Modern Physics

The Photoelectric Effect

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to define the photoelectric effect and explain its key observations.
  • By the end of this lesson students will be able to explain the terms work function and threshold frequency.
  • By the end of this lesson students will be able to apply the equation E = hf to calculate photon energy.
  • By the end of this lesson students will be able to apply Einstein's photoelectric equation, hf = φ + ½mv²_max, to solve problems.
  • By the end of this lesson students will be able to explain why the wave theory of light could not explain the photoelectric effect.

Key concepts

The Photoelectric Effect

The photoelectric effect is the emission of electrons from the surface of a metal when electromagnetic radiation of a suitable frequency shines on it. Key observations that challenged the classical wave theory of light include: 1. **Instantaneous Emission**: Electrons are emitted almost instantaneously (within 10⁻⁹ s) once the light is switched on, provided the frequency is above the threshold. 2. **Threshold Frequency (f₀)**: For each metal, there is a minimum frequency of incident radiation, called the threshold frequency, below which no electrons are emitted, regardless of the intensity of the light. 3. **Intensity and Number of Electrons**: If the frequency is above the threshold, the number of electrons emitted per second is directly proportional to the intensity of the incident radiation. 4. **Frequency and Kinetic Energy**: The maximum kinetic energy of the emitted electrons (photoelectrons) is proportional to the frequency of the incident radiation, but is independent of its intensity.

Photon Energy

In 1900, Max Planck proposed that energy is quantised, meaning it can only exist in discrete packets. Albert Einstein later applied this idea to light, suggesting that light consists of discrete packets of energy called photons. The energy of a single photon is directly proportional to its frequency.

E = hf where: E = energy of the photon (Joules, J) h = Planck's constant (6.626 × 10⁻³⁴ J s) f = frequency of the radiation (Hertz, Hz) Since the speed of light (c) is related to frequency (f) and wavelength (λ) by c = fλ, the energy equation can also be written as E = hc/λ.
Work Function (φ)

The work function (φ) of a metal is the minimum energy required to remove an electron from the surface of that particular metal. It is a characteristic property of the metal. Electrons are held within the metal by attractive forces, so energy must be supplied to overcome these forces and allow an electron to escape. The work function is typically measured in Joules (J) or electronvolts (eV), where 1 eV = 1.602 × 10⁻¹⁹ J.

φ = hf₀ where: φ = work function (J) h = Planck's constant (J s) f₀ = threshold frequency (Hz)
Threshold Frequency (f₀)

The threshold frequency (f₀) is the minimum frequency of incident radiation required to cause photoelectric emission from a particular metal. If the incident light has a frequency below the threshold frequency, no electrons will be emitted, no matter how intense the light or how long it shines on the metal. This is because the individual photons do not have enough energy to overcome the work function of the metal.

f₀ = φ/h
Einstein's Photoelectric Equation

Einstein explained the photoelectric effect by applying the principle of conservation of energy to the interaction between a photon and an electron. When a photon of energy hf strikes the surface of a metal, some of its energy (φ) is used to overcome the work function and release an electron. Any remaining energy is converted into the kinetic energy of the emitted electron. The maximum kinetic energy (½mv²_max) occurs when the electron is emitted from the very surface of the metal without any energy loss due to collisions.

hf = φ + ½mv²_max where: hf = energy of the incident photon (J) φ = work function of the metal (J) ½mv²_max = maximum kinetic energy of the emitted photoelectron (J) m = mass of the electron (9.109 × 10⁻³¹ kg) v_max = maximum speed of the emitted electron (m s⁻¹) This equation can also be written as KE_max = hf - φ.
Stopping Potential (V_s)

The stopping potential (V_s) is the minimum negative potential difference applied to the collector electrode that is just sufficient to stop the most energetic photoelectrons from reaching it. At the stopping potential, the work done by the electric field on the electron (eV_s) is equal to the maximum kinetic energy of the electron.

KE_max = eV_s where: KE_max = maximum kinetic energy of the photoelectron (J) e = elementary charge (1.602 × 10⁻¹⁹ C) V_s = stopping potential (Volts, V)

Key facts to remember

  • 1The photoelectric effect is the emission of electrons from a metal surface when light of suitable frequency shines on it.
  • 2Light consists of discrete packets of energy called photons, with energy E = hf.
  • 3The work function (φ) is the minimum energy required to remove an electron from a metal surface.
  • 4The threshold frequency (f₀) is the minimum frequency of light required for photoelectric emission (φ = hf₀).
  • 5Einstein's photoelectric equation is hf = φ + ½mv²_max, representing the conservation of energy.
  • 6The maximum kinetic energy of photoelectrons is proportional to the frequency of the incident light, not its intensity.
  • 7The number of photoelectrons emitted is proportional to the intensity of the incident light (above threshold frequency).
  • 8The wave theory of light cannot explain the instantaneous emission or the existence of a threshold frequency.

