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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

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?

A photon with wavelength \(\lambda\) = 0.0980 nm is incident on an electron that is initially at rest. If the photon scatters in the backward direction, what is the magnitude of the linear momentum of the electron just after the collision with the photon?

An x ray with a wavelength of 0.100 nm collides with an electron that is initially at rest. The x ray's final wavelength is 0.110 nm. What is the final kinetic energy of the electron?

A 75-W light source consumes 75 W of electrical power. Assume all this energy goes into emitted light of wavelength 600 nm. (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit? (c) Are the answers to parts (a) and (b) the same? Is the frequency of the light the same thing as the number of photons emitted per second? Explain.

An x-ray photon is scattered from a free electron (mass \(m\)) at rest. The wavelength of the scattered photon is \(\lambda'\), and the final speed of the struck electron is \(v\). (a) What was the initial wavelength \(\lambda\) of the photon? Express your answer in terms of \(\lambda\), \(v\), and \(m\). (\(Hint\): Use the relativistic expression for the electron kinetic energy.) (b) Through what angle \(\phi\) is the photon scattered? Express your answer in terms of \(\lambda\), \(\lambda'\), and \(m\). (c) Evaluate your results in parts (a) and (b) for a wavelength of 5.10 \(\times\) 10\(^{-3}\) nm for the scattered photon and a final electron speed of 1.80 \(\times\) 10\(^8\) m/s. Give \(\phi\) in degrees.

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