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The photoelectric threshold wavelength of a tungsten surface is 272 nm. Calculate the maximum kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency 1.45 × 1015 Hz. Express the answer in electron volts.

Short Answer

Expert verified
The maximum kinetic energy is 1.45 eV.

Step by step solution

01

Understand the Problem

We're given the threshold wavelength of tungsten and asked to calculate the maximum kinetic energy of electrons ejected when the surface is illuminated by UV light with a specified frequency. We'll use the photoelectric effect equations to find this energy.
02

Convert Threshold Wavelength to Threshold Frequency

We know that frequency u and wavelength λ are related by the speed of light c: c=uλ. Using this, the threshold frequency (u0) can be calculated: u0=cλ0 where λ0=272 nm=272×109 m and c=3×108 m/s.
03

Calculate the Threshold Frequency

Using the formula from Step 2, calculate u0: u0=3×108272×109=1.10×1015 Hz
04

Find the Energy of Incident Photons

The energy of the incident photons E is given by Planck's equation: E=hu. Here, u=1.45×1015 Hz, and h=6.63×1034 J s.
05

Calculate the Energy of Incident Photons

Substitute the values into Planck's equation: E=6.63×1034×1.45×1015=9.61×1019 J
06

Calculate the Work Function

The work function ϕ is the minimum energy required to eject an electron, calculated using ϕ=hu0, where u0=1.10×1015 Hz.
07

Determine the Work Function

Substitute into the equation: ϕ=6.63×1034×1.10×1015=7.29×1019 J
08

Calculate the Maximum Kinetic Energy

The maximum kinetic energy (KE) of the ejected electrons is given by KE=Eϕ. Substitute the calculated values: KE=9.61×10197.29×1019=2.32×1019 J
09

Convert Joules to Electron Volts

To convert the kinetic energy from joules to electron volts, use the conversion 1 eV=1.602×1019 J. Thus, KE=2.32×10191.602×1019 eV=1.45 eV
10

Final Answer

The maximum kinetic energy of the electrons ejected from the tungsten surface is 1.45 eV.

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Key Concepts

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

Threshold Wavelength
In the realm of photoelectric effect, the threshold wavelength plays a pivotal role. It refers to the longest wavelength of light that is capable of ejecting electrons from a material's surface.

When a light of wavelength longer than this threshold strikes the substance, no electrons are ejected, no matter how intense the light may be. This happens because the energy of the photons is simply not enough to liberate an electron from the material's atomic structure.

To calculate the threshold frequency, we use the equation relating the speed of light (c) to frequency (u) and wavelength (λ): = u \lambda.
  • Convert the given threshold wavelength to meters.
  • Use the speed of light in meters per second (3×108 m/s).
  • Determine the threshold frequency (u0), using the formula: u0=cλ0.
This is crucial because the threshold frequency helps in determining the work function of the material, which is essential for understanding the photoelectric effect.
Kinetic Energy Calculation
The calculation of kinetic energy of electrons plays a central role in understanding the photoelectric effect. When light hits the surface of a material, it ejects electrons with specific kinetic energy.

The energy of each photon is given by Planck's equation E=hu, where E is the energy, h is Planck's constant (6.63×1034 J s), and u is the frequency of the light.

Once the incident energy is known, the next step is to calculate the work function, which we will discuss later. Subtract the work function from the photon energy to find the kinetic energy of the ejected electrons:
  • Find the incident photon energy.
  • Use the work function to get the effective energy of electrons.
  • Calculate maximum kinetic energy using KE=Eϕ.
Understanding this relationship is vital as it gives insights into how energy is transferred from light to electrons.
Work Function
The work function (ϕ) is a key concept when examining the photoelectric effect. It is defined as the minimum energy required to dislodge an electron from the surface of a material.

In most cases, the work function is tied to the intrinsic properties of the material itself. To calculate it, we multiply Planck’s constant (h) by the threshold frequency (u0), which was derived from the threshold wavelength: ϕ=hu0.

This value serves as a threshold, representing the barrier that photon energy must overcome to emit electrons.
  • Calculate using the threshold frequency and Planck's constant.
  • Represents the energy barrier for electron ejection.
  • A crucial parameter in predicting electron emissions.
Understanding the work function helps us to predict how different materials will behave under light, assisting in the design and analysis of devices like solar panels and photo detectors.

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Most popular questions from this chapter

If a photon of wavelength 0.04250 nm strikes a free electron and is scattered at an angle of 35.0 from its original direction, find (a) the change in the wavelength of this photon; (b) the wavelength of the scattered light; (c) the change in energy of the photon (is it a loss or a gain?); (d) the energy gained by the electron.

The human eye is most sensitive to green light of wavelength 505 nm. Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina. (a) What is the frequency of this photon? (b) How much energy (in joules and electron volts) does it deliver to the receptor cells? (c) To appreciate what a small amount of energy this is, calculate how fast a typical bacterium of mass 9.5 × 1012 g would move if it had that much energy.

Higher-energy photons might be desirable for the treatment of certain tumors. Which of these actions would generate higher-energy photons in this linear accelerator? (a) Increasing the number of electrons that hit the tungsten target; (b) accelerating the electrons through a higher potential difference; (c) both (a) and (b); (d) none of these.

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about 1 MeV (106 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 1026 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 ϕ1ϕ2/2, which is valid for ϕ 1. Note that ϕ is in radians in this expression.) (c) It is estimated that a photon takes about 106 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.)

Protons are accelerated from rest by a potential difference of 4.00 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-keV electrons are used instead? Why do x-ray tubes use electrons rather than protons to produce x rays?

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