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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 \(\times\) 10\(^{15}\) 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 \( \lambda \) are related by the speed of light \( c \): \( c = u \lambda \). Using this, the threshold frequency (\( u_0 \)) can be calculated: \[ u_0 = \frac{c}{\lambda_0} \] where \( \lambda_0 = 272 \text{ nm} = 272 \times 10^{-9} \text{ m} \) and \( c = 3 \times 10^8 \text{ m/s} \).
03

Calculate the Threshold Frequency

Using the formula from Step 2, calculate \( u_0 \): \[ u_0 = \frac{3 \times 10^8}{272 \times 10^{-9}} = 1.10 \times 10^{15} \text{ Hz} \]
04

Find the Energy of Incident Photons

The energy of the incident photons \( E \) is given by Planck's equation: \( E = h u \). Here, \( u = 1.45 \times 10^{15} \text{ Hz} \), and \( h = 6.63 \times 10^{-34} \text{ J s} \).
05

Calculate the Energy of Incident Photons

Substitute the values into Planck's equation: \[ E = 6.63 \times 10^{-34} \times 1.45 \times 10^{15} = 9.61 \times 10^{-19} \text{ J} \]
06

Calculate the Work Function

The work function \( \phi \) is the minimum energy required to eject an electron, calculated using \( \phi = h u_0 \), where \( u_0 = 1.10 \times 10^{15} \text{ Hz} \).
07

Determine the Work Function

Substitute into the equation: \[ \phi = 6.63 \times 10^{-34} \times 1.10 \times 10^{15} = 7.29 \times 10^{-19} \text{ J} \]
08

Calculate the Maximum Kinetic Energy

The maximum kinetic energy (KE) of the ejected electrons is given by \( KE = E - \phi \). Substitute the calculated values: \[ KE = 9.61 \times 10^{-19} - 7.29 \times 10^{-19} = 2.32 \times 10^{-19} \text{ J} \]
09

Convert Joules to Electron Volts

To convert the kinetic energy from joules to electron volts, use the conversion \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Thus, \[ KE = \frac{2.32 \times 10^{-19}}{1.602 \times 10^{-19}} \text{ eV} = 1.45 \text{ eV} \]
10

Final Answer

The maximum kinetic energy of the electrons ejected from the tungsten surface is \( 1.45 \text{ eV} \).

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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 (\( \lambda \)): = u \lambda.
  • Convert the given threshold wavelength to meters.
  • Use the speed of light in meters per second (\(3 \times 10^8 \) m/s).
  • Determine the threshold frequency (\( u_0 \)), using the formula: \( u_0 = \frac{c}{\lambda_0}\).
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 \(E = h u \), where \(E\) is the energy, \(h\) is Planck's constant (\(6.63 \times 10^{-34} \text{ J s}\)), 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 \(KE = E - \phi \).
Understanding this relationship is vital as it gives insights into how energy is transferred from light to electrons.
Work Function
The work function (\(\phi\)) 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 (\(u_0\)), which was derived from the threshold wavelength: \(\phi = h u_0\).

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

In developing night-vision equipment, you need to measure the work function for a metal surface, so you perform a photoelectric-effect experiment. You measure the stopping potential \(V_0\) as a function of the wavelength \(\lambda\) of the light that is incident on the surface. You get the results in the table. In your analysis, you use \(c\) = 2.998 \(\times\) 10\(^8\) m/s and \(e\) = 1.602 \(\times\) 10\(^{-19}\) C, which are values obtained in other experiments. (a) Select a way to plot your results so that the data points fall close to a straight line. Using that plot, find the slope and y-intercept of the best-fit straight line to the data. (b) Use the results of part (a) to calculate Planck's constant \(h\) (as a test of your data) and the work function (in eV) of the surface. (c) What is the longest wavelength of light that will produce photoelectrons from this surface? (d) What wavelength of light is required to produce photoelectrons with kinetic energy 10.0 eV?

Consider Compton scattering of a photon by a \(moving\) electron. Before the collision the photon has wavelength \(\lambda\) and is moving in the +\(x\)-direction, and the electron is moving in the -\(x\)-direction with total energy \(E\) (including its rest energy \(mc^2\)). The photon and electron collide head-on. After the collision, both are moving in the -\(x\)-direction (that is, the photon has been scattered by 180\(^\circ\)). (a) Derive an expression for the wavelength \(\lambda'\) of the scattered photon. Show that if \(E \gg mc^2\), where m is the rest mass of the electron, your result reduces to $$\lambda' = {hc \over E} (1 + {m^2c^4\lambda \over 4hcE}) $$ (b) A beam of infrared radiation from a CO\(_2\) laser (\(\lambda = 10.6 \mu{m}\)) collides head-on with a beam of electrons, each of total energy \(E\) = 10.0 GeV (1 GeV = 10\(^9\) eV). Calculate the wavelength \(\lambda'\) of the scattered photons, assuming a 180\(^\circ\) scattering angle. (c) What kind of scattered photons are these (infrared, microwave, ultraviolet, etc.)? Can you think of an application of this effect?

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

A photon with wavelength \(\lambda\) = 0.1385 nm scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is 8.90 \(\times\) 10\(^6\) m/s?

A photon of green light has a wavelength of 520 nm. Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

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