Sure, here's an AP Physics 1 exam question related to the photoelectric effect:

Question: A metal surface is illuminated with light of different frequencies. For a given metal, the following data is obtained:

• Frequency of incident light (f1): 5.0 x 10^14 Hz
• Stopping potential (V1): 1.5 V
• Frequency of incident light (f2): 7.0 x 10^14 Hz
• Stopping potential (V2): 2.0 V

a) Calculate the work function of the metal. b) Determine the threshold frequency of the metal. c) Explain the observed relationship between the frequency of the incident light and the stopping potential.

Answer:

a) The work function (Φ) can be determined by using the equation:

Φ = hf - eV

Where:

• h = Planck's constant (6.626 x 10^-34 Js)
• f = frequency of incident light
• e = elementary charge (1.602 x 10^-19 C)
• V = stopping potential

Using the first set of data: Φ = (6.626 x 10^-34 Js)(5.0 x 10^14 Hz) - (1.602 x 10^-19 C)(1.5 V) Φ ≈ 3.313 x 10^-19 J - 2.403 x 10^-19 J Φ ≈ 0.91 x 10^-19 J

b) The threshold frequency (f₀) can be calculated using the equation:

Φ = hf₀

Rearranging the equation gives: f₀ = Φ / h

Using the value of Φ: f₀ = (0.91 x 10^-19 J) / (6.626 x 10^-34 Js) f₀ ≈ 1.37 x 10^15 Hz

c) The observed relationship between the frequency of the incident light and the stopping potential can be explained by the photoelectric effect. According to the photoelectric effect, the kinetic energy of emitted electrons is proportional to the frequency of the incident light. Increasing the frequency of the incident light leads to an increase in the maximum kinetic energy of emitted electrons, which in turn causes an increase in the stopping potential required to prevent the emission of electrons. This is consistent with the observed relationship between the frequency of incident light and the stopping potential in the given data.

Hope this helps.