Cambridge AS & A Level9702

Quantum physics

Physics 9702 Chapter Notes

What this chapter covers

Quantum physics - Energy and momentum of a photonQuantum physics - Photoelectric effectQuantum physics - Wave-particle dualityQuantum physics - Energy levels in atoms and line spectra
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1. Photons and Quantised Energy

At the start of the 20th century, classical physics viewed light purely as a wave. However, some phenomena, like the photoelectric effect, couldn't be explained this way. Max Planck proposed a revolutionary idea: electromagnetic energy is not continuous but is emitted and absorbed in discrete packets, or 'quanta'. Albert Einstein later named these packets 'photons'. The energy of a single photon is directly proportional to the frequency of the electromagnetic radiation. A higher frequency (like UV light) means higher energy photons, while a lower frequency (like red light) means lower energy photons. Because energy comes in these distinct chunks, we say it is 'quantised'. A common unit of energy at this scale is the electronvolt (eV), which is the energy gained by an electron when it is accelerated through a potential difference of 1 volt.

E = hf

E = hc/λ

1 eV = 1.60 x 10⁻¹⁹ J

Key term

Photon: A discrete packet, or quantum, of electromagnetic energy.

Examiner insight

Examiners expect you to be fluent in converting between joules and electronvolts. Always show your conversion step clearly.

Fun fact

The concept of quantised energy is so fundamental it forms the 'Quantum' in Quantum Physics. It's the basis for everything from lasers to modern computers.

Worked example 14 marks

A photon of blue light has a wavelength of 450 nm. Calculate its energy in(a) joules and(b) electronvolts. (Planck constant h = 6.63 x 10⁻³⁴ J s, speed of light c = 3.00 x 10⁸ m s⁻¹)

  1. 1

    Step 1: Convert the wavelength from nanometres to metres. λ = 450 nm = 450 x 10⁻⁹ m.

  2. 2

    Step 2: Use the formula E = hc/λ to find the energy in joules. E = (6.63 x 10⁻³⁴ Js) x (3.00 x 10⁸ m s⁻¹) / (450 x 10⁻⁹ m).

  3. 3

    Step 3: Calculate the result for part (a). E = 4.42 x 10⁻¹⁹ J.

  4. 4

    Step 4: To convert from joules to electronvolts, divide by the charge of an electron. E (in eV) = (4.42 x 10⁻¹⁹ J) / (1.60 x 10⁻¹⁹ J/eV).

  5. 5

    Step 5: Calculate the result for part (b). E = 2.76 eV.

Recap

  • Electromagnetic energy is quantised into packets called photons.
  • The energy of a photon is given by E = hf, where h is the Planck constant.
  • Photon energy can also be calculated using E = hc/λ.
  • The electronvolt (eV) is a unit of energy useful on the atomic scale.
  • To convert from joules to eV, divide by 1.60 x 10⁻¹⁹.

Quick check

  1. What is the relationship between a photon's energy and its frequency?1 mark
  2. How much energy in joules is 5.0 eV?1 mark

2. The Photoelectric Effect

The photoelectric effect is the emission of electrons from a material (usually a metal) when light shines on it. These emitted electrons are called photoelectrons. Classical wave theory predicted that any frequency of light, if intense enough, should eventually provide enough energy to free an electron. However, experiments showed something different. For a given metal, there is a specific minimum frequency, called the 'threshold frequency' (f₀), below which no photoelectrons are emitted, no matter how intense the light is. The minimum energy required to free an electron from the surface is a property of the metal called the 'work function' (Φ). A single photon must have enough energy to overcome this work function. If the photon's energy (hf) is less than the work function (Φ), no electron is emitted.

Φ = hf₀

Φ = hc/λ₀

Key term

Threshold Frequency (f₀): The minimum frequency of incident electromagnetic radiation required to cause photoelectric emission from a particular metal surface.

