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Chapter 38: Problem 33

The work function of tungsten is \(4.50 \mathrm{eV}\). Calculate the speed of the fastest electrons ejected from a tungsten surface when light whose photon energy is \(5.80 \mathrm{eV}\) shines on the surface.

Short Answer

Expert verified
The speed of the fastest ejected electrons is approximately \(5.98 \times 10^5 \mathrm{m/s}\).

Step by step solution

01

Understand the Concepts

The problem involves the photoelectric effect, where electrons are ejected from a metal surface by photons. The work function, \( \Phi \), is the minimum energy required to remove an electron from the surface. The kinetic energy of the ejected electron is the photon energy minus the work function.
02

Set Up the Equation

Use the equation for the kinetic energy of photoelectrons, which is:\[ K = E_{photon} - \Phi \]where \( K \) is the maximum kinetic energy of the ejected electrons, \( E_{photon} = 5.80 \mathrm{eV} \) is the energy of the photons, and \( \Phi = 4.50 \mathrm{eV} \) is the work function of tungsten.
03

Calculate the Kinetic Energy

Substitute the given values into the equation:\[ K = 5.80 \mathrm{eV} - 4.50 \mathrm{eV} = 1.30 \mathrm{eV} \]This result is the kinetic energy of the fastest ejected electrons.
04

Convert Kinetic Energy to Joules

Convert the kinetic energy from electronvolts to joules using the conversion factor \( 1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J} \):\[ K = 1.30 \times 1.602 \times 10^{-19} \mathrm{J} = 2.083 \times 10^{-19} \mathrm{J} \]
05

Find the Speed of the Electrons

Use the kinetic energy formula, \( K = \frac{1}{2} mv^2 \), to find the speed \( v \) of the electrons, where \( m = 9.11 \times 10^{-31} \mathrm{kg} \) is the mass of an electron:\[ 2.083 \times 10^{-19} \mathrm{J} = \frac{1}{2} \times 9.11 \times 10^{-31} \mathrm{kg} \times v^2 \]Solve for \( v \):\[ v^2 = \frac{2 \times 2.083 \times 10^{-19}}{9.11 \times 10^{-31}} \]\[ v = \sqrt{\frac{4.166 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 5.98 \times 10^5 \mathrm{m/s} \]

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Key Concepts

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

Work Function
In the realm of the photoelectric effect, the work function is a crucial concept. It refers to the minimum amount of energy that is necessary to eject an electron from the surface of a material. Think of it as the energy barrier that must be overcome to release an electron from its atomic bond.
In this exercise, the work function of tungsten is provided as \(4.50 \ \mathrm{eV}\). This means that for any electron to be freed from the tungsten surface, a photon must supply energy that meets or exceeds this threshold. If the photon鈥檚 energy is less than the work function, no electrons will be emitted.
  • Energy above the work function: Only the excess energy, which surpasses the work function, contributes to an electron鈥檚 kinetic energy.
  • Material variable: Different materials have different work functions, affecting how easily they emit electrons.
Photon Energy
Photon energy is the energy carried by a single photon. In the context of the photoelectric effect, it is the primary source of energy that facilitates the ejection of electrons from a material's surface. The energy of a photon is determined by its frequency and is calculated using the formula \(E = h \times f\), where \(E\) is the energy, \(h\) is Planck's constant, and \(f\) is the frequency of the photon.
In the given problem, the photon energy is \(5.80 \ \mathrm{eV}\). This energy is greater than the work function of tungsten (\(4.50 \ \mathrm{eV}\)), meaning that the photons have sufficient energy to not only knock electrons free but also provide them with kinetic energy.
  • Conversion to kinetic energy: Any energy from the photon that exceeds the work function turns into kinetic energy for the ejected electron.
  • Spectrum dependency: Higher frequency (or shorter wavelength) light has higher energy photons.
Kinetic Energy of Electrons
When a photon hits the surface of a material and ejects an electron, the electron gains kinetic energy. This kinetic energy is essentially the leftover energy from the photon after the work function has been accounted for. It can be calculated using the equation \(K = E_{\text{photon}} - \Phi\), where \(K\) is the kinetic energy, \(E_{\text{photon}}\) is the energy of the incoming photon, and \(\Phi\) is the work function.
In the example exercise, after doing the math, we found that the maximum kinetic energy of the ejected electrons is \(1.30 \ \mathrm{eV}\). This leftover energy transforms into the speed at which the electrons leave the material's surface.
  • Relation to velocity: The kinetic energy attained, \(K\), can then be used to calculate the velocity of the electrons using the formula \(K = \frac{1}{2} mv^2\), where \(m\) is the mass of the electron.
  • Impact of photon energy: Higher energy photons result in greater kinetic energy of ejected electrons, if the work function is constant.

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Most popular questions from this chapter

\( \mathrm{X}\) rays with a wavelength of \(71 \mathrm{pm}\) are directed onto a gold foil and eject tightly bound electrons from the gold atoms. The ejected electrons then move in circular paths of radius \(r\) in a region of uniform magnetic field \(\vec{B}\). For the fastest of the ejected electrons, the product \(B r\) is equal to \(1.88 \times 10^{-4} \mathrm{~T} \cdot \mathrm{m}\). Find (a) the maximum kinetic energy of those electrons and (b) the work done in removing them from the gold atoms.

The beam emerging from a \(1.5 \mathrm{~W}\) argon laser \((\lambda=515 \mathrm{~nm})\) has a diameter \(d\) of \(3.5 \mathrm{~mm}\). The beam is focused by a lens system with an effective focal length \(f_{\mathrm{L}}\) of \(2.5 \mathrm{~mm}\). The focused beam strikes a totally absorbing screen, where it forms a circular diffraction pattern whose central disk has a radius \(R\) given by \(1.22 f_{\mathrm{L}} \lambda / d\). It can be shown that \(84 \%\) of the incident energy ends up within this central disk. At what rate are photons absorbed by the screen in the central disk of the diffraction pattern?

Light of wavelength \(200 \mathrm{~nm}\) shines on an aluminum surface; \(4.20 \mathrm{eV}\) is required to eject an electron. What is the kinetic energy of (a) the fastest and (b) the slowest ejected electrons? (c) What is the stopping potential for this situation? (d) What is the cutoff wavelength for aluminum?

At what rate does the Sun emit photons? For simplicity, assume that the Sun's entire emission at the rate of \(3.9 \times 10^{26} \mathrm{~W}\) is at the single wavelength of \(550 \mathrm{~nm}\).

A special kind of lightbulb emits monochromatic light of wavelength \(630 \mathrm{~nm}\). Electrical energy is supplied to it at the rate of \(60 \mathrm{~W}\), and the bulb is \(93 \%\) efficient at converting that energy to light energy. How many photons are emitted by the bulb during its lifetime of \(730 \mathrm{~h}\) ?
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