Worked examples

Example 1

The work function of sodium is 2.3 eV. (i) Calculate the threshold frequency for sodium. (ii) If light of frequency 7.0 × 10¹⁴ Hz shines on sodium, calculate the maximum kinetic energy of the emitted photoelectrons in Joules.

I**Part (i): Calculate the threshold frequency (f₀)**
II1. Convert the work function from electronvolts (eV) to Joules (J):
III φ = 2.3 eV × (1.602 × 10⁻¹⁹ J/eV) = 3.6846 × 10⁻¹⁹ J
IV2. Use the formula φ = hf₀ to find f₀:
V f₀ = φ / h
VI f₀ = (3.6846 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J s)
VII f₀ = 5.560 × 10¹⁴ Hz
VIII**Part (ii): Calculate the maximum kinetic energy (KE_max)**
91. Calculate the energy of the incident photon (hf):
10 hf = (6.626 × 10⁻³⁴ J s) × (7.0 × 10¹⁴ Hz)
11 hf = 4.6382 × 10⁻¹⁹ J
122. Use Einstein's photoelectric equation: hf = φ + KE_max
13 Rearrange to find KE_max: KE_max = hf - φ
14 KE_max = (4.6382 × 10⁻¹⁹ J) - (3.6846 × 10⁻¹⁹ J)
15 KE_max = 9.536 × 10⁻²⁰ J

Answer

(i) Threshold frequency (f₀) = 5.56 × 10¹⁴ Hz (ii) Maximum kinetic energy (KE_max) = 9.54 × 10⁻²⁰ J

Remember to use consistent units (Joules) throughout the calculation before converting if necessary.

Example 2

When ultraviolet light of wavelength 200 nm falls on a metal surface, the maximum kinetic energy of the emitted electrons is 2.5 eV. (i) Calculate the work function of the metal in eV. (ii) Calculate the maximum speed of the emitted electrons.

I**Part (i): Calculate the work function (φ) in eV**
II1. Convert the wavelength from nanometres (nm) to metres (m):
III λ = 200 nm = 200 × 10⁻⁹ m = 2.00 × 10⁻⁷ m
IV2. Calculate the energy of the incident photon (hf) using E = hc/λ:
V hf = (6.626 × 10⁻³⁴ J s) × (3.0 × 10⁸ m s⁻¹) / (2.00 × 10⁻⁷ m)
VI hf = 9.939 × 10⁻¹⁹ J
VII3. Convert the photon energy from Joules to electronvolts:
VIII hf (in eV) = (9.939 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ J/eV) = 6.204 eV
94. Use Einstein's photoelectric equation: hf = φ + KE_max
10 Rearrange to find φ: φ = hf - KE_max
11 φ = 6.204 eV - 2.5 eV
12 φ = 3.704 eV
13**Part (ii): Calculate the maximum speed (v_max) of the emitted electrons**
141. Convert the maximum kinetic energy from electronvolts (eV) to Joules (J):
15 KE_max = 2.5 eV × (1.602 × 10⁻¹⁹ J/eV) = 4.005 × 10⁻¹⁹ J
162. Use the formula for kinetic energy: KE_max = ½mv²_max
17 Rearrange to find v_max: v_max = √(2 × KE_max / m)
18 v_max = √(2 × 4.005 × 10⁻¹⁹ J / 9.109 × 10⁻³¹ kg)
19 v_max = √(8.7935 × 10¹¹ m² s⁻²)
20 v_max = 9.377 × 10⁵ m s⁻¹

Answer

(i) Work function (φ) = 3.70 eV (ii) Maximum speed (v_max) = 9.38 × 10⁵ m s⁻¹

The mass of an electron (m_e) is 9.109 × 10⁻³¹ kg.

Common mistakes

  • Confusing the roles of intensity and frequency: Intensity affects the *number* of electrons emitted, while frequency affects their *maximum kinetic energy* (above threshold).
  • Failing to convert units correctly, especially between Joules (J) and electronvolts (eV), or nanometres (nm) to metres (m).
  • Using the incorrect mass for the electron (e.g., using the mass of a proton or neutron).
  • Incorrectly applying Einstein's photoelectric equation, for example, subtracting hf from φ instead of φ from hf.
  • Not recognising that if the incident frequency is below the threshold frequency, no photoelectrons will be emitted, regardless of intensity.

Exam tips

  • Always list the given values and relevant physical constants (h, c, e, m_e) at the start of your solution.
  • Pay close attention to units and ensure consistency. It's often safest to convert all energy values to Joules at the start of a calculation and only convert to electronvolts if the final answer specifically requests it.
  • Clearly state the formula being used for each step of your calculation.
  • Be prepared to explain the experimental observations of the photoelectric effect and why the classical wave theory of light fails to explain them, providing evidence for the particle nature of light (photons).

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