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Chapter 38: Problem 56

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 photoclectron. (a) What is the work function (in electron volts of this metal? (b) How many photoclectrons 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
(a) Work function is 5.84 eV. (b) 1.56 脳 10鹿鈦� electrons/s. (c) 7.81 脳 10鹿鈦� electrons/s. (d) 7.81 脳 10鹿鈦� electrons/s.

Step by step solution

01

Determine Photon Energy

To find the energy of a photon, use the formula \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ Js} \) is Planck's constant, \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 124 \text{ nm} = 124 \times 10^{-9} \text{ m} \) is the wavelength. Calculating this, we get:\[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{124 \times 10^{-9}} \approx 1.60 \times 10^{-18} \text{ J/photon} \].Convert this energy from joules to electronvolts (1 eV = \( 1.602 \times 10^{-19} \text{ J} \)):\[ E = \frac{1.60 \times 10^{-18}}{1.602 \times 10^{-19}} \approx 10.00 \text{ eV/photon} \].
02

Calculate Work Function

The maximum kinetic energy of ejected electrons is given as 4.16 eV. According to the photoelectric equation, \( E_{\text{photon}} = \phi + K_{\text{max}} \), where \( \phi \) is the work function and \( K_{\text{max}} = 4.16 \text{ eV} \).Rearranging the formula gives \( \phi = E_{\text{photon}} - K_{\text{max}} \). Substituting the values, we have:\[ \phi = 10.00 \text{ eV} - 4.16 \text{ eV} = 5.84 \text{ eV} \].
03

Calculate Number of Photons Produced Per Second

The power of the light beam is given as 2.50 W, which is equivalent to 2.50 J/s. Each photon has an energy of 1.60 \( \times 10^{-18} \) J.The number of photons produced per second is given by \( \text{Number of photons} = \frac{\text{Power}}{\text{Energy per photon}} \).Substituting the values:\[ \frac{2.50 \text{ J/s}}{1.60 \times 10^{-18} \text{ J/photon}} \approx 1.56 \times 10^{18} \text{ photons/s} \].Since each photon ejects one photoelectron, this is also the number of photoelectrons ejected per second.
04

Effect of Halving Power (Part c)

If the power of the light beam is reduced by half (i.e., to 1.25 W), the number of photons produced per second will also be halved, because the number of photons is directly proportional to the power.Thus, the number of photoelectrons ejected each second is:\[ \frac{1.25 \text{ J/s}}{1.60 \times 10^{-18} \text{ J/photon}} \approx 7.81 \times 10^{17} \text{ electrons/s} \].
05

Effect of Halving Wavelength (Part d)

Halving the wavelength will double the energy per photon since \( E = \frac{hc}{\lambda} \). Therefore, the new energy per photon is:\[ E_{\text{new}} = 2 \times 10.00 \text{ eV} = 20.00 \text{ eV} \].Converting this to joules gives \( 20.00 \text{ eV} \approx 3.204 \times 10^{-18} \text{ J} \).Use the same power to find the number of photons:\[ \frac{2.50 \text{ J/s}}{3.204 \times 10^{-18} \text{ J/photon}} \approx 7.81 \times 10^{17} \text{ photons/s} \].So, 7.81 \( \times 10^{17} \) photoelectrons are ejected each second.

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Key Concepts

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

Photon Energy
In the photoelectric effect, light is made up of particles called photons. Each photon carries a specific amount of energy that depends on its wavelength. The shorter the wavelength, the higher the energy of the photon. To calculate the energy of a photon, we use the formula:
  • \( E = \frac{hc}{\lambda} \)
Here,
  • \( h = 6.626 \times 10^{-34} \text{ Js} \) is Planck's constant.
  • \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light.
  • \( \lambda \) is the wavelength of the light.
By substituting the values, we can find the energy of a photon in joules. However, converting it to electron volts (eV) is common in physics problems related to the photoelectric effect. Here, 1 eV equals \( 1.602 \times 10^{-19} \) joules. So, for our example, the energy of a photon was calculated to be 10.00 eV. This energy is what gives the photons the ability to release electrons upon hitting a metal surface.
Work Function
The work function, denoted by \( \phi \), is a key concept in the study of the photoelectric effect. It represents the minimum amount of energy required to remove an electron from the surface of a material. Different materials require different work functions due to the variation in atomic structures.To calculate the work function, we use the equation from the photoelectric effect:
  • \( E_{\text{photon}} = \phi + K_{\text{max}} \)
Where:
  • \( E_{\text{photon}} \) is the energy of the incoming photon.
  • \( K_{\text{max}} \) is the maximum kinetic energy of the ejected electron.
By rearranging the formula, \( \phi = E_{\text{photon}} - K_{\text{max}} \), we can find the work function once we know both the photon's energy and the ejected electron's kinetic energy. In our example, the calculated work function was 5.84 eV. This value tells us how resistant the material is to losing electrons when exposed to light.
Kinetic Energy of Electrons
When a photon hits a metal surface and ejects an electron, the electron moves with a certain kinetic energy. In the context of the photoelectric effect, the maximum kinetic energy of an ejected electron is given by the difference between the energy of the incoming photon and the material's work function.The equation is:
  • \( K_{\text{max}} = E_{\text{photon}} - \phi \)
This kinetic energy determines how fast the electrons move after being ejected. In general, if the photon energy exceeds the work function by a large margin, the electrons will have higher kinetic energy.In experiments, measuring \( K_{\text{max}} \) gives us insight into the efficiency of electron ejection and the material's electronic properties. For the example provided, the maximum kinetic energy was computed to be 4.16 eV. This shows how much energy the electrons carry after overcoming the material's resistance to electron ejection, as defined by its work function.

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Most popular questions from this chapter

An x-ray tube is operating at voltage \(V\) and current \(I\) . (a) If only a fraction \(p\) of the electric power supplied is converted into \(\mathbf{x}\) . rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat capacity \(c(\text { in } \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}),\) at what average rate would its temperature rise if there were no thermal losses?(c) Evaluate your results from parts (a) and (b) for an x-ray tube operating at 18.0 \(\mathrm{kV}\) and 60.0 \(\mathrm{mA}\) that converts 1.0\(\%\) of the electric power into \(\mathrm{x}\) rays. Assume that the \(0.250-\mathrm{kg}\) target is made of lead \((c=130 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}) .\) (d) What must the physical properties of a practical target material be? What would be some suitable target elements?

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(a) A proton is moving at a speed much slower than the speed of light. It has kinetic energy \(K_{1}\) and momentum \(p_{1}\) . If the momentum of the proton is doubled, so \(p_{2}=2 p_{1},\) how is its new kinetic energy \(K_{2}\) related to \(K_{1} ?(b)\) A photon with energy \(E_{1}\) has momentum \(p_{1} .\) If another photon has momentum \(p_{2}\) that is twice \(p_{1},\) how is the energy \(E_{2}\) of the second photon related to \(E_{1} ?\)

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