Common pitfall

Confusing threshold frequency with threshold wavelength. Remember that a minimum frequency corresponds to a maximum wavelength (λ₀ = c/f₀) because of their inverse relationship.

Worked example 15 marks

The work function of zinc is 4.3 eV.(a) Calculate the threshold frequency for zinc.(b) Explain whether photoelectrons will be emitted if light of wavelength 300 nm is shone on the zinc. (h = 6.63 x 10⁻³⁴ J s)

  1. 1

    Step 1 (a): Convert the work function from eV to joules. Φ = 4.3 eV x (1.60 x 10⁻¹⁹ J/eV) = 6.88 x 10⁻¹⁹ J.

  2. 2

    Step 2 (a): Use the work function formula Φ = hf₀ and rearrange for the threshold frequency, f₀ = Φ/h.

  3. 3

    Step 3 (a): Calculate f₀. f₀ = (6.88 x 10⁻¹⁹ J) / (6.63 x 10⁻³⁴ Js) = 1.04 x 10¹⁵ Hz.

  4. 4

    Step 4 (b): Calculate the energy of a photon of the incident light using E = hc/λ. E = (6.63 x 10⁻³⁴ x 3.00 x 10⁸) / (300 x 10⁻⁹) = 6.63 x 10⁻¹⁹ J.

  5. 5

    Step 5 (b): Compare the photon energy with the work function. Photon energy E (6.63 x 10⁻¹⁹ J) is less than the work function Φ (6.88 x 10⁻¹⁹ J).

  6. 6

    Step 6 (b): Conclude that since the energy of an incident photon is not sufficient to overcome the work function, no photoelectrons will be emitted.

Recap

  • The photoelectric effect is the emission of electrons from a metal surface when illuminated by EM radiation.
  • No electrons are emitted if the radiation's frequency is below the threshold frequency (f₀).
  • The work function (Φ) is the minimum energy needed to release an electron from the surface.
  • The threshold frequency is related to the work function by Φ = hf₀.
  • A single photon must have energy E ≥ Φ to cause emission.

Quick check

  1. What happens if a photon's energy is exactly equal to the work function of a metal?2 marks

3. Einstein's Photoelectric Equation

Einstein's photoelectric equation is a beautiful application of the principle of conservation of energy to the photoelectric effect. It states that the energy of the incoming photon (hf) is used for two things: first, to provide the energy needed to free the electron from the metal (the work function, Φ), and second, any remaining energy is given to the emitted electron as kinetic energy. The equation refers to the *maximum* kinetic energy (KE_max) because some electrons may be liberated from deeper within the metal and lose some energy in collisions on their way out. The electrons that escape with the most KE are those from the very surface.

hf = Φ + KE_max

hf = Φ + (1/2)mv_max²

KE_max = eVs

Key term

Work Function (Φ): The minimum energy required to remove an electron from the surface of a particular metal.

Examiner insight

Marks are often awarded for rearranging the photoelectric equation correctly. Write the equation down first, then show the rearrangement clearly before substituting values.

Worked example 14 marks

Light of frequency 7.5 x 10¹⁴ Hz is shone onto a metal surface with a work function of 2.1 eV. Calculate the maximum kinetic energy of the emitted photoelectrons in joules. (h = 6.63 x 10⁻³⁴ J s)

  1. 1

    Step 1: Calculate the energy of the incident photons in joules. E = hf = (6.63 x 10⁻³⁴ Js) x (7.5 x 10¹⁴ Hz) = 4.97 x 10⁻¹⁹ J.

  2. 2

    Step 2: Convert the work function from eV to joules. Φ = 2.1 eV x (1.60 x 10⁻¹⁹ J/eV) = 3.36 x 10⁻¹⁹ J.

  3. 3

    Step 3: Rearrange Einstein's photoelectric equation to find KE_max. KE_max = hf - Φ.

  4. 4

    Step 4: Substitute the values and calculate the result. KE_max = 4.97 x 10⁻¹⁹ J - 3.36 x 10⁻¹⁹ J = 1.61 x 10⁻¹⁹ J.

