Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define the photoelectric effect and its significance.
• Describe the relationship between light frequency and electron energy.
• Calculate threshold frequency, work function, and kinetic energy.
• Explain the effect of light intensity and stopping potential.
• Relate the effect to practical applications like solar cells.

Lesson summary

The photoelectric effect is a cornerstone of quantum mechanics, providing evidence that light, which we often perceive as a wave, can also behave as a particle (photon). Understanding this duality is crucial for comprehending the nature of light and matter. In South Africa, this knowledge is essential for advancements in fields like solar energy, medical imaging, and telecommunications. This lesson will explore the fundamentals of the photoelectric effect, focusing on its underlying principles, mathematical relationships, and real-world applications.

Lesson notes

2.1 The Photoelectric Effect The photoelectric effect is the emission of electrons from a metal surface when light of a sufficiently high frequency shines on it. These emitted electrons are called photoelectrons. A crucial observation is that this emission only occurs when the frequency of the incident light is above a certain threshold frequency (f₀), which is specific to each metal. Increasing the intensity of light above this threshold frequency increases the number of photoelectrons emitted, but does not increase their kinetic energy. 2.2 Key Terms and Definitions Photon: A quantum of electromagnetic radiation; a "packet" of energy. The energy of a photon is given by E = hf, where h is Planck's constant (6.63 x 10⁻³⁴ J.s) and f is the frequency of the radiation. Work Function (W₀): The minimum energy required to remove an electron from the surface of a particular metal. This is a characteristic property of the metal. Threshold Frequency (f₀): The minimum frequency of light required to eject photoelectrons from a metal surface.

This corresponds to the work function: W₀ = hf₀.

Kinetic Energy (Ek): The energy of motion of the emitted photoelectrons. The maximum kinetic energy of the photoelectrons is given by: Ek(max) = hf - W₀.

Stopping Potential (Vs): The potential difference required to stop the most energetic photoelectrons from reaching the anode in a photoelectric experiment. The work done by the stopping potential equals the maximum kinetic energy of the photoelectrons: eVs = Ek(max), where 'e' is the elementary charge (1.6 x 10⁻¹⁹ C).

Intensity: The power of the light per unit area. The intensity of light is proportional to the number of photons incident on the surface per unit time. 2.3 Einstein's Photoelectric Equation Einstein explained the photoelectric effect by proposing that light is quantized into photons. The energy of a single photon is given by E = hf. When a photon strikes a metal surface, it transfers its energy to an electron. If the photon's energy (hf) is greater than the work function (W₀) of the metal, the electron is ejected with a kinetic energy Ek(max). The excess energy, i.e., the energy exceeding the work function, becomes the kinetic energy of the emitted electron.

The equation is: Ek(max) = hf - W₀ This equation highlights the following: Kinetic energy is directly proportional to the frequency of incident light. There is a minimum frequency (threshold frequency) below which no electrons are emitted, regardless of the intensity of the light. The intensity of light determines the number of photoelectrons emitted (current), not their kinetic energy. 2.4 Worked Examples Example 1: A metal has a work function of 3.0 eV. Light with a wavelength of 400 nm is shone on the metal. Calculate the energy of the incident photons in Joules. Calculate the maximum kinetic energy of the emitted photoelectrons in Joules. Calculate the stopping potential required to stop the emitted photoelectrons.

Solution: Energy of the photon (E = hf): First, calculate the frequency of the light: c = fλ, so f = c/λ f = (3 x 10⁸ m/s) / (400 x 10⁻⁹ m) = 7.5 x 10¹⁴ Hz E = hf = (6.63 x 10⁻³⁴ J.s) x (7.5 x 10¹⁴ Hz) = 4.97 x 10⁻¹⁹ J Maximum kinetic energy (Ek(max) = hf - W₀): Convert the work function from eV to Joules: W₀ = 3.0 eV x (1.6 x 10⁻¹⁹ J/eV) = 4.8 x 10⁻¹⁹ J Ek(max) = 4.97 x 10⁻¹⁹ J - 4.8 x 10⁻¹⁹ J = 0.17 x 10⁻¹⁹ J Stopping Potential (eVs = Ek(max)): Vs = Ek(max) / e = (0.17 x 10⁻¹⁹ J) / (1.6 x 10⁻¹⁹ C) = 0.106 V Example 2: A certain metal has a threshold frequency of 5.0 x 10¹⁴ Hz. Calculate the work function of the metal in Joules. If light with a frequency of 8.0 x 10¹⁴ Hz is shone on the metal, what is the maximum velocity of the emitted photoelectrons?

Solution: Work Function (W₀ = hf₀): W₀ = (6.63 x 10⁻³⁴ J.s) x (5.0 x 10¹⁴ Hz) = 3.315 x 10⁻¹⁹ J Maximum Velocity (Ek(max) = ½mv²): First, calculate the maximum kinetic energy: Ek(max) = hf - W₀ Ek(max) = (6.63 x 10⁻³⁴ J.s) x (8.0 x 10¹⁴ Hz) - 3.315 x 10⁻¹⁹ J Ek(max) = 5.304 x 10⁻¹⁹ J - 3.315 x 10⁻¹⁹ J = 1.989 x 10⁻¹⁹ J Now, use Ek(max) = ½mv² to find v: v = √(2Ek(max) / m) where m is the mass of an electron (9.11 x 10⁻³¹ kg) v = √(2 x 1.989 x 10⁻¹⁹ J / 9.11 x 10⁻³¹ kg) = √(4.367 x 10¹¹) = 6.61 x 10⁵ m/s Example 3: The work function of potassium is 2.3 eV. Calculate the longest wavelength of light that will cause photoelectric emission from potassium.

Solution: Relationship between wavelength and frequency: We know that W₀ = hf₀ and c = f₀λ₀, where λ₀ is the threshold wavelength.

Therefore, f₀ = c/λ₀. Substituting for f₀, we get W₀ = hc/λ₀. We need to find λ₀. Solve for λ₀: λ₀ = hc/W₀ First convert work function to Joules: W₀ = 2.3 eV * 1.6 x 10⁻¹⁹ J/eV = 3.68 x 10⁻¹⁹ J λ₀ = (6.63 x 10⁻³⁴ J.s * 3 x 10⁸ m/s) / 3.68 x 10⁻¹⁹ J = 5.4 x 10⁻⁷ m = 540 nm Guided Practice (With Solutions)

Question 1: A metal has a work function of 2.5 e

V. What is the threshold frequency for this metal?

Solution: We know that W₀ = hf₀. Rearranging for f₀, we get f₀ = W₀/h.