/*! 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 13 When ultraviolet light with a wa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When ultraviolet light with a wavelength of 254 nm falls on a clean copper surface, the stopping potential necessary to stop emission of photoelectrons is 0.181 \(\mathrm{V}\) (a) What is the photoelectric threshold wavelength for this copper surface? (b) What is the work function for this surface, and how does your calculated value compare with that given in Table 38.1\(?\)

Short Answer

Expert verified
Threshold wavelength is 267 nm, and work function is 4.6 eV, consistent with reference values.

Step by step solution

01

Understanding the Problem

We need to find the threshold wavelength for a copper surface and calculate the work function using the photoelectric effect principles. Given: Wavelength \( \lambda = 254 \) nm and stopping potential \( V_s = 0.181 \) V.
02

Calculate the Energy of the Incident Photon

Calculate the energy \( E \) of the photon using \( E = \frac{hc}{\lambda} \). Here, \( h = 6.626 \times 10^{-34} \; \text{J s} \), \( c = 3 \times 10^8 \; \text{m/s} \), and \( \lambda = 254 \times 10^{-9} \; \text{m} \).
03

Convert the Energy to Electron Volts

Convert the energy from joules to electron volts (1 eV = \( 1.602 \times 10^{-19} \) J) to align with the stopping potential given in volts.
04

Use Stopping Potential to Find Work Function

Use the photoelectric equation \( \frac{hc}{\lambda} = \text{Work function} + eV_s \) to solve for the work function \( \phi \).
05

Calculate the Threshold Wavelength

The threshold wavelength \( \lambda_0 \) is found using the work function: \( \phi = \frac{hc}{\lambda_0} \). Rearrange to solve for \( \lambda_0 \).
06

Compare Calculated Work Function

Compare the calculated work function \( \phi \) with known data to check consistency.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Threshold Wavelength
The threshold wavelength is a concept in the photoelectric effect that refers to the longest wavelength of light that can still cause the removal of an electron from a material. When light hits the surface of a material, its energy is absorbed by electrons. If the energy is high enough, an electron can be ejected from the surface.- This longest allowable wavelength is known as the threshold wavelength.- Beyond this wavelength, the energy of the photons is too low to liberate any electrons.To calculate the threshold wavelength, we use the equation involving the work function \[\phi = \frac{hc}{\lambda_0}\] where \( \phi \) is the work function, \( h \) is Planck's constant, and \( c \) is the speed of light. Rearranging this formula allows us to solve for \( \lambda_0 \), which is the threshold wavelength:\[\lambda_0 = \frac{hc}{\phi}\]Understanding the threshold wavelength helps us predict whether certain light sources can cause photoelectron emission.
Work Function
The work function is a fundamental property of a material surface that indicates the minimum energy required to remove an electron from the surface.- It's typically measured in electron volts (eV).- The value varies between different materials due to their unique atomic structure.In the context of the photoelectric effect, the work function represents the barrier that photons must overcome to eject electrons. The photoelectric effect is governed by the equation:\[\frac{hc}{\lambda} = \phi + eV_s\]Here, \( \phi \) symbolizes the work function, and \( eV_s \) is the energy due to the stopping potential. By rearranging this, we can solve for \( \phi \):\[\phi = \frac{hc}{\lambda} - eV_s\]Thus, to calculate the work function, we subtract the energy provided by the stopping potential from the total energy of the incident photon. This approach helps us understand how different materials interact with light differently.
Stopping Potential
Stopping potential, often denoted as \( V_s \), is a critical component in the study of the photoelectric effect. It represents the potential difference needed to halt the most energetic electrons emitted from a surface.- The stopping potential directly measures the maximum kinetic energy of the ejected electrons.- Higher stopping potentials indicate more energetic photoelectrons are being emitted.In the formula related to the photoelectric effect \[\frac{hc}{\lambda} = \phi + eV_s\]the term \( eV_s \) (where \( e \) is the elementary charge) quantifies this energy required to stop the photoelectrons. The stopping potential provides insight into the energy characteristics of electrons freed by photons, and it is crucial for calculating the work function. Understanding stopping potential helps clarify how energy from light converts into kinetic energy in the photoelectric effect.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 75 -W light source consumes 75 W of electrical power. Assume all this energy goes into emitted light of wavelength 600 \(\mathrm{nm}\) . (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit? (c) Are the answers to parts (a) and (b) the same? Is the frequency of the light the same thing as the number of photons emitted per second? Explain.

A photon has momentum of magnitude \(8.24 \times\) \(10^{-28} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

Protons are accelerated from rest by a potential difference of 4.00 \(\mathrm{kV}\) and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting \(x\) rays? How does your answer compare to the minimum wave-length if \(4.00-\mathrm{keV}\) electrons are used instead? Why do x-ray tubes use electrons rather than protons to produce x rays?

A 2.50 -W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 \(\mathrm{cV}\) . Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts of this metal? (b) How many photoelectrons are ejected eachsecond from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

BIO Response of the Eve. The human eve is most sensitive to green light of wavelength 505 \(\mathrm{nm}\) . Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina. (a) What is the frequency of this photon? (b) How much energy (in joules and electron volts) does it deliver to the receptor cells? (c) To appreciate what a smount of energy this is, calculate how fast a typical bacterium of mass \(9.5 \times 10^{-12}\) g would move if it had that much energy.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.