Worked example 24 marks

When light of wavelength 420 nm falls on a metal surface, the maximum speed of the photoelectrons is 5.2 x 10⁵ m s⁻¹. Calculate the work function of the metal. (Mass of electron mₑ = 9.11 x 10⁻³¹ kg)

  1. 1

    Step 1: Calculate the energy of the incident photons. E = hc/λ = (6.63 x 10⁻³⁴ x 3.00 x 10⁸) / (420 x 10⁻⁹) = 4.74 x 10⁻¹⁹ J.

  2. 2

    Step 2: Calculate the maximum kinetic energy of the photoelectrons. KE_max = (1/2)mv² = 0.5 x (9.11 x 10⁻³¹ kg) x (5.2 x 10⁵ m s⁻¹)² = 1.23 x 10⁻¹⁹ J.

  3. 3

    Step 3: Rearrange Einstein's photoelectric equation to find the work function. Φ = hf - KE_max (or E_photon - KE_max).

  4. 4

    Step 4: Substitute the values and calculate Φ. Φ = 4.74 x 10⁻¹⁹ J - 1.23 x 10⁻¹⁹ J = 3.51 x 10⁻¹⁹ J.

Recap

  • Einstein's photoelectric equation is hf = Φ + KE_max.
  • It is an expression of energy conservation for a single photon-electron interaction.
  • Photon energy (hf) is split between the work function (Φ) and the electron's kinetic energy.
  • KE_max is the kinetic energy of electrons emitted from the surface with no energy loss.
  • The equation can be used to find photon energy, work function, or the maximum speed of photoelectrons.

Quick check

  1. In the photoelectric equation, what does KE_max represent?1 mark
  2. If you plot a graph of KE_max against frequency (f), what does the y-intercept represent?1 mark

4. Photoelectric Effect as Evidence for Particles

The photoelectric effect provides compelling evidence for the particle nature of light. The classical wave model fails to explain key experimental observations.

  1. Existence of a Threshold Frequency: The wave model suggests any frequency of light, if intense enough, would eventually transfer enough energy to an electron. The particle model explains this perfectly: a single photon must have energy hf > Φ. If not, no emission occurs, regardless of how many photons (intensity) arrive.
  2. Instantaneous Emission: Experiments show electrons are emitted the moment light of sufficient frequency hits the metal. The wave model predicts a time delay, as energy would need to accumulate over a region of the surface. The particle model explains this as a one-to-one, instantaneous interaction between a photon and an electron.
  3. KE vs. Intensity: Increasing the light intensity (brightness) increases the number of photoelectrons emitted per second (the photoelectric current), but does not change their maximum kinetic energy. The particle model explains this because higher intensity means more photons arriving per second, leading to more one-to-one interactions, but the energy of each individual photon (and thus the KE_max) remains the same as long as frequency is constant.

Key term

Photoelectric Current: The rate of flow of charge due to the emitted photoelectrons, which is directly proportional to the intensity of the incident radiation (above the threshold frequency).

Examiner insight

Examiners look for clear, concise explanations that contrast the predictions of the wave model with the observations explained by the photon model. Use phrases like 'one-to-one interaction' and 'energy of a single photon'.

Worked example 14 marks

A physicist observes the photoelectric effect using a source of monochromatic green light. Explain, in terms of photons, what happens to(a) the rate of emission of photoelectrons and(b) the maximum kinetic energy of the photoelectrons, if the intensity of the green light is doubled.

  1. 1

    Step 1 (a): State the effect of intensity on photons. Doubling the intensity means doubling the number of photons arriving at the metal surface per second.

  2. 2

    Step 2 (a): Link photons to electrons. Since each photon interacts with a single electron (in a one-to-one interaction), more photons per second will cause more electrons to be emitted per second.

  3. 3

    Step 3 (a): Conclude for the rate. Therefore, the rate of emission of photoelectrons doubles.

