Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define the photoelectric effect in terms of photons and electron ejection.
• Use the equation E = hf to calculate photon energy.
• Apply the work function equation W₀ = hf₀.
• Calculate the kinetic energy of photoelectrons using E = W₀ + KEmax.
• Explain the relationship between light intensity and the number of photoelectrons.

Lesson summary

This week, we delve into the fascinating world of optical phenomena and the photoelectric effect, fundamental concepts bridging the wave and particle nature of light. Understanding these principles is crucial as they underpin many modern technologies impacting our daily lives, from solar energy powering homes in rural South Africa to the digital cameras in our phones.

Furthermore, these concepts are essential for understanding more advanced physics topics later in your studies. The photoelectric effect, in particular, provides a powerful demonstration of quantum mechanics and how energy behaves at the atomic level.

Lesson notes

2.1 The Photoelectric Effect The photoelectric effect is the phenomenon where electrons are emitted from a metal surface when light (electromagnetic radiation) shines on it. These emitted electrons are called photoelectrons. Classically, one would expect that any frequency of light, given sufficient intensity, would be able to eject electrons.

However, experiments showed that this was not the case.

Key Observations: Threshold Frequency (f₀): Electrons are only emitted if the frequency of the incident light is above a certain threshold frequency specific to each metal. Below this frequency, no electrons are emitted, no matter how intense the light.

Instantaneous Emission: Electrons are emitted almost instantaneously when light above the threshold frequency shines on the metal. There is no time delay, which contradicts classical wave theory expectations.

Kinetic Energy Dependence: The kinetic energy of the emitted electrons increases with the frequency of the incident light, not with the intensity.

Intensity and Photoelectron Number: The intensity of the light is directly proportional to the number of photoelectrons emitted if the frequency is above the threshold frequency. Higher intensity means more photons striking the metal surface, leading to more electrons being ejected.

Explanation Using Quantum Theory: Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons.

The energy of a photon is given by: E = hf Where: E is the energy of the photon (in Joules, J) h is Planck's constant (6.63 x 10⁻³⁴ J.s) f is the frequency of the light (in Hertz, Hz) When a photon strikes the metal surface, it transfers its energy to an electron. To escape the metal, an electron needs to overcome the attractive forces holding it within the metal. This minimum energy required is called the work function (W₀) of the metal. The work function is also related to the threshold frequency by: W₀ = hf₀ Where: W₀ is the work function of the metal (in Joules, J) h is Planck's constant (6.63 x 10⁻³⁴ J.s) f₀ is the threshold frequency (in Hertz, Hz) If the energy of the photon (E = hf) is greater than the work function (W₀), the electron is ejected with a certain kinetic energy (KEmax). The remaining energy from the photon is converted into the kinetic energy of the emitted electron. This can be expressed by Einstein's photoelectric equation: E = W₀ + KEmax Therefore: hf = hf₀ + 1/2 mv²max Where: KEmax is the maximum kinetic energy of the emitted electron (in Joules, J) m is the mass of the electron (9.11 x 10⁻³¹ kg) vmax is the maximum velocity of the emitted electron (in m/s) 2.2 Worked Examples Example 1: A metal has a work function of 3.2 x 10⁻¹⁹ J. Light with a frequency of 8.0 x 10¹⁴ Hz shines on the metal. a) Calculate the energy of a photon of the incident light. b) Calculate the maximum kinetic energy of the emitted photoelectrons. c) Calculate the maximum velocity of the emitted photoelectrons.

Solution: a) E = hf = (6.63 x 10⁻³⁴ J.s)(8.0 x 10¹⁴ Hz) = 5.30 x 10⁻¹⁹ J b) KEmax = E - W₀ = 5.30 x 10⁻¹⁹ J - 3.2 x 10⁻¹⁹ J = 2.10 x 10⁻¹⁹ J c) KEmax = 1/2 mv²max 2.10 x 10⁻¹⁹ J = 1/2 (9.11 x 10⁻³¹ kg) v²max v²max = (2 x 2.10 x 10⁻¹⁹ J) / (9.11 x 10⁻³¹ kg) = 4.61 x 10¹¹ m²/s² vmax = √(4.61 x 10¹¹) m²/s² = 6.79 x 10⁵ m/s Example 2: The threshold frequency for a certain metal is 5.0 x 10¹⁴ Hz. What is the work function of the metal?

Solution: W₀ = hf₀ = (6.63 x 10⁻³⁴ J.s)(5.0 x 10¹⁴ Hz) = 3.32 x 10⁻¹⁹ J Example 3: Light with a wavelength of 400 nm shines on a metal surface. The work function of the metal is 2.5 eV. (1 eV = 1.6 x 10⁻¹⁹ J) a) Calculate the energy of a photon of the incident light (in Joules). b) Calculate the maximum kinetic energy of the emitted photoelectrons (in Joules).

Solution: First, we need to convert the wavelength to frequency using the relationship c = fλ, where c is the speed of light (3.0 x 10⁸ m/s). f = c / λ = (3.0 x 10⁸ m/s) / (400 x 10⁻⁹ m) = 7.5 x 10¹⁴ Hz a) E = hf = (6.63 x 10⁻³⁴ J.s)(7.5 x 10¹⁴ Hz) = 4.97 x 10⁻¹⁹ J b) First, convert the work function to Joules: W₀ = 2.5 eV * 1.6 x 10⁻¹⁹ J/eV = 4.0 x 10⁻¹⁹ J KEmax = E - W₀ = 4.97 x 10⁻¹⁹ J - 4.0 x 10⁻¹⁹ J = 9.7 x 10⁻²⁰ J Guided Practice (With Solutions)

Question 1: A certain metal has a work function of 4.0 x 10⁻¹⁹

J. What is the threshold frequency for this metal?

Solution: W₀ = hf₀ f₀ = W₀ / h = (4.0 x 10⁻¹⁹ J) / (6.63 x 10⁻³⁴ J.s) = 6.03 x 10¹⁴ Hz

Commentary: This question directly applies the work function equation to calculate the threshold frequency. Ensure students understand the units and the relationship between work function and threshold frequency.

Question 2: Light with a frequency of 9.0 x 10¹⁴ Hz shines on a metal surface. The maximum kinetic energy of the emitted photoelectrons is 1.5 x 10⁻¹⁹ J. What is the work function of the metal?