/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Multiple-Concept Example 3 revie... [FREE SOLUTION] | 91Ó°ÊÓ

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Multiple-Concept Example 3 reviews the concepts necessary to solve this problem. Light is incident on the surface of metallic sodium, whose work function is \(2.3 \mathrm{eV}\). The maximum speed of the photoelectrons emitted by the surface is \(1.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the wavelength of the light?

Short Answer

Expert verified
The wavelength of the light is 194 nm.

Step by step solution

01

Understand the Problem

We are given the work function \( W_0 = 2.3 \, \text{eV} \) for metallic sodium and the maximum speed \( v_{max} = 1.2 \times 10^6 \, \text{m/s} \) of the photoelectrons emitted. We need to find the wavelength \( \lambda \) of the incident light that causes these photoelectrons to be emitted.
02

Convert Work Function to Joules

The work function is provided in electron volts (eV). Since \(1\, \text{eV} = 1.6 \times 10^{-19} \text{J}\), we convert the work function: \( W_0 = 2.3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.68 \times 10^{-19} \, \text{J} \).
03

Calculate Maximum Kinetic Energy

The maximum kinetic energy \( KE_{max} \) of the ejected photoelectrons can be calculated using the expression \( KE_{max} = \frac{1}{2} m v_{max}^2 \) where \( m \) is the electron mass \( 9.11 \times 10^{-31} \, \text{kg} \). Substitute the values: \( KE_{max} = \frac{1}{2} \times 9.11 \times 10^{-31} \times (1.2 \times 10^6)^2 = 6.55 \times 10^{-19} \, \text{J} \).
04

Use Einstein’s Photoelectric Equation

Use Einstein's photoelectric equation: \( E = W_0 + KE_{max} \), which implies \( E = 3.68 \times 10^{-19} \text{J} + 6.55 \times 10^{-19} \text{J} = 1.023 \times 10^{-18} \text{J} \).
05

Relate Energy to Wavelength

The energy of the light \( E \) and its wavelength \( \lambda \) are related by the equation \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \). Rearrange to solve for \( \lambda \): \( \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.023 \times 10^{-18}} = 1.94 \times 10^{-7} \, \text{m} \).
06

Convert to Nanometers

Finally, convert the wavelength from meters to nanometers: \( 1.94 \times 10^{-7} \, \text{m} = 194 \, \text{nm} \).

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Key Concepts

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

Work Function
In the context of the photoelectric effect, the term 'work function' refers to the minimum energy required to eject an electron from the surface of a metal. For metallic sodium, this energy is provided in units of electron volts (eV). The work function is a property that varies from one metal to another and is represented by the symbol \( W_0 \). If the energy of the incoming photon is less than the work function, no photoelectrons are emitted regardless of its intensity.
This foundational concept is key to understanding why only photons with sufficient energy can cause the emission of electrons. In our example, metallic sodium has a work function of 2.3 eV. To use this in our calculations, it's important to convert the value into joules, the SI unit of energy. Remember,\(1 \, \text{eV} = 1.6 \times 10^{-19} \text{J}\), thus making the work function in joules:\( W_0 = 3.68 \times 10^{-19} \text{J}\).
Understanding the work function is crucial because it sets the threshold energy necessary for the photoelectric effect to occur.
Kinetic Energy
In the photoelectric effect, the kinetic energy of the emitted photoelectrons measures how much energy they carry after being knocked free from the metal. When a photon strikes the surface, it imparts energy to the electron, and once the work function energy is covered, any remaining energy becomes the kinetic energy of the electron.
The kinetic energy can be calculated using the classical equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its velocity. For electrons, \( m \) is approximately \( 9.11 \times 10^{-31} \text{kg} \), and in our example, they have a velocity of \( 1.2 \times 10^6 \text{m/s} \). Plug these values into the equation to obtain the kinetic energy:\( KE_{max} = 6.55 \times 10^{-19} \text{J}\).
This kinetic energy is essential because it directly influences the characteristics of the emitted electrons, ultimately allowing us to derive the energy of the incident light.
Wavelength Calculation
Calculating the wavelength of light involves understanding the relationship between energy and wavelength. In physics, this relationship is given by \( E = \frac{hc}{\lambda} \), where \( E \) is the energy, \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)), \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)), and \( \lambda \) is the wavelength.
By rearranging this equation, we solve for the wavelength: \( \lambda = \frac{hc}{E} \). Using the total energy calculated from the work function and kinetic energy (\(1.023 \times 10^{-18} \, \text{J} \)), we find the wavelength to be \(1.94 \times 10^{-7} \, \text{m} \).
Finally, to make the wavelength more interpretable, we convert it to nanometers by recognizing that \(1 \, \text{m} = 10^9 \, \text{nm}\), yielding a final result of \(194 \, \text{nm}\). This step helps connect the properties of the electromagnetic spectrum to real-world observations.
Einstein’s Photoelectric Equation
Einstein's photoelectric equation is fundamental for explaining how light interacts with matter to produce the photoelectric effect. It is expressed as \( E = W_0 + KE_{max} \), where \( E \) is the energy of the incident photon, \( W_0 \) is the work function, and \( KE_{max} \) represents the maximum kinetic energy of the emitted electrons.
This equation highlights one of the pivotal breakthroughs in physics by showing that light must have both particle and wave qualities. Through rearrangement and solving for \( E \), we can derive the kinetic energy and work function, affirming that only specific light frequencies can cause electron emission.
The application of this equation enables us to determine the energy and hence the wavelength required for an electron to be liberated from a metal's surface. In our example, the equation corroborates the combined energy value needed to surpass the work function of sodium and provide kinetic energy to the electrons.

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Most popular questions from this chapter

A photon of known wavelength strikes a free electron that is initially at rest, and the photon is scattered straight backward. (a) How is the kinetic energy KE of the recoil electron related to the energies \(E\) and \(E_{\prime}\) of the incident and scattered photons, respectively? (b) If the wavelength \(\lambda\) of the incident photon is known, how can the photon's energy \(E\) be determined? (c) How can the wavelength \(\mathcal{N}\) of the scattered photon and, hence, its energy \(E^{\prime}\) be found if we know the wavelength of the incident photon and the fact that the photon is scattered straight backward? A photon of wavelength \(0.45000 \mathrm{~nm}\) strikes a free electron that is initially at rest. The photon is scattered straight backward. What is the speed of the recoil electron after the collision?

Light is shining perpendicularly on the surface of the earth with an intensity of \(680 \mathrm{~W} / \mathrm{m}^{2}\) Assuming all the photons in the light have the same wavelength (in vacuum) of \(730 \mathrm{nm}\), determine the number of photons per second per square meter that reach the earth.

Consult Interactive LearningWare 29.1 at for background material relating to this problem. An owl has good night vision because its eyes can detect a light inten sity as small as \(5.0 \times 10^{-13} \mathrm{~W} / \mathrm{m}^{2}\). What is the minimum number of photons per second that an owl eye can detect if its pupil has a diameter of \(8.5 \mathrm{~mm}\) and the light has a wavelength of \(510 \mathrm{nm} ?\)

(a) How many photons (wavelength \(=620 \mathrm{nm}\) ) must be absorbed to melt a \(2.0-\mathrm{kg}\) block of ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C} ?\) (b) On the average, how many \(\mathrm{H}_{2} \mathrm{O}\) molecules does one photon convert from the ice phase to the water phase?

An electron, starting from rest, accelerates through a potential difference of \(418 \mathrm{~V}\). What is the final de Broglie wavelength of the electron, assuming that its final speed is much less than the speed of light?

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