  4. 4

    Step 4 (b): State the effect of intensity on photon energy. The intensity of the light does not affect the frequency, so the energy of each individual photon (E = hf) remains unchanged.

  5. 5

    Step 5 (b): Link photon energy to kinetic energy. According to hf = Φ + KE_max, since hf and Φ are constant, the maximum kinetic energy (KE_max) of the photoelectrons must also remain unchanged.

  6. 6

    Step 6 (b): Conclude for the kinetic energy. Therefore, the maximum kinetic energy of the photoelectrons does not change.

Recap

  • The photoelectric effect provides strong evidence for the particle nature of light.
  • The existence of a threshold frequency cannot be explained by wave theory.
  • The instantaneous emission of electrons supports the one-to-one photon-electron interaction model.
  • Maximum KE of photoelectrons depends on frequency, not intensity.
  • Photoelectric current is proportional to intensity (for f > f₀).

Quick check

  1. According to the wave model of light, would you expect a threshold frequency for the photoelectric effect? Explain why or why not.2 marks

5. Wave-Particle Duality and de Broglie Wavelength

Quantum physics reveals a strange truth: entities like light and electrons exhibit both wave-like and particle-like properties. This is called wave-particle duality. While the photoelectric effect shows light behaving as particles (photons), phenomena like diffraction and interference show it behaving as a wave. In 1924, Louis de Broglie proposed that this duality wasn't just for light. He suggested that all matter has wave-like properties. A moving particle, like an electron, has an associated wavelength, now called the de Broglie wavelength (λ). This wavelength is inversely proportional to the particle's momentum (p). This was a radical idea, but it was confirmed experimentally by observing electron diffraction. When a beam of electrons is passed through a thin crystal lattice (like graphite), they produce a diffraction pattern of concentric rings, just as waves would. This confirmed that particles can behave like waves. A photon, despite having no rest mass, also has momentum, which is related to its energy.

λ = h/p

p = mv

p = E/c (for photons only)

Key term

de Broglie Wavelength (λ): The wavelength associated with a moving particle, given by the equation λ = h/p, where h is the Planck constant and p is the particle's momentum.

Common pitfall

Applying the photon momentum formula p = E/c to particles with mass like electrons. This formula is only for massless photons. For massive particles, always use p = mv.

Fun fact

The de Broglie wavelength of a thrown baseball is incredibly tiny, about 10⁻³⁴ m, which is why we never observe macroscopic objects diffracting. You need a very small mass and momentum for the wavelength to be significant.

Worked example 15 marks

An electron is accelerated from rest through a potential difference of 2500 V. Calculate its de Broglie wavelength. (h = 6.63 x 10⁻³⁴ J s, mₑ = 9.11 x 10⁻³¹ kg, e = 1.60 x 10⁻¹⁹ C)

  1. 1

    Step 1: Calculate the kinetic energy gained by the electron in joules. KE = eV = (1.60 x 10⁻¹⁹ C) x (2500 V) = 4.00 x 10⁻¹⁶ J.

  2. 2

    Step 2: Relate kinetic energy to momentum. KE = p²/2m, so p = √(2mKE).

  3. 3

    Step 3: Calculate the momentum of the electron. p = √(2 x 9.11 x 10⁻³¹ kg x 4.00 x 10⁻¹⁶ J) = 2.70 x 10⁻²³ kg m s⁻¹.

  4. 4

    Step 4: Use the de Broglie equation to find the wavelength. λ = h/p.

  5. 5

    Step 5: Substitute values and calculate λ. λ = (6.63 x 10⁻³⁴ Js) / (2.70 x 10⁻²³ kg m s⁻¹) = 2.46 x 10⁻¹¹ m.

Recap

  • Wave-particle duality means that entities can show both wave and particle properties.
  • The photoelectric effect shows light as a particle; diffraction shows light as a wave.
  • Electron diffraction provides evidence for the wave nature of particles.
  • Any moving particle has a de Broglie wavelength given by λ = h/p.
  • A photon has momentum given by p = E/c.

