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A photoelectric cell is illuminated with green light. How will moving the light source farther away affect (a) the stopping potential; (b) the current through the cell?

Short Answer

Expert verified
(a) The stopping potential remains unchanged; (b) the current decreases.

Step by step solution

01

Understanding the Photoelectric Effect

The photoelectric effect occurs when light of a certain frequency shines on a material and ejects electrons from its surface. These electrons are freed with a kinetic energy given by the difference between the energy of the photons and the work function of the material. The stopping potential is the potential needed to stop these electrons from reaching the other side of the cell.
02

Analyzing Stopping Potential

The stopping potential is determined by the maximum kinetic energy of the photoelectrons, calculated by the equation: \[ KE_{ ext{max}} = e imes V_{ ext{stop}} = hf - ext{work function} \]where \(h\) is Planck's constant, \(f\) is the frequency of the light, and \(e\) is the electron charge. Since moving the light source affects the intensity but not the wavelength (or frequency), the stopping potential remains unchanged because it depends on the frequency, not the intensity.
03

Examining the Effect on Current

The current through the cell depends on the number of photoelectrons ejected, which in turn depends on the intensity of light. As the light source is moved farther away, the light intensity decreases, resulting in fewer photoelectrons and thus a reduced current. Therefore, the current through the cell will decrease if the light source is moved away.

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Key Concepts

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

Stopping Potential
When light shines onto a material in a photoelectric cell, electrons can be ejected, depending on the energy of the light. To understand stopping potential, picture a tiny barrier that prevents these ejected electrons from reaching the other side of the cell. This potential, known as the stopping potential, represents the minimum energy needed to stop every ejected electron.

It's calculated using the formula:\[KE_{\text{max}} = e \times V_{\text{stop}} = hf - \text{work function}\]Here, \(KE_{\text{max}}\) is the maximum kinetic energy of the photoelectrons, \(hf\) is the energy of the incident photons, and \(e\) is the electron charge. While intensity changes may impact how many electrons are ejected, the stopping potential remains unchanged because it is determined by the light's frequency. Thus, even if you move the light source further away, the stopping potential does not change.
Kinetic Energy of Photoelectrons
The kinetic energy of photoelectrons is crucial in understanding the photoelectric effect. When photons from light hit the material, they transfer energy to electrons, allowing them to escape. The energy transferred is defined by the equation:\[KE = hf - \text{work function}\]where \(KE\) is the kinetic energy of the photoelectrons.

In this equation, \(hf\) represents the energy of the incoming photons, and the term 'work function' is the minimum energy needed to free an electron from the material. This means the kinetic energy of photoelectrons relies heavily on the frequency of the light, which impacts their velocity once ejected. The change in distance from the light source doesn't alter their kinetic energy because it does not change the frequency of light.
Light Intensity
Light intensity relates to the number of photons hitting a surface within a specific timeframe. It's about how bright the light source appears to the material. A key aspect of the photoelectric effect is how this intensity affects the number of photoelectrons ejected from the material.
  • Increased intensity means more photons are hitting the material, thus freeing more electrons.
  • Decreased intensity results in fewer electrons being ejected because fewer photons are available.
If the light source moves farther away, the intensity decreases due to the spreading of light over a larger area. Consequently, this leads to a reduced current in the photoelectric cell because fewer electrons are ejected.
Work Function
The work function is a fundamental concept to grasp in the study of the photoelectric effect. It refers to the minimum energy necessary to eject an electron from the surface of a material. This value is intrinsic to each material and affects how easily electrons can be emitted when exposed to light.

It's important to note that the work function provides a threshold for the photon energy required to liberate an electron. If the energy of the incoming photons, denoted as \(hf\), is less than the work function, no photoelectrons are emitted. On the flip side, if \(hf\) exceeds the work function, then photoelectrons are released, and any extra energy after overcoming the work function is transferred into the kinetic energy of these electrons. Understanding the work function helps explain why different materials respond differently under identical lighting conditions.

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Most popular questions from this chapter

The work function of sodium is \(3.12 \cdot 10^{-19} \mathrm{~J}\). A sodium lamp emits yellow light with a wavelength of \(590 \mathrm{~nm}\). Calculate the stopping potential of the electrons emitted from a sodium plate illuminated by a sodium lamp.

When an electron penetrates the surface of a metal, its energy changes as a result of the metal's work function. a. Does an electron's energy increase or decrease when it penetrates a metal's surface? b. In the Davisson and Germer experiment, the electrons were accelerated through low potentials; \(54 \mathrm{~V}\) was a typical value. Calculate the velocity of the electrons before they entered the nickel crystal and after entering on the assumption that the work function of nickel is \(2.6 \mathrm{eV}\). c. Calculate the wavelengths of the electrons before and after they penetrate the surface.

Explain the following: a. An insulated metal plate that is illuminated with ultra-violet light emits electrons for just a short time; the emission then ceases. b. Distant stars and galaxies could not be photographed were light not of particle character.

A \(100 \mathrm{~mW}\) argon laser emits a beam of light of wavelength \(488 \mathrm{~nm}\) in vacuo. a. How many photons does the laser emit each second b. What current will flow through a photoelectric cell whose cathode (emitting. plate) is illuminated by the laser's light beam if \(10 \%\) of the photons release an electron from it. c. What potential will be required between the cell's electrodes in order to stop the current, if the threshold frequency of the cathode is \(5.2 \cdot 10^{14} \mathrm{~Hz}\).

A beam of X-rays is scattered by free electrons. At a scattering angle of \(45^{\circ}\) the wavelength of the scattered beam is \(2.2 \cdot 10^{-12} \mathrm{~m}\). Calculate the wavelength of the incident beam.

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