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Chapter 38: Problem 8

What would the minimum work function for a metal have to be for visible light (380-750 nm) to eject photoelectrons?

Short Answer

Expert verified
The minimum work function must be less than 3.26 eV.

Step by step solution

01

Understand the Problem

The problem is asking for the minimum work function needed for photoelectrons to be ejected by visible light. This involves the photoelectric effect, where light must have enough energy, given by the photon's energy, to overcome the work function of the metal.
02

Identify Key Formulas

The energy of a photon is given by the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant (\(6.626 \times 10^{-34} \text{ J s}\)), \( c \) is the speed of light (\(3 \times 10^8 \text{ m/s}\)), and \( \lambda \) is the wavelength of light.
03

Calculate the Maximum Photon Energy

For the shortest wavelength of visible light (380 nm), the energy of the photon is maximum. Convert 380 nm to meters: \(380 \text{ nm} = 380 \times 10^{-9} \text{ m}\). Substitute into the equation: \( E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{380 \times 10^{-9}}\). Calculate to find \( E \).
04

Solve for Minimum Work Function

The minimum work function must be equal to or less than the maximum photon energy calculated in Step 3. Convert the energy from Joules to electronvolts (1 eV = \(1.602 \times 10^{-19}\) J), to express the work function in eV. Compute the value.

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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 the context of the photoelectric effect. It represents the minimum energy required to dislodge an electron from the surface of a material, typically a metal. In simple terms, it's like the energy "threshold" that must be crossed for electrons to escape from the material when light hits it.
The work function is crucial because only photons with energy equal to or greater than this value can successfully eject electrons. For metals that interact with visible light, determining the minimum work function is essential for applications like photoelectric sensors and solar cells.
  • This energy is usually measured in electronvolts (eV), a more convenient unit than Joules when working on the atomic scale.
  • When photons from visible light strike a metal, their energy must surpass the work function for electrons to be emitted.
  • The photoelectric effect reveals how light can be thought of as particles (photons) with distinct energy levels based on their wavelength.
Photon Energy
Photon energy is a key factor in the photoelectric effect. It dictates whether or not an electron will be ejected from the surface of a metal. The energy of a photon is dependent on its wavelength and can be calculated using the formula:\[ E = \frac{hc}{\lambda} \]Here, \( E \) represents the photon鈥檚 energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon.
Understanding photon energy helps us predict which wavelengths are capable of ejecting electrons. Shorter wavelengths have higher frequencies and therefore more energy. Consequently, they have a better chance of overcoming the work function of a metal.
  • Photon energy is directly proportional to frequency and inversely proportional to wavelength.
  • The higher the energy of a photon, the more likely it is to eject electrons from a metal surface.
  • In visible light, photons from the violet end (shorter wavelength) possess more energy than those from the red end.
Visible Light Spectrum
The visible light spectrum is the portion of the electromagnetic spectrum that the human eye can see. It ranges from approximately 380 nm to 750 nm in wavelength. Each color within this spectrum corresponds to a different wavelength, with violet having the shortest wavelength and red having the longest.
When considering the photoelectric effect, it's important to understand how the visible light spectrum relates to photon energy. Shorter wavelengths (like violet) carry higher energy photons than longer wavelengths (like red).
  • The visible spectrum usually appears as a continuous range of colors, from violet to red.
  • Only photons within specific energy ranges can overcome the work function of certain metals.
  • This is why, in our problem, we calculate the minimum work function by considering the shortest wavelength (380 nm) since it has the highest energy potential.
Planck's Constant
Planck's constant, denoted as \( h \), is a fundamental quantity in quantum mechanics that plays a key role in the calculation of photon energy. Its value is approximately \( 6.626 \times 10^{-34} \text{ Js} \). Planck's constant is pivotal because it bridges wave and particle models of light, showing light as quantized packets of energy called photons.
In the formula for photon energy \( E = \frac{hc}{\lambda} \), Planck's constant is multiplied by the speed of light \( c \) and divided by wavelength \( \lambda \) to find the energy of a photon.
  • Planck鈥檚 constant is a fundamental constant used in various quantum mechanical equations, not just in calculations involving photons.
  • This constant underpins the quantization of energy, limiting energy exchanges to discrete amounts.
  • By understanding this constant, we gain deeper insights into the dual nature of electromagnetic radiation.

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Most popular questions from this chapter

An ultrashort pulse has a duration of 9.00 fs and produces light at a wavelength of 556 nm. What are the momentum and momentum uncertainty of a single photon in the pulse?

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?

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about 1 MeV (10\(^6\) eV). By contrast, what we see emanating from the sun's surface are visiblelight photons with wavelengths of about 500 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.)

A photon scatters in the backward direction (\(\phi = 180^\circ\)) from a free proton that is initially at rest. What must the wavelength of the incident photon be if it is to undergo a 10.0\(\%\) change in wavelength as a result of the scattering?

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 x rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat \(c\) (in J/kg \(\bullet\) 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 kV and 60.0 mA that converts 1.0\(\%\) of the electric power into x rays. Assume that the 0.250-kg target is made of lead (\(c\) = 130 J/kg \(\bullet\) K). (d) What must the physical properties of a practical target material be? What would be some suitable target elements?
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