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Chapter 38: Problem 22

The wavelength associated with the cutoff frequency for silver is \(325 \mathrm{~nm}\). Find the maximum kinetic energy of electrons ejected from a silver surface by ultraviolet light of wavelength \(254 \mathrm{~nm}\).

Short Answer

Expert verified
The maximum kinetic energy is approximately 1.07 eV.

Step by step solution

01

Understand the Problem

We need to find the maximum kinetic energy of electrons ejected from a silver surface. For this, we'll use the photoelectric effect, where the kinetic energy of ejected electrons is given by the difference between the photon energy and the work function.
02

Convert Wavelengths to Energy

To find the energy associated with the wavelengths, use the formula for photon energy: \[ E = \frac{hc}{\lambda} \]where \(h\) is Planck's constant \(6.626 \times 10^{-34} J \cdot s\), \(c\) is the speed of light \(3.00 \times 10^8 m/s\), and \(\lambda\) is the wavelength.
03

Calculate Work Function

First, find the energy at the cutoff wavelength for silver, \(\lambda_0 = 325 \mathrm{nm}\):\[ \phi = \frac{hc}{\lambda_0} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{325 \times 10^{-9}} = 6.12 \times 10^{-19} \, J \]
04

Calculate Photon Energy of Incident Light

Now, calculate the energy of the ultraviolet light with wavelength \(\lambda = 254 \mathrm{~nm}\):\[ E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{254 \times 10^{-9}} = 7.83 \times 10^{-19} \, J \]
05

Determine Maximum Kinetic Energy

The maximum kinetic energy \( KE_{max} \) of the ejected electrons is given by:\[ KE_{max} = E - \phi = 7.83 \times 10^{-19} \,-\, 6.12 \times 10^{-19} \KE_{max} = 1.71 \times 10^{-19} \, J \]
06

Convert Energy to More Practical Units

Convert the energy from joules to electronvolts (eV), knowing that \(1 \, eV = 1.602 \times 10^{-19} \, J\):\[ KE_{max} = \frac{1.71 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 1.07 \, eV \]

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Key Concepts

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

Wavelength in the Photoelectric Effect
Wavelength is a critical parameter in understanding the photoelectric effect. It is the distance between successive peaks of a wave, and in this context, it determines the energy of photons hitting a material surface.
The shorter the wavelength of light, the higher the energy of the photons. To convert a wavelength to energy, we use the formula:
  • \[ E = \frac{hc}{\lambda} \]
where:
  • \(E\) is the energy,
  • \(h\) is Planck's constant \(6.626 \times 10^{-34} J \cdot s\),
  • \(c\) is the speed of light \(3.00 \times 10^8 m/s\),
  • \(\lambda\) is the wavelength.
By calculating the energy from a specific wavelength, we gain insight into whether the photons have enough energy to eject electrons from the surface of a material, like silver in this case. Understanding this conversion helps us predict which wavelengths are most effective for different materials.
Kinetic Energy of Ejected Electrons
Kinetic energy refers to the energy that an object possesses due to its motion. In the scenario of the photoelectric effect, kinetic energy is of particular interest because it describes the energy of the electrons that are ejected from a material when hit by high-energy photons. The maximum kinetic energy of these electrons can be calculated using the formula:
  • \[ KE_{max} = E_{photon} - \phi \]
where:
  • \(KE_{max}\) is the maximum kinetic energy of the ejected electrons,
  • \(E_{photon}\) is the energy of the incident photon,
  • \(\phi\) is the work function of the material.
This equation highlights how much of the photon's energy is converted into electron movement, as opposed to being consumed by the work function.
Work Function and its Role
The work function is an intrinsic property of a material and is a measure of the minimum energy needed to eject an electron from the surface of that material. It plays a crucial role in the photoelectric effect as it defines the threshold energy below which no electron ejection occurs.The formula for calculating the work function when you know the cutoff wavelength is:
  • \[ \phi = \frac{hc}{\lambda_{0}} \]
where:
  • \(\phi\) is the work function,
  • \(\lambda_{0}\) is the cutoff wavelength for the material.
The work function is crucial in determining whether a given photon will be able to remove an electron from a surface. If the photon's energy is less than the work function, no electrons are released.
Understanding Planck's Constant
Planck's constant is a fundamental constant in quantum mechanics that links the energy carried by a photon to its frequency. This constant is essential in the formula used to compute the energy of a photon based on its wavelength in the context of the photoelectric effect.Its value is:
  • \( h = 6.626 \times 10^{-34} J \cdot s \)
Its inclusion in the equation \(E = \frac{hc}{\lambda}\) ties together the wave and particle nature of light. Planck's constant helps us to quantify the energy levels that light of various frequencies can transfer, making it indispensable for calculations involving quantum phenomena. Understanding it allows us to comprehend how varied frequencies and wavelengths affect electron ejection.

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Most popular questions from this chapter

If the de Broglie wavelength of a proton is \(100 \mathrm{fm}\), (a) what is the speed of the proton and (b) through what electric potential would the proton have to be accelerated to acquire this speed?

Neutrons in thermal equilibrium with matter have an average kinetic energy of \((3 / 2) k T\), where \(k\) is the Boltzmann constant and \(T\), which may be taken to be \(300 \mathrm{~K}\), is the temperature of the environment of the neutrons. (a) What is the average kinetic energy of such a neutron? (b) What is the corresponding de Broglie wavelength?

Wuppose the fractional efficiency of a cesium surface (with work function \(1.80 \mathrm{eV}\) ) is \(1.0 \times 10^{-16}\); that is, on average one electron is ejected for every \(10^{16}\) photons that reach the surface. What would be the current of electrons ejected from such a surface if it were illuminated with \(600 \mathrm{~nm}\) light from a \(2.00 \mathrm{~mW}\) laser and all the ejected electrons took part in the charge flow?

A 3.0 MeV proton is incident on a potential energy barrier of thickness \(10 \mathrm{fm}\) and height \(10 \mathrm{MeV}\). What are (a) the transmission coefficient \(T,(\mathrm{~b})\) the kinetic energy \(K_{t}\) the proton will have on the other side of the barrier if it tunnels through the barrier, and (c) the kinetic energy \(K_{r}\) it will have if it reflects from the barrier? A \(3.0 \mathrm{MeV}\) deuteron (the same charge but twice the mass as a proton) is incident on the same barrier. What are (d) \(T\), (e) \(K_{t}\), and (f) \(K_{r}\) ?

What (a) frequency, (b) photon energy, and (c) photon momentum magnitude (in \(\mathrm{keV} / \mathrm{c}\) ) are associated with \(\mathrm{x}\) rays having wavelength \(35.0 \mathrm{pm}\) ?
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