The work function of a certain material is \(5.8 \mathrm{eV}\). What is the photoelectric threshold for this material?

Short Answer

Expert verified
Answer: The photoelectric threshold frequency for the material is approximately \(1.403 \times 10^{15} \mathrm{Hz}\).

Step by step solution

01

Express the work function as energy in Joules

To find the threshold frequency, we will first need to convert the given work function from electron volts (eV) to Joules (J). We can use the conversion factor, which is \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\).
02

Calculate the energy in Joules

Now, let's multiply the given work function (\(5.8 \mathrm{eV}\)) by the conversion factor to find the energy in Joules. Energy (J) = Work function (eV) × Conversion factor Energy (J) = \(5.8 \times 1.602 \times 10^{-19} \mathrm{J/eV}\)
03

Find the photoelectric threshold frequency

Using Planck's equation, we can find the photoelectric threshold frequency, where \(E = h\nu\). Here, \(E\) is the energy in Joules, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), and \(\nu\) is the threshold frequency. We need to find the value of \(\nu\), so we can rearrange Planck's equation to solve for \(\nu\): \(\nu = \frac{E}{h}\) Now, we can substitute the values of \(E\) and \(h\) and compute the threshold frequency: \(\nu = \frac{5.8 \times 1.602 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}\)
04

Solve for the threshold frequency

After calculating the expression, we obtain the photoelectric threshold frequency for the material: \(\nu \approx 1.403 \times 10^{15} \mathrm{Hz}\) This is the minimum frequency of incident light that can release electrons from the material.

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