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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stopping Voltage
The stopping voltage in the photoelectric effect is a crucial concept used to halt the flow of photoelectric current. When light shines on a metal surface, it can eject electrons if the light's energy is sufficient. These ejected electrons create a current. To stop this current, a voltage, specifically the stopping voltage, is applied in the opposite direction of the current.

It's important to note that if the energy of the incident light is less than the work function of the metal, as in the example provided, no electrons are ejected, and hence no stopping voltage is needed. In this case, the stopping voltage is zero. This particular example illustrates the situation where the photon energy is less than the work function of zinc, showing that stopping voltage calculations depend on these conditions.

Understanding stopping voltage helps us delve into the heart of how energy interactions on a subatomic level can affect observable electrical phenomena.
Work Function
The work function is a key concept in the study of the photoelectric effect. It represents the minimum energy required to eject an electron from a metal's surface when it's illuminated by light. Different metals have different work functions, which is a fixed value for each material.
  • The work function of zinc as given in the problem is 4.3 eV.
  • This value shows the energy threshold that the incident photons must overcome to release electrons from the zinc surface.

If the light’s energy, calculated from its frequency and Planck’s equation, is below this threshold, no electrons will be emitted. This leads to no photoelectric current, as seen in this example. The concept of work function is critical as it’s directly related to the characteristics and quality of the material and plays a vital role in applications involving photoelectric devices.
Planck's Equation
Planck's equation is fundamental in understanding the photoelectric effect as it connects the frequency of incident light to its energy. The equation is expressed as:

\[ E = h \times f \]

Where:
  • \( E \) is the energy of a photon,
  • \( h \) is Planck's constant \( (6.63 \times 10^{-34} \, Js) \),
  • \( f \) is the frequency of the light.
By using Planck's equation, we can determine whether the light shining on a metal surface has enough energy to eject an electron (overcoming the work function). In the given problem, we found that the energy calculated from the wavelength of 550 nm was approximately 2.26 eV.

This energy was found to be insufficient to overcome the work function of zinc, which was 4.3 eV. This calculation clearly demonstrates how Planck's equation provides the framework for connecting light properties to the energy interactions observed in the photoelectric effect. Understanding and applying Planck’s equation is essential for predicting and analyzing phenomena in quantum and classical physics.

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Most popular questions from this chapter

Consider an electron whose de Broglie wavelength is equal to the wavelength of green light (about \(550 \mathrm{nm}\) ). a) Treating the electron nonrelativistically, what is its speed? b) Does your calculation confirm that a nonrelativistic treatment is sufficient? c) Calculate the kinetic energy of the electron in \(\mathrm{eV}\).

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?

If I look in a mirror while wearing a blue shirt, I see a blue shirt in my reflection, not a red shirt. But according to the Compton effect, the photons that bounce back should have a lower energy and therefore a longer wavelength. Explain why my reflection shows the same color shirt as am wearing.

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Consider an object at room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right)\) and the radiation it emits. For radiation at the peak of the spectral energy density, calculate a) the wavelength, c) the energy of one b) the frequency, and photon.

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