Modern Physics

Atomic Structure and Spectra

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to describe the Rutherford model of the atom and its limitations.
  • By the end of this lesson students will be able to outline the postulates of the Bohr model of the atom and explain how it addresses Rutherford's limitations.
  • By the end of this lesson students will be able to explain the concept of discrete energy levels in atoms, including ground state, excited states, and ionisation.
  • By the end of this lesson students will be able to distinguish between emission and absorption spectra and explain their formation in terms of electron transitions.
  • By the end of this lesson students will be able to solve problems involving energy level transitions and photon energy using E = hf = hc/λ.

Key concepts

Rutherford Model of the Atom

Based on the alpha-particle scattering experiment, Rutherford proposed that an atom consists of a tiny, dense, positively charged nucleus at its centre, with electrons orbiting it in a vast empty space. This is often referred to as the 'nuclear model' or 'planetary model'. Limitations: 1. Atomic Stability: Classical physics predicts that orbiting electrons, being accelerating charges, should continuously radiate energy and spiral into the nucleus, causing the atom to collapse. This contradicts the observed stability of atoms. 2. Atomic Spectra: The model predicts a continuous spectrum of light emitted by atoms, whereas experiments show discrete line spectra (specific, distinct wavelengths of light).

Bohr Model of the Atom

Niels Bohr proposed a model for the hydrogen atom that addressed the limitations of Rutherford's model. His model introduced the concept of quantised energy levels. Postulates of the Bohr Model: 1. Electrons orbit the nucleus in certain stable, circular orbits (called stationary states) without radiating energy. These orbits correspond to discrete, quantised energy levels. 2. Electrons in these stable orbits have discrete, quantised angular momentum, L = n(h/2π), where n is an integer (principal quantum number) and h is Planck's constant. 3. An electron can only jump from one stationary state to another by absorbing or emitting a photon of specific energy. The energy of the photon (hf) must be exactly equal to the energy difference between the two states: hf = E₂ - E₁. How it addresses Rutherford's limitations: 1. Atomic Stability: Postulate 1 states that electrons do not radiate energy when in stable orbits, thus preventing the atom from collapsing. 2. Atomic Spectra: Postulate 3 explains the discrete line spectra, as only specific energy differences (and thus specific photon frequencies/wavelengths) are possible when electrons transition between quantised energy levels.

hf = E₂ - E₁
Energy Levels

Electrons in an atom can only occupy specific, discrete energy states or levels. These energy levels are quantised, meaning electrons cannot exist at energies between these levels. * Ground State: The lowest possible energy level an electron can occupy (n=1). * Excited States: Higher energy levels (n=2, 3, 4, etc.) that an electron can occupy if it gains sufficient energy. * Ionisation: If an electron gains enough energy to completely escape the atom, the atom becomes an ion. The ionisation energy is the minimum energy required to remove an electron from the ground state to infinity (where its energy is considered 0 eV).

Emission Spectrum

An emission spectrum is produced when electrons in excited atoms (e.g., in a gas discharge tube) fall from higher energy levels to lower energy levels. As they transition, they emit photons. Since only specific energy differences are allowed, only photons of specific frequencies (and thus specific wavelengths) are emitted. This results in a spectrum of bright lines on a dark background. Each element has a unique emission spectrum, acting like a 'fingerprint'.

E = hf = hc/λ
Absorption Spectrum

An absorption spectrum is produced when white light (containing a continuous range of wavelengths) passes through a cool gas. Electrons in the gas atoms absorb photons that have energies exactly matching the energy differences between their allowed energy levels. These absorbed photons cause electrons to jump to higher energy levels. The wavelengths corresponding to these absorbed photons are removed from the continuous spectrum, appearing as dark lines on a bright background. The dark lines in an absorption spectrum occur at the same wavelengths as the bright lines in the emission spectrum of the same element.

E = hf = hc/λ

Key facts to remember

  • 1Rutherford's model proposed a dense, positive nucleus with electrons orbiting, but it couldn't explain atomic stability or discrete spectra.
  • 2Bohr's model introduced quantised energy levels and orbits where electrons do not radiate energy, explaining atomic stability and line spectra.
  • 3Electrons in atoms can only occupy specific, discrete energy levels (quantisation of energy).
  • 4The ground state is the lowest energy level; excited states are higher energy levels.
  • 5Ionisation energy is the minimum energy required to remove an electron from the ground state of an atom.
  • 6Emission spectra consist of bright lines on a dark background, formed when electrons fall to lower energy levels, emitting photons.
  • 7Absorption spectra consist of dark lines on a continuous background, formed when electrons absorb photons of specific energies to jump to higher levels.
  • 8The energy of a photon is given by E = hf, and since c = fλ, it can also be expressed as E = hc/λ.

