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Chapter 36: Problem 26

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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Most popular questions from this chapter

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?

Consider a system made up of \(N\) particles. The average energy per particle is given by \(\langle E\rangle=\left(\sum E_{i} e^{-E_{i} / k_{B} T}\right) / Z\) where \(Z\) is the partition function defined in equation \(36.29 .\) If this is a two-state system with \(E_{1}=0\) and \(E_{2}=E\) and \(g_{1}=\) \(g_{2}=1,\) calculate the heat capacity of the system, defined as \(N(d\langle E\rangle / d T)\) and approximate its behavior at very high and very low temperatures (that is, \(k_{\mathrm{B}} T \gg 1\) and \(k_{\mathrm{B}} T \ll 1\) ).

You are performing a photoelectric effect experiment. Using a photocathode made of cesium, you first illuminate it with a green laser beam \((\lambda=514.5 \mathrm{nm})\) of power \(100 \mathrm{~mW}\). Next, you double the power of your laser beam, to \(200 \mathrm{~mW}\). How will the energies per electron of the electrons emitted by the cathode compare for the two cases?

Now consider de Broglie waves for a (relativistic) particle of mass \(m\), momentum \(p=m v \gamma\), and total energy \(E=m c^{2} \gamma\), with \(\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}\). The waves have wavelength \(\lambda=h / p\) and frequency \(f=E / h\) as before, but with the relativistic momentum and energy. a) Calculate the dispersion relation for these waves. b) Calculate the phase and group velocities of these waves. Now which corresponds to the classical velocity of the particle?

What is the minimum uncertainty in the velocity of a 1.0 -nanogram particle that is at rest on the head of a 1.0 -mm-wide pin?
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