/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 The photoelectric work functions... [FREE SOLUTION] | 91Ó°ÊÓ

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The photoelectric work functions for particular samples of certain metals are as follows: cesium, 2.1 eV; copper, 4.7 eV; potassium, \(2.3 \mathrm{eV} ;\) and \(\mathrm{zinc}, 4.3\) eV. (a) What is the threshold wavelength for each metal surface? (b) Which of these metals could not emit photoelectrons when irradiated with visible light \((400-700 \mathrm{nm}) ?\)

Short Answer

Expert verified
Copper and Zinc cannot emit photoelectrons with visible light.

Step by step solution

01

Understanding Work Function and Energy

The work function (\(\Phi\)) is the minimum energy needed to remove an electron from the surface of a metal. It's given in electron volts (eV). To find the threshold wavelength (\(\lambda_0\)), which is the longest wavelength that can cause electrons to eject, we need the photon's energy to match this work function.
02

Using the Energy-Wavelength Formula

The energy of a photon is related to its wavelength by the equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{Js})\) and \(c\) is the speed of light \((3.00 \times 10^8 \, \text{m/s})\). In eV, this equation becomes \(E \text{(eV)} = \frac{hc}{\lambda_0} \times \frac{1.602 \times 10^{-19} \text{Joule}}{1\,\text{eV}}\). Rearrange to find the threshold wavelength: \(\lambda_0 = \frac{hc}{E \cdot 1.602 \times 10^{-19}}\).
03

Calculating Threshold Wavelength for Each Metal

Using \(hc = 1240 \text{ eV nm}\), we calculate \(\lambda_0\) for each metal:- **Cesium**: \(\lambda_0 = \frac{1240}{2.1} = 590.5\, \text{nm}\)- **Copper**: \(\lambda_0 = \frac{1240}{4.7} = 263.8\, \text{nm}\)- **Potassium**: \(\lambda_0 = \frac{1240}{2.3} = 539.1\, \text{nm}\)- **Zinc**: \(\lambda_0 = \frac{1240}{4.3} = 288.4\, \text{nm}\)
04

Analyzing Visible Light Range

Visible light ranges from 400 nm to 700 nm. A metal can emit photoelectrons if its threshold wavelength \(\lambda_0\) is within or shorter than this range, meaning the photon energy is high enough to surpass the work function.
05

Determining Metals Inactive Under Visible Light

Comparing the threshold wavelengths with visible light:- **Cesium** and **Potassium** have threshold wavelengths within 400-700 nm range; thus, they can emit photoelectrons with visible light.- **Copper** and **Zinc** have \(\lambda_0\) shorter than 400 nm; thus, they cannot emit photoelectrons under visible light.

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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 photoelectric effect. It refers to the minimum amount of energy required to remove an electron from the surface of a metal. Think of it like the energy needed to "free" an electron. This energy is usually measured in electron volts (eV), a unit of energy commonly used in atomic and particle physics.

Each metal requires a different amount of energy to perform this action due to its unique atomic structure. For example, in our provided exercise, cesium has a work function of 2.1 eV, copper 4.7 eV, potassium 2.3 eV, and zinc 4.3 eV. The smaller the work function, the easier it is to emit an electron from the metal surface when exposed to light. This is why metals such as cesium and potassium are more responsive to photoelectron emission than metals like copper or zinc.

Understanding work function helps in determining how metals react to different wavelengths of light, which is crucial for applications in photoelectric devices and solar panels.
Threshold Wavelength
The threshold wavelength is crucial for understanding when photoelectron emission will occur. It is the longest wavelength of light that has enough energy to eject an electron from a metal surface. The shorter the wavelength, the higher the energy of the light, according to the equation:
  • \(E = \frac{hc}{\lambda}\)
where:
  • \(E\) is the energy of the photon
  • \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{Js})\)
  • \(c\) is the speed of light \((3.00 \times 10^8 \, \text{m/s})\)
  • \(\lambda\) is the wavelength

Once the photon's energy matches or exceeds the work function, it can eject an electron. For instance, if a light's wavelength is shorter than a metal's threshold wavelength, it has enough energy for photoelectron emission. In the example with cesium, potassium, copper, and zinc, we've calculated these values:
- Cesium: 590.5 nm- Copper: 263.8 nm- Potassium: 539.1 nm- Zinc: 288.4 nm

Recognizing threshold wavelengths can help determine which metals are utilized in devices sensitive to light energy, such as solar cells.
Photoelectron Emission
Photoelectron emission describes the process by which electrons are ejected from a material's surface when exposed to light of sufficient energy. This phenomenon is grounded in the photoelectric effect, famously explained by Albert Einstein, for which he won the Nobel Prize.

When light strikes a metal surface with energy equal to or greater than the work function, it provides the electrons with enough energy to be liberated. These freed electrons are termed "photoelectrons."

In practical scenarios, not every metal will emit photoelectrons under visible light conditions. The visible light spectrum ranges from 400 nm to 700 nm. For example, in our exercise, only cesium and potassium possess threshold wavelengths that fall within this range, making them capable of emitting photoelectrons when exposed to visible light. On the other hand, metals like copper and zinc, which have shorter threshold wavelengths, require ultraviolet light or more energetic wavelengths than those found in visible light to emit photoelectrons.

Understanding the criterion for photoelectron emission is essential for developing technologies like photo-detectors, enhancing our grasp of materials responsive to different light energies.

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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?

What would the minimum work function for a metal have to be for visible light \((400-700 \mathrm{mn})\) to eject photoelectrons?

(a) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the \(\mathrm{H}_{a}\) line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the \(n=3\) level and eventually ends up in the ground level? Calculate the wavelength of the emilted photon in each case.

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

(a) Show that, as \(n\) gets very large, the energy levels of the hydrogen atom get closer and closer together in energy. (b) Do the radii of these energy levels also get closer together?

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