/*! 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 10 The photoelectric work function ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The photoelectric work function of potassium is 2.3 eV. If light that has a wavelength of 190 nm falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speed of these electrons.

Short Answer

Expert verified
(a) 4.26 V; (b) 4.26 eV; (c) \(9.73 \times 10^5\) m/s.

Step by step solution

01

Convert Wavelength to Energy

First, we need to convert the wavelength of the incoming light to energy using the formula \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \) J\cdot s is the Planck's constant, \( c = 3 \times 10^8 \) m/s is the speed of light, and \( \lambda = 190 \) nm is the wavelength. Convert the wavelength from nanometers to meters by multiplying by \( 10^{-9} \). Substitute the values:\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{190 \times 10^{-9}} \] This gives the energy in joules. To convert this energy to electron volts (eV), divide by the charge of an electron (~\(1.6 \times 10^{-19}\) C).
02

Calculate Energy in Electron Volts

Calculate the energy (\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{190 \times 10^{-9}} = 1.05 \times 10^{-18} \text{ J} \] Convert energy from joules to electron volts (1 eV = \(1.6 \times 10^{-19} \) J):\[ E = \frac{1.05 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 6.56 \text{ eV} \] This is the energy of the incoming photons in eV.
03

Determine the Kinetic Energy of Electrons

Using the photoelectric equation \( KE_{max} = E - \phi \), where \( E \) is the energy of the photons and \( \phi \) is the work function in eV (2.3 eV for potassium), we can find the maximum kinetic energy of the ejected electrons:\[ KE_{max} = 6.56 - 2.3 = 4.26 \text{ eV} \] This is the kinetic energy of the most energetic electrons ejected.
04

Calculate the Stopping Potential

The stopping potential \( V_s \) is simply the maximum kinetic energy divided by the electron charge to ensure electrons stop moving, given by:\[ V_s = KE_{max} \] Substitute the values:\[ V_s = 4.26 \text{ V} \] This is the stopping potential in volts.
05

Find the Speed of the Ejected Electrons

The kinetic energy \( KE \) is also given by the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is the electron mass \(9.11 \times 10^{-31}\) kg and \( v \) is the speed. Solving for \( v \):\[ v = \sqrt{\frac{2 \times KE}{m}} \] Convert the kinetic energy back to joules (4.26 eV, so \( 4.26 \times 1.6 \times 10^{-19} \) J). Substitute in \( v = \sqrt{\frac{2 \times 4.26 \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31}}} \).Calculate \( v \) to find the speed of these electrons.

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 fundamental concept in the study of the photoelectric effect. It refers to the minimum energy required to remove an electron from the surface of a material. In the context of the exercise, potassium has a work function of 2.3 eV, meaning that any incident photon must have at least this amount of energy to eject an electron.

Understanding the work function is crucial because it sets the threshold for the photoelectric effect to occur. When light hits the surface of a metal, like potassium in this case, only photons with energy equal to or greater than the work function can liberate electrons. If the photon's energy is less, no electrons will be emitted, and the photoelectric effect cannot be observed.

The work function can vary from one material to another, making some materials more susceptible to the photoelectric effect than others.
Kinetic Energy of Electrons
When light strikes a metal surface and electrons are ejected due to the photoelectric effect, they gain kinetic energy. The kinetic energy (KE) of these electrons is what allows us to measure their speed and movement. In the given exercise, the energy of the incoming photon was calculated to be 6.56 eV, which is greater than the work function of 2.3 eV.

The formula to find the maximum kinetic energy of the electrons is:
  • \( KE_{max} = E - \phi \)
Here, \( E \) is the energy of the photons (6.56 eV in this case) and \( \phi \) is the work function (2.3 eV). By substituting these values, we find that the maximum kinetic energy of the ejected electrons is 4.26 eV.

This calculation is important as it tells us about the excess energy the electrons have after overcoming the work function needed to escape the metal surface. This energy contributes to their velocity as they leave the surface.
Stopping Potential
The stopping potential, denoted as \( V_s \), is a critical concept used to measure the kinetic energy of ejected electrons in the photoelectric effect. It is defined as the electric potential needed to stop the most energetic photoelectrons from reaching the opposite electrode.

In simpler terms, it's the voltage needed to bring the fastest ejected electrons to a halt. In the given problem, since the maximum kinetic energy of electrons was determined to be 4.26 eV, directly the stopping potential required is also 4.26 V, since 1 eV corresponds to 1 V.

By applying this stopping potential, we create a reverse field that precisely cancels out the electrons' kinetic energy, effectively stopping their motion. Stopping potential experiments are instrumental in confirming Einstein's photoelectric equation and understanding the energy dynamics of photoelectrons.
Planck's Constant
Planck's constant, represented by the symbol \( h \), is a fundamental constant in quantum mechanics. It quantifies the relationship between the energy of a photon and the frequency of the electromagnetic wave from which the photon originates. The constant is approximately equal to \(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\).

In the exercise, Planck's constant plays a pivotal role in determining the energy contained in photons of a particular wavelength. Using the equation:
  • \( E = \frac{hc}{\lambda} \)
We see that the energy \( E \) of a photon is directly proportional to Planck's constant \( h \) and inversely proportional to the wavelength \( \lambda \). Here \( c \) represents the speed of light.

Thus, Planck's constant is the fee for nature's exchange between wave-like and particle-like behaviors of light. Its utilization in calculating photon energy allows us to bridge the gap between macroscopic electromagnetic waves and their microscopic photoelectric effects.

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 photon has momentum of magnitude 8.24 \(\times\) 10\(^{-28}\) kg \(\bullet\) m/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?

The cathode-ray tubes that generated the picture in early color televisions were sources of x rays. If the acceleration voltage in a television tube is 15.0 kV, what are the shortest-wavelength x rays produced by the television?

X rays with initial wavelength 0.0665 nm undergo Compton scattering. What is the longest wavelength found in the scattered x rays? At which scattering angle is this wavelength observed

A laser produces light of wavelength 625 nm in an ultrashort pulse. What is the minimum duration of the pulse if the minimum uncertainty in the energy of the photons is 1.0\(\%\)?

A pulsed dye laser emits light of wavelength 585 nm in 450-\(\mu\)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 port-wine-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 (4190 J / kg \(\bullet\) K, 2.256 \(\times\) 10\(^6\) J / kg). Suppose that each pulse must remove 2.0 mg of blood by evaporating it, starting at 33\(^\circ\)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?

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.