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Chapter 36: Problem 30

You illuminate a zinc surface with 550 -nm light. How high do you have to turn up the stopping voltage to squelch the photoelectric current completely?

Short Answer

Expert verified
Answer: The stopping voltage required to completely stop the photoelectric current is 0 volts.

Step by step solution

01

Identify the known values and required value

We are given the following information: - Wavelength of the light (\(\lambda\)) = 550 nm - Work function of zinc (\(\phi\)) = 4.3 eV (This value can be found in reference tables) We are asked to find the stopping voltage (\(V_s\)).
02

Convert wavelength to frequency

To convert the wavelength of the light to its frequency (\(f\)), we can use the following equation: \(f = \frac{c}{\lambda}\) Where \(c\) is the speed of light (\(3.00 \times 10^8 \, m/s\)) and \(\lambda\) is the wavelength of the light. Note that we need to convert the wavelength to meters: \(\lambda = 550 \, nm = 550 \times 10^{-9} \, m\) Now, we can calculate the frequency, \(f\): \(f = \frac{3.00 \times 10^8 \, m/s}{550 \times 10^{-9} \, m} = 5.45 \times 10^{14} \, Hz\)
03

Calculate the energy of the incident photons

Now we can calculate the energy of the incident photons (\(E\)) using the Planck's equation: \(E = h \times f\) Where \(h\) is the Planck's constant (\(6.63 \times 10^{-34} \, Js\)) and \(f\) is the frequency of the light. Now we can calculate \(E\): \(E = (6.63 \times 10^{-34} \, Js)(5.45 \times 10^{14} \, Hz) = 3.61 \times 10^{-19} \, J\) To make our calculations easier, we can convert this energy to electron volts (eV): \(E = 3.61 \times 10^{-19} \, J \times \frac{1 \, eV}{1.60 \times 10^{-19} \, J} = 2.26 \, eV\)
04

Calculate the maximum kinetic energy of the ejected electrons

The photoelectric effect equation relates the work function of zinc, the energy of the incident photons, and the maximum kinetic energy of the ejected electrons (\(K_{max}\)): \(K_{max} = E - \phi\) Now, we plug in the values for \(E\) and \(\phi\): \(K_{max} = 2.26 \, eV - 4.3 \, eV = -2.04 \, eV\) The negative value indicates that the energy of the incident photons is not sufficient to eject electrons from the zinc surface, hence no photoelectric current is generated.
05

Stopping voltage calculation

Since there is no photoelectric current, we do not need to apply any stopping voltage to squelch the photoelectric current. Therefore, the stopping voltage required is: \(V_s = 0\) So, the stopping voltage required to squelch the photoelectric current completely is 0 volts.

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Most popular questions from this chapter

What would a classical physicist expect would be the result of shining a brighter UV lamp on a metal surface, in terms of the energy of emitted electrons? How does this differ from what the theory of the photoelectric effect predicts?

To have a larger photocurrent, which of the following should occur? (select all the correct changes) a) brighter light c) higher frequency b) dimmer light d) lower frequency

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Scintillation detectors for gamma rays transfer the energy of a gamma-ray photon to an electron within a crystal, via the photoelectric effect or Compton scattering. The electron transfers its energy to atoms in the crystal, which re-emit it as a light flash detected by a photomultiplier tube. The charge pulse produced by the photomultiplier tube is proportional to the energy originally deposited in the crystal; this can be measured so an energy spectrum can be displayed. Gamma rays absorbed by the photoelectric effect are recorded as a photopeak in the spectrum, at the full energy of the gammas. The Compton-scattered electrons are also recorded, at a range of lower energies known as the Compton plateau. The highest-energy of these form the Compton edge of the plateau. Gamma-ray photons scattered \(180 .^{\circ}\) by the Compton effect appear as a backscatter peak in the spectrum. For gamma-ray photons of energy \(511 \mathrm{KeV}\) calculate the energies of the Compton edge and the backscatter peak in the spectrum.
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