/*! 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 35 In developing night-vision equip... [FREE SOLUTION] | 91Ó°ÊÓ

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In developing night-vision equipment, you need to measure the work function for a metal surface, so you perform a photoelectric-effect experiment. You measure the stopping potential \(V_{0}\) as a function of the wavelength \(\lambda\) of the light that is incident on the surface. You get the results in the table. $$ \begin{array}{l|llllll} \boldsymbol{\lambda}(\mathbf{n m}) & 100 & 120 & 140 & 160 & 180 & 200 \\ \hline \boldsymbol{V}_{0}(\mathbf{V}) & 7.53 & 5.59 & 3.98 & 2.92 & 2.06 & 1.43 \end{array} $$ In your analysis, you use \(c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\) and \(e=1.602 \times 10^{-19} \mathrm{C}\) which are values obtained in other experiments. (a) Select a way to plot your results so that the data points fall close to a straight line. Using that plot, find the slope and \(y\) -intercept of the best-fit straight line to the data. (b) Use the results of part (a) to calculate Planck's constant \(h\) (as a test of your data) and the work function (in eV) of the surface. (c) What is the longest wavelength of light that will produce photoelectrons from this surface? (d) What wavelength of light is required to produce photoelectrons with kinetic energy \(10.0 \mathrm{eV} ?\)

Short Answer

Expert verified
The exact answer varies based on the values obtained from the graph. However, the methodology described above will accurately solve the problem, providing values for \( h \), \( \Phi \), \( \lambda_{max} \) and the wavelength required for kinetic energy of 10eV.

Step by step solution

01

Data Arrangement and Graph Plotting

Arrange the given wavelengths \( \lambda \) and the corresponding stopping potentials \( V_0 \) in two columns, and calculate values for \( 1/\lambda \) in another column. Plot a graph of stopping potentials \( V_0 \) against \( 1/\lambda \), which ought to result in a straight line.
02

Determining the Slope and Intercept

The slope and y-intercept of the best-fit line for the data can be found using any line of best fit method, for example, the least squares method. The slope corresponds to the value of \( h \) and the y-intercept corresponds to \( -\Phi / e \).
03

Calculation of Planck's constant and the Work function

Planck's Constant \( h \) is determined by the slope from the plot, keeping in mind that you'll have to unconventionally use units of electronvolts for the elementary charge \( e \) so that \( h \) will be in Joule-seconds. Multiply the negative y-intercept by the charge of an electron \( e \) to obtain the work function \( \Phi \) in electronvolts.
04

Calculation of Longest Wavelength of Light

The longest wavelength of light that will produce photoelectrons from the surface can be found using the formula \( \lambda_{max} = h/(\Phi c) \). Substituting the values of Planck's constant \( h \), speed of light \( c \) and work function \( \Phi \) in this formula will yield the longest wavelength \( \lambda_{max} \).
05

Wavelength Required for Kinetic Energy of 10eV

The wavelength of light that is necessary to produce photoelectrons with a kinetic energy of 10.0eV can be determined via the formula \( \lambda = h/(c(E_{k} + \Phi)) \). Substitute the values of Planck's constant \( h \), speed of light \( c \), given kinetic energy \( E_{k} \), and work function \( \Phi \) into this formula to find the required wavelength.

