causes the scattering, they do not need to physically collide.

Fun fact

If an atom were the size of a cathedral, the nucleus would be about the size of a fly buzzing around the centre.

Worked example 16 marks

Explain how the results of the Geiger-Marsden alpha scattering experiment led to the rejection of the 'plum pudding' model and the development of the nuclear model of the atom.

  1. 1
    1. State the 'plum pudding' model's prediction: In this model, charge and mass are evenly distributed. It predicted that all alpha particles would pass through with only very minor deflections.
  2. 2
    1. Describe the key experimental observation that contradicted this: A very small number of alpha particles were scattered through large angles (>90°).
  3. 3
    1. Explain the conclusion from this observation: Such a large deflection requires a very strong repulsive force. This could only occur if the atom's positive charge and mass were concentrated in a tiny, dense core (the nucleus).
  4. 4
    1. Describe the second key observation: Most alpha particles passed straight through undeflected.
  5. 5
    1. Explain the conclusion from this second observation: This implies that the atom must be mostly empty space, with the nucleus being extremely small compared to the overall size of the atom.
  6. 6
    1. Conclude: These results were incompatible with the plum pudding model but were perfectly explained by the new nuclear model, which has a small, dense, positive nucleus orbited by electrons.

Recap

  • The Geiger-Marsden experiment fired alpha particles at thin gold foil.
  • Most alpha particles passed straight through, showing the atom is mostly empty space.
  • A few particles were deflected, and a tiny number bounced back, revealing a small, dense, positive nucleus.
  • The experiment replaced the 'plum pudding' model with the Rutherford nuclear model.
  • The electrostatic force holds the negative electrons in orbit around the positive nucleus.

Quick check

  1. State the three main observations from the alpha particle scattering experiment.3 marks
  2. What is the charge of the nucleus and what is the charge of an alpha particle?2 marks

2. Atomic Spectra and Discrete Energy Levels

When a gas at low pressure is heated or has a high voltage applied across it, it emits light. If this light is passed through a prism or diffraction grating, it doesn't form a continuous rainbow. Instead, it produces a series of sharp, bright lines of specific colours on a dark background. This is called a line emission spectrum. Conversely, if white light (containing all wavelengths) is passed through a cool sample of the same gas, the resulting spectrum is a continuous rainbow but with dark lines at exactly the same positions as the bright lines in the emission spectrum. This is a line absorption spectrum. These spectra provide direct evidence that electrons in an atom cannot have just any amount of energy. They are restricted to specific, fixed energy values called discrete energy levels. An electron is normally in the lowest possible energy level, known as the ground state. If the atom absorbs energy (e.g., from heat or a collision), an electron can jump up to a higher energy level, an excited state. This state is unstable, and the electron will quickly fall back down to a lower level, emitting the extra energy as a single packet of light, a photon. Because the energy levels are discrete, the energy difference between any two levels is also a fixed value, so the emitted photons have specific energies, corresponding to the specific frequencies and wavelengths seen in the line spectra.

Key term

Discrete Energy Levels: Specific, allowed energy values that an electron in an atom can possess, between which it can transition by absorbing or emitting photons.

Examiner insight

Marks are often awarded for clearly explaining that each line in a spectrum corresponds to a unique electron transition between two specific energy levels.

Common pitfall

Confusing emission spectra (produced by hot, excited gases) with absorption spectra (produced by cool gases absorbing light). Remember: Emission is 'giving out' light, Absorption is 'taking in' light.

Worked example 15 marks

Explain how line emission spectra provide evidence for the existence of discrete energy levels in atoms.

  1. 1
    1. Define a line emission spectrum: It consists of a series of discrete bright lines of specific wavelengths/colours against a dark background.
  2. 2
    1. Explain the origin of the light: The light is emitted by atoms in a gas that have been excited (e.g., by heating).
  3. 3
    1. Link light emission to electron transitions: This light is emitted when excited electrons in the atoms fall from a higher energy level to a lower energy level.
  4. 4
    1. Explain the role of photons: During this transition, a photon is emitted with energy exactly equal to the energy difference between the two levels (ΔE = E_higher - E_lower).
  5. 5
    1. Connect photon energy to wavelength: The energy of the photon determines its wavelength (and thus its colour), via the relation E = hc/λ.
  6. 6
    1. Make the final link: Since only specific wavelengths are observed, it means only photons of specific energies are being emitted. This implies that the energy differences between electron orbits are fixed, and therefore the energy levels themselves must be discrete and not continuous.

Recap

  • A line emission spectrum has bright lines on a dark background, created by excited atoms emitting photons.
  • A line absorption spectrum has dark lines on a continuous background, created when atoms absorb photons from white light.
  • The lines in both spectra for a given element occur at the same wavelengths.
  • These spectra are evidence that electrons can only exist in discrete energy levels.
  • The ground state is the lowest energy level, while higher levels are called excited states.