Worked examples

Example 1

The energy levels for a hydrogen atom are given by the formula E_n = -13.6/n² eV, where n is the principal quantum number. Calculate the energy and wavelength of the photon emitted when an electron transitions from the n=3 energy level to the n=2 energy level. (Planck's constant h = 6.626 × 10⁻³⁴ J s, speed of light c = 3.00 × 10⁸ m s⁻¹, elementary charge e = 1.602 × 10⁻¹⁹ C)

I1. Calculate the energy of the n=3 level (E₃):
II E₃ = -13.6 / 3² = -13.6 / 9 = -1.511 eV (to 3 decimal places)
III2. Calculate the energy of the n=2 level (E₂):
IV E₂ = -13.6 / 2² = -13.6 / 4 = -3.400 eV
V3. Calculate the energy difference (ΔE) between the two levels, which is the energy of the emitted photon:
VI ΔE = E_final - E_initial = E₃ - E₂ = (-1.511 eV) - (-3.400 eV) = 1.889 eV
VII4. Convert the photon energy from electronvolts (eV) to Joules (J):
VIII ΔE = 1.889 eV × (1.602 × 10⁻¹⁹ J/eV) = 3.026078 × 10⁻¹⁹ J
95. Use the formula E = hc/λ to find the wavelength (λ) of the photon:
10 λ = hc / ΔE
11 λ = (6.626 × 10⁻³⁴ J s × 3.00 × 10⁸ m s⁻¹) / (3.026078 × 10⁻¹⁹ J)
12 λ = (1.9878 × 10⁻²⁵ J m) / (3.026078 × 10⁻¹⁹ J)
13 λ = 6.569 × 10⁻⁷ m
146. Convert the wavelength to nanometres (nm) for convenience:
15 λ = 6.569 × 10⁻⁷ m × (10⁹ nm / 1 m) = 656.9 nm

Answer

The energy of the emitted photon is 1.89 eV (or 3.03 × 10⁻¹⁹ J), and its wavelength is 657 nm.

This transition corresponds to the H-alpha line in the Balmer series, which is visible red light.

Example 2

An atom has the following energy levels: E₁ = -12.0 eV (ground state), E₂ = -5.0 eV, E₃ = -2.0 eV. (a) What is the ionisation energy of this atom? (b) Can a photon of energy 6.5 eV excite an electron from the ground state to an excited state? Explain your answer. (c) What is the maximum wavelength of radiation that can be absorbed by an electron in the ground state?

I(a) Calculate the ionisation energy:
II Ionisation energy is the energy required to remove an electron from the ground state (E₁) to infinity (0 eV).
III Ionisation Energy = 0 eV - E₁ = 0 eV - (-12.0 eV) = 12.0 eV
IV(b) Determine if a 6.5 eV photon can excite an electron from the ground state:
V Possible energy transitions from the ground state (E₁):
VI To E₂: ΔE₁₂ = E₂ - E₁ = (-5.0 eV) - (-12.0 eV) = 7.0 eV
VII To E₃: ΔE₁₃ = E₃ - E₁ = (-2.0 eV) - (-12.0 eV) = 10.0 eV
VIII A photon of 6.5 eV cannot excite the electron because its energy does not exactly match any of the allowed energy differences (7.0 eV or 10.0 eV) from the ground state. Electrons can only absorb photons with energies that precisely match the energy difference between two allowed energy levels.
9(c) Calculate the maximum wavelength of radiation that can be absorbed by an electron in the ground state:
10 Maximum wavelength corresponds to the minimum energy required for absorption from the ground state. This is the transition from E₁ to E₂ (ΔE₁₂ = 7.0 eV).
11 1. Convert the minimum energy difference to Joules:
12 ΔE = 7.0 eV × (1.602 × 10⁻¹⁹ J/eV) = 1.1214 × 10⁻¹⁸ J
13 2. Use the formula λ = hc/ΔE:
14 λ = (6.626 × 10⁻³⁴ J s × 3.00 × 10⁸ m s⁻¹) / (1.1214 × 10⁻¹⁸ J)
15 λ = (1.9878 × 10⁻²⁵ J m) / (1.1214 × 10⁻¹⁸ J)
16 λ = 1.7726 × 10⁻⁷ m
17 3. Convert to nanometres:
18 λ = 177.3 nm

Answer

(a) The ionisation energy is 12.0 eV. (b) No, a 6.5 eV photon cannot excite the electron as its energy does not match an allowed energy transition from the ground state. (c) The maximum wavelength of radiation that can be absorbed is 177 nm.

Remember that for absorption, the photon energy must be an exact match to an energy level difference. For ionisation, any energy equal to or greater than the ionisation energy can remove the electron, but the excess energy becomes kinetic energy of the free electron.

Common mistakes

  • Confusing emission spectra (bright lines) with absorption spectra (dark lines) and their formation mechanisms.
  • Incorrectly calculating energy differences for transitions (e.g., E_initial - E_final instead of E_final - E_initial).
  • Failing to convert energy units between electronvolts (eV) and Joules (J) when using Planck's constant (h) or the speed of light (c).
  • Assuming electrons can exist at energies between allowed energy levels.
  • Believing that any photon energy can be absorbed; only photons with energies exactly matching an energy level difference can be absorbed to cause a transition.

Exam tips

  • Always draw an energy level diagram if not provided; it helps visualise transitions and energy differences.
  • Memorise the values of fundamental constants: Planck's constant (h), speed of light (c), and the elementary charge (e) for eV to J conversion.
  • Pay close attention to units throughout your calculations, especially when converting between eV and J.
  • Clearly state the limitations of the Rutherford model and how each postulate of the Bohr model addresses these limitations.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.