Quick check

  1. What happens to the de Broglie wavelength of a particle if its speed is doubled?1 mark
  2. Name an experiment that demonstrates the wave nature of electrons.1 mark

6. Discrete Atomic Energy Levels

In an isolated atom, electrons can't just have any amount of energy. They are restricted to specific, discrete energy levels. Think of it like a staircase: you can stand on step 1, or step 2, but not on step 1.5. You can't be 'in between' levels. The lowest possible energy level is called the ground state (n=1). Higher levels are called excited states (n=2, n=3, etc.). By convention, the energy of a free electron (one that has escaped the atom completely, n=∞) is defined as 0 eV. Since energy must be supplied to an electron to move it to a higher level and eventually free it, all the energy levels within the atom have negative values. For example, the ground state of a hydrogen atom is -13.6 eV. An electron can 'jump' to a higher energy level if it absorbs the exact amount of energy corresponding to the difference between the levels. This energy is typically supplied by absorbing a photon or through a collision.

Key term

Energy Level: A specific, discrete amount of energy that an electron is allowed to have within an isolated atom.

Examiner insight

Examiners reward answers that explicitly state that the absorbed photon's energy must be *exactly* equal to the energy difference between two levels for excitation to occur.

Fun fact

The idea of discrete energy levels was a key part of the Bohr model of the atom, one of the first successful quantum descriptions of how atoms work.

Worked example 13 marks

The diagram shows some of the energy levels of a hydrogen atom. An electron is in the ground state (n=1). A photon with energy 12.09 eV is incident on the atom. Explain what happens to the electron.

  1. 1

    Step 1: Identify the initial state of the electron. The electron is in the ground state, E₁ = -13.6 eV.

  2. 2

    Step 2: Calculate the energy the electron would have if it absorbed the photon. E_new = -13.6 eV + 12.09 eV = -1.51 eV.

  3. 3

    Step 3: Check if this new energy corresponds to an allowed energy level. The diagram shows an energy level at n=3 with energy E₃ = -1.51 eV.

  4. 4

    Step 4: Conclude what happens. Because the photon's energy is exactly equal to the difference in energy between the n=1 and n=3 levels (E₃ - E₁), the electron absorbs the photon and is excited (jumps) from the ground state to the n=3 energy level.

Recap

  • Electrons in isolated atoms can only exist in discrete energy levels.
  • The lowest energy level is the ground state; higher levels are excited states.
  • Energy levels are assigned negative values, with 0 eV representing a free electron.
  • An electron can move to a higher level by absorbing a photon with the exact energy difference.
  • This process is called excitation.

Quick check

  1. Why are atomic energy levels given negative values?2 marks

7. Emission and Absorption Spectra

Atomic energy levels provide a perfect explanation for line spectra. When a gas is heated or has a high voltage passed through it, its atoms become excited. Electrons jump to higher energy levels but are unstable there. They quickly fall back down to lower, more stable levels. As an electron de-excites, it loses a specific amount of energy (ΔE = E_initial - E_final). This energy is released as a single photon. The photon's energy is exactly equal to the energy lost by the electron, so hf = E_initial - E_final. Since only certain energy transitions are possible, only photons of specific frequencies (and therefore specific colours) are emitted. When viewed through a spectrometer, this produces a series of bright coloured lines on a dark background, known as an emission line spectrum. Conversely, if white light (containing all frequencies) is passed through a cool gas, atoms will absorb photons that have the exact energies needed to excite their electrons to higher levels. These specific frequencies are then missing from the light that passes through. This creates a continuous spectrum with dark lines on it, known as an absorption line spectrum. The dark lines in a gas's absorption spectrum are at the exact same frequencies as the bright lines in its emission spectrum.

ΔE = E_initial - E_final

hf = E_initial - E_final

Key term

Line Spectrum: A spectrum consisting of a series of discrete bright or dark lines, corresponding to specific frequencies of light emitted or absorbed by atoms.

