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\(\bullet\) In the photoelectric effect, what is the relationship between the threshold frequency \(f_{0}\) and the work function \(\phi ?\)

Short Answer

Expert verified
The relationship is \(\phi = hf_0\).

Step by step solution

01

Understanding Key Concepts

The photoelectric effect involves the ejection of electrons from a metal surface when light shines on it. Two important concepts here are the threshold frequency \(f_{0}\) and the work function \(\phi\). The threshold frequency \(f_0\) is the minimum frequency of light required to eject an electron. The work function \(\phi\) is the minimum energy needed to remove an electron from the surface of the metal.
02

Relating Energy and Frequency

According to the formula for the energy of a photon, \(E = hf\), where \(h\) is Planck's constant and \(f\) is the frequency. For the photoelectric effect to occur, the energy of the incoming photon \(hf\) must be at least equal to the work function \(\phi\). At the threshold frequency \(f_0\), the energy of the photon is exactly the work function, so \(hf_0 = \phi\).
03

Formulating the Relationship

Given that the energy of a photon at the threshold frequency is equal to the work function, we can write the equation \(\phi = hf_0\). Thus, the work function is directly proportional to the threshold frequency and can be expressed as the product of Planck's constant and the threshold frequency.

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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, denoted by \( \phi \), is a crucial concept in understanding the photoelectric effect. It represents the minimum energy required to eject an electron from the surface of a metal. This energy
  • is unique to each material, meaning different metals will have different work functions,
  • depends on the nature of the material and its surface properties,
  • dictates how easily the material can emit electrons when exposed to light.
Electrons in the metal are held by the attractive forces of the nuclei. For an electron to be released, it must overcome these forces with the energy equal to or greater than the work function. The concept of the work function is also important in applications beyond the photoelectric effect, such as in solar cells and semiconductor devices.Understanding the work function helps in predicting and explaining whether light of a given frequency can lead to the photoelectric emission.
Threshold Frequency
The threshold frequency \( f_0 \) is the minimum frequency of light that can cause electrons to be emitted from a metal surface. It is intrinsic to the metal and directly linked to its work function. The relationship between the threshold frequency and the energy is given by
  • the equation \( E = hf_0 \), with \( E \) being the energy of the photon and \( h \) being Planck's constant.
  • If the frequency of the incoming light is below this threshold, no electrons will be emitted, regardless of the light's intensity,
  • showing that energy and frequency, not intensity, determine electron emission.
This concept was key in the acceptance of quantum mechanics, as it offered an explanation beyond classical wave theories which could not account for the occurrence of a minimal frequency requirement.Knowing the threshold frequency helps in determining the minimum energy photons need to have to cause a photoelectric effect. This defines the boundary between no electron emission and possible emission by a photon.
Planck's Constant
Planck's constant \( h \) is a fundamental constant in physics, essential for quantifying the relationship between energy and frequency in quantum theory. It appears in the equation \( E = hf \), which is central to the understanding of the photoelectric effect. Planck's constant has a value of approximately \( 6.626 \times 10^{-34} \text{ Js} \), and it:
  • connects the energy of photons (quantum of light) to their frequency,
  • plays a pivotal role in the field of quantum mechanics, influencing how we understand the behavior of particles on a small scale.
This constant emphasizes the quantized nature of energy transfer. The discovery and application of Planck’s constant were instrumental in shifting from classical physics to modern quantum physics, and it also helps explain phenomena that classical physics could not adequately address, like the work function an electron must overcome.Understanding Planck's constant is vital for anyone studying phenomena linked to energy quantization and electromagnetic waves.

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Most popular questions from this chapter

\(\cdot\) Removing birthmarks. Pulsed dye lasers emit light of wavelength 585 \(\mathrm{nm}\) in 0.45 \(\mathrm{ms}\) pulses to remove skin blemishes such as birthmarks. The beam is usually focused onto a circular spot 5.0 \(\mathrm{mm}\) in diameter. Suppose that the output of one such laser is 20.0 \(\mathrm{W}\) . (a) What is the energy of each photon, in eV? (b) How many photons per square millimeter are delivered to the blemish during each pulse?

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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?

When ultraviolet light with a wavelength of 254 nm falls upon a clean metal surface, the stopping potential necessary to terminate the emission of photoelectrons is 0.181 \(\mathrm{V}\) . (a) What is the photoelectric threshold wavelength for this metal? (b) What is the work function for the metal?

\(\bullet\) 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 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)?

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