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Key Concepts

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

Work Function
The concept of the work function is pivotal in understanding the photoelectric effect. It refers to the minimum amount of energy needed to remove an electron from the surface of a metal. This energy is specific to each material due to its unique atomic structure and bonding. In the context of a photoelectric effect experiment, the work function is symbolized as \( \Phi \), and its value is usually measured in electronvolts (eV). The relationship between the work function, the energy of the incident photons, and the kinetic energy of the emitted electrons is given by the equation:
  • \( h u = \Phi + KE \), where \( h u \) is the energy of the incident light, and \( KE \) is the kinetic energy of the ejected electrons.
By plotting stopping potential \( V_0 \) against the inverse of wavelength \( 1/\lambda \), one can determine the work function. The y-intercept of the line derived from such a plot provides \( -\Phi/e \), allowing calculation of \( \Phi \) after multiplying by the elementary charge \( e \). This step is critical for determining the material's propensities in photoelectric interactions.
Planck's Constant
Planck's constant, denoted by \( h \), is a fundamental constant in physics that relates the energy of a photon to its frequency. It plays a crucial role in quantum mechanics and the photoelectric effect, providing a bridge between wave and particle theories of light. The formula \( E = h u \) is used to calculate the energy \( E \) carried by a photon when its frequency \( u \) is known. In practice, the slope of the line when plotting stopping potential \( V_0 \) against \( 1/\lambda \) provides a value for \( hc/e \), making it possible to calculate \( h \) by rearranging terms.
Once the slope is determined from the best-fit line in a photoelectric experiment, we can solve for Planck's constant given that:
  • \( \text{Slope} = \frac{hc}{e} \)
Multiplying the slope by the elementary charge \( e \) gives the product \( hc \), from which \( h \) is isolated. Understanding Planck's constant is fundamental for anyone studying photoelectric phenomena and quantum physics overall.
Stopping Potential
Stopping potential is a key concept in the study of the photoelectric effect. It represents the minimum voltage needed to stop the most energetic photoelectrons emitted from a metal surface when light shines on it. This potential, denoted as \( V_0 \), is directly related to the maximum kinetic energy \( KE_{max} \) of these photoelectrons by the equation:
  • \( eV_0 = KE_{max} \)
By measuring \( V_0 \), we can infer important aspects about the energy levels of the incident photons and the material's work function. The stopping potential is practically determined through experiments by adjusting an external voltage supply until photoelectrons cease to flow.
Analyzing the plot of \( V_0 \) against \( 1/\lambda \) provides critical insight into the interaction of light and matter, confirming theories established in quantum mechanics. In particular, it's through this analysis that concepts like Planck's constant and the work function are experimentally verified.
Wavelength
Wavelength is an essential term when discussing light and other forms of electromagnetic radiation. It describes the distance between successive crests of a wave. In the context of the photoelectric effect, the wavelength \( \lambda \) of incident light is inversely related to its frequency and, consequently, its energy. As per the equation \( c = \lambda u \), where \( c \) is the speed of light, any change in wavelength affects the photon's energy, impacting electron ejection from a surface.
  • The shortest wavelength translates to the highest photon energy.
  • The longest wavelength sufficient to release photoelectrons corresponds to meeting the work function \( \Phi \).
To determine the longest wavelength of light that can still produce photoelectrons, we use the formula \( \lambda_{\text{max}} = \frac{hc}{\Phi} \). For any application involving the photoelectric effect, understanding and manipulating the wavelength of light is vital to achieving desired experimental outcomes.

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Most popular questions from this chapter

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.)

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?

A pulsed dye laser emits light of wavelength \(585 \mathrm{nm}\) in \(450 \mu \mathrm{s}\) pulses. Because this wavelength is strongly absorbed by the hemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood, such as portwine-colored birthmarks. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and heat of vaporization as water \(\left(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, 2.256 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right)\) Suppose that each pulse must remove \(2.0 \mu \mathrm{g}\) of blood by evaporating it, starting at \(33^{\circ} \mathrm{C}\). (a) How much energy must each pulse deliver to the blemish? (b) What must be the power output of this laser? (c) How many photons does each pulse deliver to the blemish?

A photon of green light has a wavelength of \(520 \mathrm{nm}\). Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

A horizontal beam of laser light of wavelength \(585 \mathrm{nm}\) passes through a narrow slit that has width \(0.0620 \mathrm{~mm}\). The intensity of the light is measured on a vertical screen that is \(2.00 \mathrm{~m}\) from the slit. (a) What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? (b) Use the result of part (a) to estimate the width of the central diffraction maximum that is observed on the screen.

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