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Chapter 38: Problem 6

Find the maximum kinetic energy of electrons ejected from a certain material if the material's work function is \(2.3 \mathrm{eV}\) and the frequency of the incident radiation is \(2.5 \times 10^{15} \mathrm{~Hz}\).

Short Answer

Expert verified
The maximum kinetic energy of ejected electrons is approximately 1.902 eV.

Step by step solution

01

Recall the Photoelectric Equation

In the photoelectric effect, the maximum kinetic energy of ejected electrons is given by the equation: \( KE_{max} = h u - \phi \), where \( h \) is Planck's constant, \( u \) is the frequency of the incident radiation, and \( \phi \) is the work function of the material.
02

Convert the Work Function to Joules

The work function is given in electron volts (eV) and needs to be converted to Joules (J). The conversion factor is \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \). Thus, \( \phi = 2.3 \, \text{eV} \) is \( \phi = 2.3 \times 1.602 \times 10^{-19} \, \text{J} \).
03

Calculate the Energy of the Incident Photons

Use Planck's constant \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) to calculate the energy of the incident photons, which is \( E = h u = 6.626 \times 10^{-34} \times 2.5 \times 10^{15} \).
04

Calculate the Maximum Kinetic Energy

Substitute for \( h u \) and \( \phi \) into the photoelectric equation \( KE_{max} = h u - \phi \). After calculating \( h u \) and \( \phi \), solve for \( KE_{max} \).
05

Simplify and Present the Result

Simplify the expression to get the maximum kinetic energy in Joules, and if needed, convert back to electron volts for the final answer.

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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, symbolized by \( \phi \), is the minimum energy required to release an electron from the surface of a material.
Think of it as a threshold energy鈥攐nly photons with sufficient energy can "kick out" electrons.
This energy threshold is specific to each material, measured in electron volts (eV) for convenience.
  • If the photon energy is less than the work function, electrons will not be ejected from the surface.
  • In contrast, if the photon's energy exceeds the work function, electrons are liberated, and the excess energy converts into kinetic energy.
In our exercise, the work function is given as 2.3 eV. To proceed with calculations, this is converted into Joules using the relationship: 1 eV = \( 1.602 \times 10^{-19} \) J. Thus, 2.3 eV equals \( 2.3 \times 1.602 \times 10^{-19} \) J.
Planck's Constant
Planck's constant \( h \) is a fundamental constant in quantum mechanics.
It relates the energy of photons to their frequency, crucial for understanding the photoelectric effect.
The value of Planck's constant is \( 6.626 \times 10^{-34} \;\mathrm{J \cdot s} \).
  • This constant forms a bridge between the wave and particle nature of light.
  • The equation \( E = hu \) (Energy = Planck's constant \( \times \) Frequency) allows us to compute the energy of a photon given its frequency.
In our scenario, this relationship helps calculate the energy of incoming photons striking the surface.
Kinetic Energy
Kinetic energy in the context of the photoelectric effect refers to the energy possessed by the ejected electrons post-collision with photons.
It is determined by the excess energy after subtracting the work function from the incident photon energy. The formula is:
\[ KE_{max} = h u - \phi \]
  • \( KE_{max} \) represents the maximum kinetic energy of the ejected electrons.
  • This formula highlights that only excess energy beyond the work function is converted into electron motion.
After calculating the photon's energy with its frequency and subtracting the work function, we can derive the kinetic energy. This energy characterizes how energetically electrons leave the material's surface.
Incident Photon Energy
Incident photon energy is the energy carried by each photon as it approaches a material surface.
This energy depends on the photon's frequency and is calculated using Planck's constant, with the formula \( E = hu \).
  • Higher frequency photons possess greater energy, capable of ejecting electrons from surfaces with higher work functions.
  • This energy "strikes" electrons, potentially displacing them if sufficient to overcome the material's work function.
In the problem, the photon's frequency is \( 2.5 \times 10^{15} \;\mathrm{Hz} \). Therefore, its energy is \( E = 6.626 \times 10^{-34} \times 2.5 \times 10^{15} \), giving us the energy value needed to determine the subsequent kinetic energy of released electrons.

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Most popular questions from this chapter

Just after detonation, the fireball in a nuclear blast is approximately an ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{7} \mathbf{K}\). (a) Find the wavelength at which the thermal radiation is maximum and (b) identify the type of electromagnetic wave corresponding to that wavelength. (See Fig. 33-1.) This radiation is almost immediately absorbed by the surrounding air molecules, which produces another ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{5} \mathrm{~K}\). (c) Find the wavelength at which the thermal radiation is maximum and (d) identify the type of electromagnetic wave corresponding to that wavelength.

A light detector (your eye) has an area of \(2.00 \times 10^{-6} \mathrm{~m}^{2}\) and absorbs \(80 \%\) of the incident light, which is at wavelength \(500 \mathrm{~nm}\). The detector faces an isotropic source, \(3.00 \mathrm{~m}\) from the source. If the detector absorbs photons at the rate of exactly \(4.000 \mathrm{~s}^{-1}\), at what power does the emitter emit light?

The wavelength of the yellow spectral emission line of sodium is \(590 \mathrm{~nm}\). At what kinetic energy would an electron have that wavelength as its de Broglie wavelength?

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?

An ultraviolet lamp emits light of wavelength \(400 \mathrm{~nm}\) at the rate of 400 W. An infrared lamp emits light of wavelength \(700 \mathrm{~nm}\), also at the rate of \(400 \mathrm{~W}\). (a) Which lamp emits photons at the greater rate and (b) what is that greater rate?
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