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Chapter 28: Problem 50

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 an electron. (a) What is the work function (in electronvolts) 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
(a) 5.84 eV; (b) \(1.56 \times 10^{18}\); (c) \(7.8 \times 10^{17}\); (d) \(7.81 \times 10^{17}\).

Step by step solution

01

Calculate Energy of a Photon

The energy of a photon is given by the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.63 \times 10^{-34} \mathrm{~J \cdot s} \), \( c \) is the speed of light \( 3 \times 10^8 \mathrm{~m/s} \), and \( \lambda = 124 \mathrm{~nm} = 124 \times 10^{-9} \mathrm{~m} \). Calculate the energy \( E \):\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{124 \times 10^{-9}} \mathrm{~J} \approx 1.60 \times 10^{-18} \mathrm{~J} \]Convert this energy to electronvolts \((1 \mathrm{~eV} = 1.6 \times 10^{-19} \mathrm{~J})\):\[ E \approx \frac{1.60 \times 10^{-18}}{1.6 \times 10^{-19}} = 10 \mathrm{~eV} \]
02

Calculate the Work Function

The work function \( \phi \) is the energy needed to release an electron from the metal and is related to the photon's energy and the maximum kinetic energy \( K_{\text{max}} \) of the ejected electrons by \( E = \phi + K_{\text{max}} \). Solve for the work function:\[ \phi = E - K_{\text{max}} = 10 \mathrm{~eV} - 4.16 \mathrm{~eV} = 5.84 \mathrm{~eV} \]
03

Calculate Number of Photoelectrons Ejected per Second

The power of the beam \( P = 2.50 \mathrm{~W} \). Since each photon ejects one electron, the number of photoelectrons \( n \) per second is given by the formula \( n = \frac{P}{E} \). Convert the energy of each photon back to joules:\[ E = 1.60 \times 10^{-18} \mathrm{~J} \]Then calculate \( n \):\[ n = \frac{2.50}{1.60 \times 10^{-18}} \approx 1.56 \times 10^{18} \] photoelectrons per second.
04

Effect of Reducing Power by Half

If the power is reduced by half, the new power is \( P' = 1.25 \mathrm{~W} \). The energy of a photon remains the same, so the number of photoelectrons ejected per second becomes:\[ n = \frac{1.25}{1.60 \times 10^{-18}} \approx 0.78 \times 10^{18} = 7.8 \times 10^{17} \]Therefore, if the power is reduced by half, the number of photoelectrons ejected each second is approximately \( 7.8 \times 10^{17} \).
05

Effect of Reducing Wavelength by Half

Reducing the wavelength by half changes the photon energy. New wavelength \( \lambda' = 62 \mathrm{~nm} \), so new photon energy:\[ E' = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{62 \times 10^{-9}} \approx 3.20 \times 10^{-18} \mathrm{~J} = 20 \mathrm{~eV} \]The number of photoelectrons ejected per second is:\[ n = \frac{2.50}{3.20 \times 10^{-18}} \approx 7.81 \times 10^{17} \]If the wavelength is reduced by half, approximately \( 7.81 \times 10^{17} \) 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
Photon energy is the energy carried by a single photon. It is directly proportional to the frequency of the photon and inversely proportional to its wavelength. This can be defined mathematically by the formula \( E = \frac{hc}{\lambda} \), where:
  • \( E \) is the energy of the photon
  • \( h \) is Planck's constant, a fundamental constant of nature
  • \( c \) is the speed of light
  • \( \lambda \) is the wavelength of the light
The energy of a photon explains why light can exert force or energy on objects, even though individual photons are very small. When light of a specific wavelength hits a surface, its photons transfer energy to the electrons on that surface.
If the energy transferred is sufficient, it can eject electrons from the surface, leading to the photoelectric effect. The energy of the photons in this scenario can be measured in electronvolts (eV), which is a convenient unit of energy when dealing with atomic and subatomic particles.
Work Function
The work function of a material is the minimum energy required to remove an electron from the surface of a material, often a metal. In the context of the photoelectric effect, the work function is a key component. The photoelectric effect occurs when photons hit a material and transfer their energy to electrons, potentially freeing them if their energy meets or exceeds the work function.
The equation for the photoelectric effect is given by:
  • \( E = \phi + K_{\text{max}} \)
  • where \( E \) is the photon energy
  • \( \phi \) is the work function
  • \( K_{\text{max}} \) is the maximum kinetic energy of the ejected electrons
This relationship helps in determining the work function given the energy of incoming photons and the kinetic energy of the ejected electrons. If the photon's energy is just equal to the work function, electrons are freed with no kinetic energy.
However, with higher photon energies, the excess energy is converted into kinetic energy for the ejected electrons.
Planck's Constant
Planck's constant \( h \) is a fundamental physical constant. It plays a vital role in the realm of quantum mechanics and is central to the description of the photoelectric effect. Its value is \( 6.63 \times 10^{-34} \text{ J} \cdot \text{s} \), which might seem small, but it is crucial for calculations involving quantum phenomena.
Planck's constant relates the energy of a photon to its frequency, showing that electromagnetic energy is quantized鈥攁 core principle of quantum theory. It appears in the equation for photon energy: \( E = h u \), where \( u \) is the frequency of the photon.
  • With high frequency (small wavelength) light, Planck's constant ensures detectable levels of energy per photon.
  • It's central to understanding why low-frequency light cannot eject electrons from certain materials, regardless of intensity.
This constant's importance goes beyond light and encompasses various aspects of quantum physics, governing processes on the atomic scale and ensuring stability in the universe's smallest constituents.

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Most popular questions from this chapter

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 wavelength if \(4.00 \mathrm{keV}\) electrons are used instead? Why do X-ray tubes use electrons rather than protons to produce X-rays?

(a) What accelerating potential is needed to produce electrons of wavelength \(5.00 \mathrm{nm} ?\) (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

The neutral \(\pi^{0}\) meson is an unstable particle produced in highenergy particle collisions. Its mass is about 264 times that of the electron, and it exists for an average lifetime of \(8.4 \times 10^{-17} \mathrm{~s}\) before decaying into two gamma-ray photons. Assuming that the mass and energy of the particle are related by the Einstein equation \(E=m c^{2},\) find the uncertainty in the mass of the particle and express it as a fraction of the particle's mass.

Laser surgery. Using a mixture of \(\mathrm{CO}_{2}, \mathrm{~N}_{2},\) and sometimes He, \(\mathrm{CO}_{2}\) lasers emit a wavelength of \(10.6 \mu \mathrm{m}\). At power outputs of \(0.100 \mathrm{~kW},\) such lasers are used for surgery. How many photons per second does a \(\mathrm{CO}_{2}\) laser deliver to the tissue during its use in an operation?

Light with a wavelength range of \(145-295 \mathrm{nm}\) shines on a silicon surface in a photoelectric effect apparatus, and a reversing potential of \(3.50 \mathrm{~V}\) is applied to the resulting photoelectrons. (a) What is the longest wavelength of the light that will eject electrons from the silicon surface? (b) With what maximum kinetic energy will electrons reach the anode?
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