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

Expert verified
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

01

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 \(f\): \(f = \frac{0.310e + 2.1e}{h} = \frac{2.41e}{6.626 \times 10^{-34} \mathrm{Js}} \approx 3.63 \times 10^{15}\,\mathrm{Hz}\)
02

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 \(1.90\,\mathrm{eV}\).
03

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.

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