Examiner insight

When calculating photon wavelength from an energy level transition, a common mistake is forgetting to convert the energy difference from eV to Joules before using E=hc/λ. Always check your units.

Fun fact

Astronomers use absorption spectra to determine the chemical composition of distant stars. The dark lines in the starlight reveal which elements are present in the star's outer atmosphere.

Worked example 14 marks

An electron in an excited atom falls from an energy level of -1.51 eV to the ground state at -13.6 eV. Calculate the wavelength of the photon emitted. (h = 6.63 x 10⁻³⁴ J s, c = 3.00 x 10⁸ m s⁻¹)

  1. 1

    Step 1: Calculate the energy difference between the levels in eV. ΔE = E_initial - E_final = (-1.51 eV) - (-13.6 eV) = 12.09 eV.

  2. 2

    Step 2: This energy difference is the energy of the emitted photon. Convert this energy to joules. E_photon = 12.09 eV x (1.60 x 10⁻¹⁹ J/eV) = 1.934 x 10⁻¹⁸ J.

  3. 3

    Step 3: Use the photon energy equation E = hc/λ and rearrange for wavelength, λ = hc/E.

  4. 4

    Step 4: Substitute the values and calculate the wavelength. λ = (6.63 x 10⁻³⁴ x 3.00 x 10⁸) / (1.934 x 10⁻¹⁸) = 1.03 x 10⁻⁷ m (or 103 nm).

Recap

  • An emission spectrum is produced when excited electrons de-excite, emitting photons.
  • An absorption spectrum is produced when electrons absorb photons to become excited.
  • The energy of the photon is equal to the energy difference between the two levels: hf = E₁ - E₂.
  • Each element has a unique line spectrum, which acts like a 'fingerprint'.
  • The lines in an element's emission and absorption spectra occur at the same frequencies.

Quick check

  1. What is the difference in appearance between an emission spectrum and an absorption spectrum?2 marks

End-of-chapter exercise

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

  1. UV radiation of wavelength 2.50 x 10⁻⁷ m is incident on a metal plate. The work function of the metal is 4.30 x 10⁻¹⁹ J. Calculate the maximum speed of the emitted photoelectrons. (Mass of electron = 9.11 x 10⁻³¹ kg)5 marks
  2. Explain why, for the photoelectric effect to occur, the incident light must have a frequency above a certain threshold value, and why this observation supports the particle model of light over the wave model.4 marks
  3. Calculate the de Broglie wavelength of a proton (mass = 1.67 x 10⁻²⁷ kg) travelling at 5.0% of the speed of light.3 marks
  4. A mercury vapour lamp produces an emission spectrum. One of the lines in the visible spectrum has a wavelength of 546 nm. Calculate the energy loss, in eV, of an electron in a mercury atom that gives rise to this photon.3 marks
  5. A graph of maximum kinetic energy (KE_max) of photoelectrons against frequency (f) for a certain metal is a straight line. State what is represented by (a) the gradient of the graph, and (b) the x-intercept of the graph.2 marks
  6. Describe the process of electron diffraction and explain how it provides evidence for the wave nature of matter.4 marks
  7. The energy levels in a hydrogen atom are given by the formula Eₙ = -13.6/n² eV, where n is an integer (n=1, 2, 3...). An electron makes a transition from the n=4 state to the n=2 state. Determine the frequency of the emitted photon.4 marks
  8. Distinguish between an emission line spectrum and an absorption line spectrum with reference to their appearance and formation.4 marks
  9. A laser beam with a power of 2.0 mW has a wavelength of 633 nm. Calculate the number of photons emitted by the laser per second.4 marks
  10. An electron and a photon both have a momentum of 2.00 x 10⁻²⁴ kg m s⁻¹. Calculate (a) the de Broglie wavelength of the electron and (b) the energy, in eV, of the photon.5 marks

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