/*! 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 8 What would the minimum work func... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What would the minimum work function for a metal have to be for visible light \((400-700 \mathrm{mn})\) to eject photoelectrons?

Short Answer

Expert verified
The minimum work function for the metal is approximately 3.1 eV.

Step by step solution

01

Understanding the work function

The work function is the minimum energy required to eject electrons from a metal surface. This energy is often provided by photons from light, in this case, visible light.
02

Identify photon energy equation

Photon energy can be calculated using the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \) Js, \( c \) is the speed of light \( 3 \times 10^8 \) m/s, and \( \lambda \) is the wavelength of light.
03

Determine energy for minimum wavelength

Use the shortest wavelength, 400 nm, as it has the highest energy to potentially eject electrons. Convert 400 nm to meters by multiplying by \( 1 \times 10^{-9} \) to get \( 4 \times 10^{-7} \) meters.
04

Calculate photon energy for 400 nm

Plug 400 nm into the energy equation: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} \]. Simplifying, \( E \approx 4.97 \times 10^{-19} \) joules.
05

Convert energy to electronvolts

Convert the energy from joules to electronvolts for convenient comparison, using the conversion factor \(1 \, eV = 1.602 \times 10^{-19} \) joules. Divide the computed energy by this factor, \[ E_{eV} = \frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 3.1 \, eV \].
06

Conclusion on minimum work function

The minimum work function required for visible light at 400 nm to eject photoelectrons is approximately 3.1 eV.

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.

Work Function
The work function is a crucial concept in the study of the photoelectric effect. It represents the minimum energy needed to eject an electron from the surface of a metal. Think of it as the energy hurdle that photons must leap over to free electrons.
The work function, often denoted by the Greek letter phi (\( \varphi \)), is specific to each type of metal and depends on the nature of the material.
  • If the energy of the incoming photon is greater than the work function, the metal will emit electrons.
  • If the energy is less, the photon will not have enough energy to overcome the attraction holding the electrons.
Knowing the work function is essential when designing devices like photoelectric sensors or photovoltaic cells, where light energy is converted into electric energy. This conversion efficiency largely depends on how well the work function matches the energy level of incoming photons, such as those from visible light.
Photon Energy
Photon energy is the energy carried by a single photon, essential for understanding how light interacts with matter. In the context of the photoelectric effect, photon energy determines if an electron can be ejected from a metal surface when exposed to light.
To calculate photon energy, we use the formula:

Understanding the Photon Energy Formula

\[E = \frac{hc}{\lambda}\]where
  • \( E \) is the energy of the photon.
  • \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \) Js.
  • \( c \) is the speed of light, \( 3 \times 10^8 \) m/s.
  • \( \lambda \) is the wavelength of the light.
This equation shows that shorter wavelengths (like 400 nm) carry more energy than longer wavelengths (like 700 nm). The energy from shorter wavelengths can more easily overcome the work function, making it crucial for the ejection of photoelectrons. This principle is the reason visible light can sometimes eject electrons, depending on the work function of the metal.
Visible Light
Visible light is the portion of the electromagnetic spectrum that human eyes can detect, ranging from approximately 400 nm to 700 nm in wavelength. It encompasses all the colors seen in a rainbow, from violet at the shorter wavelengths to red at the longer.
A clear understanding of visible light's properties is vital in contexts like the photoelectric effect, where light interacts with materials.
Photon energy in visible light varies across its spectrum.
  • Violet light, at about 400 nm, contains the highest photon energy in the visible range.
  • Red light, around 700 nm, has the lowest energy.
The ability of visible light to eject electrons from a metal surface hinges on the photon energy relative to the metal's work function. Metals with a lower work function might allow even the lower energy red light to cause electron ejection, whereas those with higher work functions might only react to violet light. Understanding how visible light interacts at a microscopic level leads to better designs in technologies like solar panels and light detectors, optimizing them to use this energy source efficiently.

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) If the average frequency emitted by a \(200-\mathrm{W}\) light bulb is \(5.00 \times 10^{14} \mathrm{Hz},\) and 10.0\(\%\) of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second? (b) At what distance would this correspond to \(1.00 \times 10^{11}\) visible-light photons per square centimeter per second if the light is emitted uniformly in all directlons?

PRK Surgery. Photorefractive keratectomy (PRK) is a laser-based surgical procedure that corrects near- and farsightedness by removing part on the lens of the eye to change its curvature and hence focal length. This procedure can remove layers 0.25\(\mu \mathrm{m}\) thick using pulses lasting 12.0 \(\mathrm{ns}\) from a laser beam of wavelength 193 \(\mathrm{nm}\) . Low- intensity beams can be used because each individual photon has enough energy to break the covalent bonds of the tissue. (a) In what part of the electromagnetic spectrum does this light he? (b) What is the energy of a single photon? (c) If 1.50-mW beam is used, how many photons are delivered to the lens in each pulse?

Sirius B. The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system of two stars, the smaller one (Sirius B) being a white dwarf. Spectral analysis of Sirius B indicates that its surface temperature is \(24,000 \mathrm{K}\) and that it radiates energy at a total rate of \(1.0 \times 10^{25} \mathrm{W}\) . Assume that it behaves like an ideal blackbody. (a) What is the total radiated intensity of Sirius \(\mathrm{B}\) (b) What is the peak-intensity wavelength? Is this wave-length visible to humans? (c) What is the radius of Sirius B? Express your answer in kilometers and as a fraction of our sun's radius. (d) Which star radiates more total energy per second, the hot Sirius or the (relatively) cool sun with a surface temperature of 5800 \(\mathrm{K}\) ? To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.

(a) Calculate the maximum increase in photon wavelength that can occur during Compton scattering. (b) What is the energy (in electron volts) of the lowest- energy \(x\) -ray photon for which Compton scattering could result in doubling the original wave- length?

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the \(n=4\) level. Determine the wave- length and frequency of the photon.

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.