Quick check

  1. What is the difference between the ground state and an excited state of an atom?2 marks

3. Atomic Transitions and Photon Calculations

When an electron moves from a higher energy level (E_initial) to a lower energy level (E_final), the atom de-excites and emits a photon. The energy of this photon (E_photon) is precisely the difference between the two energy levels: ΔE = E_initial - E_final. This energy is related to the photon's frequency(f) and wavelength (λ) by the Planck-Einstein relation and the wave speed equation. Energy levels are often given in electron-volts (eV), a convenient unit for atomic physics. 1 eV is the energy gained by an electron when accelerated through a potential difference of 1 volt. When performing calculations using constants like Planck's constant(h) in SI units (Joule-seconds), you must convert energies from eV to Joules (J) using the conversion: 1 eV = 1.60 x 10⁻¹⁹ J.

ΔE = E_initial - E_final

E = hf

c = fλ

E = hc/λ

Key term

Photon: A discrete quantum or packet of electromagnetic energy, which has particle-like properties.

Examiner insight

Examiners look for clear, step-by-step working. Write down the formula you are using, show the substitution of values, and give the final answer with the correct units.

Common pitfall

Forgetting to convert energy from electron-volts (eV) to Joules (J) before using Planck's constant. This is the most common source of calculation errors.

Worked example 14 marks

An electron in a hydrogen atom transitions from the n=3 energy level (-1.51 eV) to the n=2 energy level (-3.40 eV). Calculate the wavelength of the emitted photon. (h = 6.63 x 10⁻³⁴ Js, c = 3.00 x 10⁸ m/s, 1 eV = 1.60 x 10⁻¹⁹ J)

  1. 1
    1. Calculate the energy difference (ΔE) in eV: ΔE = E_initial - E_final = (-1.51 eV) - (-3.40 eV) = 1.89 eV.
  2. 2
    1. Convert the energy from eV to Joules (J): ΔE = 1.89 eV * (1.60 x 10⁻¹⁹ J/eV) = 3.024 x 10⁻¹⁹ J.
  3. 3
    1. State the formula relating energy and wavelength: E = hc/λ.
  4. 4
    1. Rearrange the formula to solve for wavelength (λ): λ = hc/E.
  5. 5
    1. Substitute the values and calculate λ: λ = (6.63 x 10⁻³⁴ Js * 3.00 x 10⁸ m/s) / (3.024 x 10⁻¹⁹ J) = 6.58 x 10⁻⁷ m.
  6. 6
    1. State the final answer with appropriate units, often in nanometres: λ = 658 nm.

Worked example 23 marks

A mercury lamp emits blue light with a wavelength of 436 nm. What is the energy of a single photon of this light, in electron-volts?

  1. 1
    1. State the formula for photon energy: E = hc/λ.
  2. 2
    1. Substitute the values in SI units (remembering 1 nm = 1 x 10⁻⁹ m): E = (6.63 x 10⁻³⁴ * 3.00 x 10⁸) / (436 x 10⁻⁹) = 4.56 x 10⁻¹⁹ J.
  3. 3
    1. Convert the energy from Joules to electron-volts: E = (4.56 x 10⁻¹⁹ J) / (1.60 x 10⁻¹⁹ J/eV) = 2.85 eV.

Recap

  • An electron transition to a lower energy level emits a photon.
  • The photon's energy equals the energy difference between the levels: ΔE = E_initial - E_final.
  • Photon energy is related to frequency by E = hf.
  • Photon energy is related to wavelength by E = hc/λ.
  • Always convert energy values from eV to Joules for calculations with h in SI units.

Quick check

  1. An electron falls between two energy levels, releasing a photon of energy 2.1 eV. What is the energy difference between the levels in Joules?2 marks

4. The Bohr Model and Quantised Angular Momentum

The Rutherford model was revolutionary, but it had a major flaw. According to classical physics, an orbiting electron is accelerating, and an accelerating charge should continuously radiate electromagnetic energy. This would cause the electron to lose energy and spiral into the nucleus in a fraction of a second, meaning atoms should not be stable. In 1913, Niels Bohr proposed a radical solution for the hydrogen atom that combined classical and early quantum ideas. He made three key postulates:

  1. Electrons exist in 'stationary states' (the discrete energy levels) and do not radiate energy while in these stable orbits.
  2. An electron can 'jump' between stationary states by absorbing or emitting a photon whose energy is equal to the energy difference between the states.
  3. The angular momentum (L) of an electron in a stationary state is quantised. It can only take on integer multiples of a fundamental value, h/2π. The formula is L = mvr = n(h/2π), where 'n' is an integer called the principal quantum number (n = 1, 2, 3, ...), 'm' is the electron mass, 'v' is its speed, 'r' is the orbit radius, and 'h' is Planck's constant.

