Question

In a photoelectric effect experiment, a laser beam of unknown wavelength is shined on a cesium cathode (work function \(\phi=2.100 \mathrm{eV}\) ). It is found that a stopping potential of \(0.310 \mathrm{~V}\) is needed to eliminate the current. Next, the same laser is shined on a cathode made of an unknown material, and a stopping potential of \(0.110 \mathrm{~V}\) is found to be needed to eliminate the current. a) What is the work function for the unknown cathode? b) What would be a possible candidate for the material of this unknown cathode?

Short answer

Answer: The work function of the unknown cathode is approximately 1.90 eV. A possible candidate for the material of the unknown cathode is potassium (K), which has a work function of approximately 1.92 eV.

Step-by-step solution

Find the frequency of the light

First, use the equation for the stopping potential and work function to find the frequency of the light: \(0.310V = \frac{(h \times f) - (2.1eV)}{e}\) Solving for \(f\): \(f = \frac{0.310e + 2.1e}{h} = \frac{2.41e}{6.626 \times 10^{-34} \mathrm{Js}} \approx 3.63 \times 10^{15}\,\mathrm{Hz}\)

Find the work function of the unknown cathode

Now that we have the frequency of the light, we can use the stopping potential for the unknown material to find its work function: \(0.110 \mathrm{V} = \frac{(h \times f) - \phi_\mathrm{unknown}}{e}\) Solving for \(\phi_\mathrm{unknown}\): \(\phi_\mathrm{unknown} = (h \times f) - (0.110e) \approx 1.90\,\mathrm{eV}\) So, the work function for the unknown cathode is approximately \(1.90\,\mathrm{eV}\).

Suggest a possible candidate for the material of the unknown cathode

Based on the work function value, a possible candidate for the material of the unknown cathode is potassium (K), which has a work function of approximately \(1.92\,\mathrm{eV}\). Keep in mind, there are other materials with similar work functions, but potassium is one possible option given its close proximity to the calculated work function value.

Key concepts

Work Function

The work function is a crucial concept in the photoelectric effect. It represents the minimum energy required to remove an electron from the surface of a material. In this context, it's often given in electron volts (eV).
For example, in our exercise, cesium has a known work function of 2.100 eV. This value is essential when calculating other properties of the system, like stopping potential and frequency of light.
This energy must be overcome for electrons to be liberated. Think of it as an energy barrier or a threshold. The higher the work function, the more energy is needed to free the electrons.

• The work function can vary depending on factors like the material's composition, surface structure, and temperature.
• This property is critical for designing devices that rely on the photoelectric effect, such as photodetectors and solar cells.

Stopping Potential

Stopping potential is another key component of the photoelectric effect experiment. This is the potential difference needed to stop the photoemitted electrons from reaching the anode. When the stopping potential reaches a certain level, the current stops.
In the given scenario, for cesium, a stopping potential of 0.310 V is required to stop the current. This value changes when the material of the cathode changes due to the difference in work functions.

• The stopping potential provides insight into the kinetic energy of the ejected electrons. This energy is directly influenced by the energy of incoming photons minus the work function.
• Practically, understanding and calculating stopping potential helps fine-tune electronic properties in devices, ensuring they function efficiently.

Frequency of Light

The frequency of light is a fundamental aspect of the photoelectric effect. It is directly related to the energy of the photons striking the material. The relationship is expressed through Planck's equation: \( E = h \times f \) where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is the frequency.
In our problem, the light's frequency was determined using the stopping potential and the known work function of cesium. This frequency is approximately \( 3.63 \times 10^{15} \) Hz.
Knowing the frequency allows for a deeper understanding of the interactions between light and matter, particularly how photons transfer energy to electrons. Remember, not all frequencies are sufficient to overcome the work function.

• Higher frequency means more energetic photons, which are more likely to eject electrons if other conditions permit.
• Calculating light's frequency helps in identifying suitable materials for specific applications based on their work functions.