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A \(2.50 \mathrm{~W}\) beam of light of wavelength \(124 \mathrm{nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is \(4.16 \mathrm{eV}\). 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 each second 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)?

Short Answer

Expert verified
The results are: (a) The work function of the metal is about 3.96 eV. (b) About 9.47 x 10^15 photoelectron are ejected per second from this metal. (c) If the power is reduced by half, then about 4.73 x 10^15 photoelectrons are ejected per second. (d) If the wavelength is reduced by half, the number of photoelectrons ejected per second remains the same.

Step by step solution

01

Calculate work function

Use Einstein's photoelectric equation \( KE = hf - Φ \) where KE is the kinetic energy of the ejected electron, h is Planck's constant (6.63 × 10^-34 J.s), f is the frequency (c/λ, where c is the speed of light and λ is the wavelength), and Φ is the work function. Solve for Φ: \( Φ = hf - KE \). Convert λ to meters, KE to Joules, calculate f, and then Φ in Joules. Lastly, convert Φ to electron volts (1 eV = 1.6 x 10^-19 J).
02

Calculate number of photoelectrons ejected per second

Knowing that each photon ejects one photoelectron, the rate of ejection of photoelectrons is equal to the rate of incident photons. The power of the beam (P) is given by \( P = nhf \), where n is the number of photons per unit time. As before, h is Planck's constant, and f is the frequency. Solving for n, we get \( n = P / hf \). Note that the power P is in watts, which are equal to Joules per second.
03

Reduced power

If the power of the light beam is reduced by half, the number of photoelectrons would also reduce by half because the number of photons is halved, and each photon ejects one photoelectron.
04

Reduced wavelength

If the wavelength is halved, the frequency would be doubled because the frequency is inversely proportional to the wavelength. However, as per Einstein's photoelectric equation, this increase in frequency does not change the number of photons, and therefore the number of photoelectrons would not change.

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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 understanding the photoelectric effect. It represents the minimum energy required to remove an electron from the surface of a metal. When light illuminates a metal surface, only photons with enough energy can eject electrons. This energy threshold is the work function, usually represented by the Greek letter \( \Phi \).

In our exercise, determining the work function is crucial for interpreting the metal's response to light. It involves using Einstein's photoelectric equation and incorporating known values such as the kinetic energy of the ejected electrons and the energy of incoming photons, which depends on the frequency of the light. After conducting the calculations, the work function is the difference between the photon energy and the kinetic energy of the electron.

To make these concepts clear to students, imagine a party with a bouncer representing the work function. The photons are guests trying to enter the club—the metal surface. If a guest (a photon) has enough energy (work function) to convince the bouncer (overcome the work function), they're allowed to leave the club (eject an electron).
Kinetic Energy of Electrons
Following the work function, the kinetic energy of electrons is another piece of the puzzle. This is the energy that electrons possess as a result of being ejected by photons in the photoelectric effect. Basically, after an electron overcomes the metal's work function barrier, any additional energy provided by the photon is converted into the electron's kinetic energy—or its motion energy.

In our exercise, we look at the maximum kinetic energy of the electrons, which is provided in electron volts (eV). This maximum kinetic energy offers insights into how the electrons behave after they have been ejected: faster-moving electrons have higher kinetic energy. This concept is crucial in applications such as electricity generation from solar panels, where the movement of these electrons generates an electric current.

Students often find the relationship between light energy and kinetic energy challenging, but it can be likened to kicking a soccer ball—the harder you kick (more energy from the photon), the faster the ball moves (greater kinetic energy of the electron), provided the ball has already left the ground (work function has been overcome).
Einstein's Photoelectric Equation
Einstein's photoelectric equation is the bedrock of our understanding of the photoelectric effect. This equation connects the dots between the energy of the incoming photons and the resulting kinetic energy of the ejected electrons. It is represented as \( KE = hf - \Phi \) where \( KE \) is the kinetic energy of the ejected electron, \( h \) is Planck's constant, \( f \) is the frequency of the light, and \( \Phi \) is the work function of the metal.

Applying this equation to our exercise provides a systematic approach to finding the work function and understanding the number of electrons ejected. It shows how altering light properties—like its power or wavelength—affects the ejection of photoelectrons. For instance, halving the light's power will halve the quantity of ejected electrons, while changing the wavelength impacts their energy but not their quantity.

It's akin to adjusting the volume or the pitch of music—the volume (power) determines how many can hear it (number of ejected electrons), whereas the pitch (wavelength) changes the experience (electron's energy) without altering the crowd size.

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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.)

The photoelectric work function of potassium is \(2.3 \mathrm{eV}\). If light that has a wavelength of \(190 \mathrm{nm}\) 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.

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?

A laser produces light of wavelength \(625 \mathrm{nm}\) 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 photon with wavelength \(0.1100 \mathrm{nm}\) collides with a free electron that is initially at rest. After the collision the wavelength is \(0.1132 \mathrm{nm}\). (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

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