This third postulate is the most fundamental. By restricting the angular momentum to discrete values, it automatically restricts the possible orbit radii and energy levels to discrete values as well. This quantisation condition provided the theoretical reason *why* the energy levels were discrete and explained why atoms were stable.

L = mvr = n(h/2π)

Key term

Quantisation: The principle that a physical quantity, such as energy or angular momentum, can only have certain discrete values rather than a continuous range of values.

Examiner insight

You are not expected to perform complex calculations with the angular momentum formula, but you must understand its significance: that quantised angular momentum is the underlying reason for discrete energy levels and atomic stability in the Bohr model.

Common pitfall

Thinking that Bohr's model is the final, correct model of the atom. It was a brilliant but ultimately limited model that works well for hydrogen but fails for more complex atoms. It has been superseded by modern quantum mechanics.

Fun fact

The quantity h/2π appears so often in quantum mechanics that it is given its own symbol, ħ, called 'h-bar'. So Bohr's condition is often written as L = nħ.

Worked example 14 marks

Explain how Niels Bohr's model of the atom addressed the instability of the Rutherford model.

  1. 1
    1. Identify the problem with the Rutherford model: According to classical electromagnetic theory, an orbiting electron is accelerating and should continuously radiate energy.
  2. 2
    1. State the consequence of this radiation: This energy loss would cause the electron to spiral into the nucleus, making the atom unstable.
  3. 3
    1. Introduce Bohr's first postulate: Bohr proposed that electrons exist in stable, stationary orbits (or energy levels) where they do not radiate energy, despite accelerating.
  4. 4
    1. Introduce Bohr's third postulate: He further proposed that these stable orbits were only allowed if the electron's angular momentum was an integer multiple of h/2π (quantised angular momentum).
  5. 5
    1. Conclude: This condition of quantised angular momentum is the fundamental reason for the existence of discrete, stable energy levels. By forbidding continuous energy radiation, Bohr's model explained why atoms are stable and do not collapse.

Recap

  • The classical Rutherford model was unstable because orbiting electrons should radiate energy and spiral inwards.
  • Bohr proposed that electrons exist in stable 'stationary states' without radiating energy.
  • Bohr's key idea was that the angular momentum of an electron is quantised: L = n(h/2π).
  • This quantisation of angular momentum is the reason for the existence of discrete energy levels.
  • Bohr's model successfully explained the stability of atoms and the line spectrum of hydrogen.

Quick check

  1. What physical quantity did Bohr propose was quantised, leading to the existence of stable energy levels?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Describe two key observations from the Geiger-Marsden alpha scattering experiment and state the conclusion drawn from each one.4 marks
  2. An electron in a hydrogen atom de-excites from the n=3 state (E = -1.51 eV) to the ground state (n=1, E = -13.6 eV). Calculate the energy of the emitted photon in Joules. (1 eV = 1.60 x 10⁻¹⁹ J)3 marks
  3. Explain the difference between a line emission spectrum and a line absorption spectrum, relating both to the concept of discrete energy levels in atoms.4 marks
  4. A photon of wavelength 486 nm is emitted from a hydrogen atom. Calculate its energy in electron-volts. (h = 6.63 x 10⁻³⁴ Js, c = 3.00 x 10⁸ m/s, e = 1.60 x 10⁻¹⁹ C)4 marks
  5. Explain why the alpha-scattering experiment had to be performed in an evacuated container.2 marks
  6. In the Bohr model of the hydrogen atom, the energy levels are given by E_n = -13.6/n² eV. An electron transitions from the n=4 level to the n=2 level. Calculate the wavelength of the light emitted.5 marks
  7. Explain how Bohr's postulate of quantised angular momentum solved the main theoretical problem with Rutherford's nuclear model of the atom.4 marks
  8. A beam of electrons, each with kinetic energy 12.8 eV, is fired through a gas of hydrogen atoms in their ground state (E₁ = -13.6 eV). The energy levels are given by E_n = -13.6/n² eV. Determine the highest energy level (n) to which a hydrogen atom can be excited by one of these electrons, and find the kinetic energy of the electron after the collision.5 marks
  9. The gold foil used in the Geiger-Marsden experiment was extremely thin, only a few hundred atoms thick. Explain why this was necessary for the experiment's conclusions to be valid.3 marks
  10. A hypothetical atom has the energy levels shown: Ground state = -12.0 eV, First excited state = -6.0 eV, Second excited state = -4.0 eV, Ionisation level = 0 eV. a) What is the ionisation energy of this atom? b) The atom is in the second excited state. Calculate all the possible wavelengths of photons that can be emitted as it de-excites back to the ground state.6 marks

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