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Chapter 38: Problem 8

The photoelectric work function of potassium is \(2.3 \mathrm{eV}\). If light that has a wavelength of \(190 \mathrm{nm}\) falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speed of these electrons.

Short Answer

Expert verified
The stopping potential of the most energetic electrons is calculated to be the difference between the energy of the incident light and the work function of potassium. The kinetic energy of the most energetic electrons is calculated from this potential difference, and the speed of those electrons is calculated from the kinetic energy. The specific values would depend on the precise values and units used for the constants in each formula.

Step by step solution

01

Calculation of Energy of the Incident Light

Determine the energy \(E_{light}\) of the incident light using the formula \(E_{light} = \frac{hc}{\lambda}\), where \(h\) is Planck's constant (\(6.626 \times 10^{-34} m^2 kg/s\)), \(c\) is the speed of light (\(3 \times 10^8 m/s\)), and \(\lambda\) is the wavelength of the light (\(190 nm = 190 \times 10^-9 m\)). Note that this energy is in Joules, to convert it to electron volts divide by \(1.602 \times 10^{-19} J/eV\).
02

Calculation of Stopping Potential and Maximum Kinetic Energy

Determine the stopping potential \(V_{stop}\) and the maximum kinetic energy of the photoelectrons using the photoelectric equation \(eV_{stop} = E_{light} - \phi\), where \(e\) is the charge of an electron (\(1.602 \times 10^{-19} Coulombs\)) and \(\phi\) is the work function given in the question \(2.3eV\). Rearrange the equation for \(V_{stop}\), and calculate it. The maximum kinetic energy is given by \(eV_{stop}\) and is equal to the energy of the light \(E_{light}\) minus the work function \(\phi\).
03

Calculation of Speed of Ejected Electrons

Determine the speed \(v\) of the most energetic ejected electrons using the formula \(K.E. = \frac{1}{2}mv^2\), where \(m\) is the mass of an electron \(9.11 \times 10^{-31}kg\) and \(K.E.\) is the kinetic energy calculated in step 2. Rearrange the formula and solve for \(v\).

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Key Concepts

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

Work Function
The work function, denoted by the Greek letter \( \phi \), is the minimum energy required to eject an electron from the surface of a material. In the context of the photoelectric effect, the work function is an essential factor.
  • It represents a threshold energy that incoming photons must exceed to release electrons from the material.
  • The work function varies between different materials, making some more suitable for photoelectric applications than others.
  • It is usually measured in electron volts (eV), providing a convenient way to express this small amount of energy.
For potassium, the work function given is \( 2.3 \, \mathrm{eV} \). This means any photon must have at least this much energy to eject electrons from a potassium surface.
Photoelectric Equation
The photoelectric equation is the cornerstone of understanding the photoelectric effect. This equation connects the energy of incoming photons with the energy needed to eject electrons from a material. It is succinctly represented by:\[E_{light} = \phi + K.E._{max}\]
  • \(E_{light}\) is the energy of the incident light photons.
  • \(\phi\) is the work function, or the energy needed to remove an electron from the material.
  • \(K.E._{max}\) is the maximum kinetic energy that the ejected electrons can have.
This equation helps to calculate not only whether photoemission will occur but also the energy surplus that translates into kinetic energy of the emitted electrons.
Kinetic Energy
Kinetic energy in the photoelectric effect refers to the leftover energy of the electrons after surpassing the work function.
  • The maximum kinetic energy \( K.E._{max} \) can be calculated using:\[K.E._{max} = E_{light} - \phi\]
  • This tells us how much energy the ejected electrons have for movement.
  • In this problem, obtaining the energy of the incident light in electron volts is crucial since the given work function is also in electron volts.
Thus, kinetic energy not only informs us about the success of photoemission but also about the behavior of electrons once they are free from their original material substate.
Wavelength
Wavelength is a crucial factor for determining the energy of the incident light in the photoelectric effect.
  • The energy of a photon is inversely proportional to its wavelength, described by the equation:\[E_{light} = \frac{hc}{\lambda}\]
  • Where \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \mathrm{m^2 kg/s}\), and \(c\) is the speed of light \(3 \times 10^8 \, \mathrm{m/s}\).
  • A shorter wavelength corresponds to higher energy and vice versa.
In this exercise, the given wavelength \(190 \, \mathrm{nm}\) is converted to meters to plug into the formula, helping to ascertain whether the photons have enough energy to overcome the work function of potassium.
Electron Volts
Electron volts (eV) are a practical unit of energy used often in atomic and subatomic physics because they are more suitable for measuring small energy scales.
  • An electron volt is defined as the amount of kinetic energy gained or lost by a single electron moving across an electric potential difference of one volt.
  • In Joules, one electron volt equals \(1.602 \times 10^{-19} \, \mathrm{J}\).
  • This conversion is crucial in photoelectric effect calculations where energies in Joules are often converted to electron volts to simplify calculations and relate to the work function provided in eV.
Using electron volts as the unit allows for a direct comparison of photon energies and the work function, facilitating a smoother, more intuitive understanding of the ejected electrons' behavior and energy dynamics.

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Most popular questions from this chapter

A laser used to weld detached retinas emits light with a wavelength of \(652 \mathrm{nm}\) in pulses that are \(20.0 \mathrm{~ms}\) in duration. The average power during each pulse is \(0.600 \mathrm{~W}\). (a) How much energy is in each pulse in joules? In electron volts? (b) What is the energy of one photon in joules? In electron volts? (c) How many photons are in each pulse?

When ultraviolet light with a wavelength of \(400.0 \mathrm{nm}\) falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be \(1.10 \mathrm{eV}\). What is the maximum kinetic energy of the photoelectrons when light of wavelength \(300.0 \mathrm{nm}\) falls on the same surface?

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about \(1 \mathrm{MeV}\left(10^{6} \mathrm{eV}\right) .\) By contrast, what we see emanating from the sun's surface are visiblelight photons with wavelengths of about \(500 \mathrm{nm}\). A simple model that explains this difference in wavelength is that a photon undergoes Compton scattering many times-in fact, about \(10^{26}\) times, as suggested by models of the solar interior - as it travels from the center of the sun to its surface. (a) Estimate the increase in wavelength of a photon in an average Compton-scattering event. (b) Find the angle in degrees through which the photon is scattered in the scattering event described in part (a). (Hint: A useful approximation is \(\cos \phi \approx 1-\phi^{2} / 2,\) which is valid for \(\phi \ll 1 .\) Note that \(\phi\) is in radians in this expression.) (c) It is estimated that a photon takes about \(10^{6}\) years to travel from the core to the surface of the sun. Find the average distance that light can travel within the interior of the sun without being scattered. (This distance is roughly equivalent to how far you could see if you were inside the sun and could survive the extreme temperatures there. As your answer shows, the interior of the sun is very opaque.)

A \(2.50 \mathrm{~W}\) beam of light of wavelength \(124 \mathrm{nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is \(4.16 \mathrm{eV}\). Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

An ultrashort pulse has a duration of \(9.00 \mathrm{fs}\) and produces light at a wavelength of \(556 \mathrm{nm}\). What are the momentum and momentum uncertainty of a single photon in the pulse?
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