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In an old-fashioned television set, electrons are accelerated through a potential difference of \(25.0 \mathrm{kV}\). What is the de Broglie wavelength of such electrons? (Relativity is not needed.)

Short Answer

Expert verified
The de Broglie wavelength is approximately \(1.10 \times 10^{-11} \, \mathrm{m}\).

Step by step solution

01

Understand the Relationship Between Kinetic Energy and Potential Difference

The kinetic energy gained by the electrons is equal to the work done on them by the electric field, which can be expressed as \( KE = eV \), where \( e \) is the charge of an electron \( (1.6 \times 10^{-19} \, \mathrm{C}) \) and \( V \) is the potential difference (given as 25.0 kV, or 25,000 V).
02

Calculate the Kinetic Energy

Use the formula from Step 1 to calculate the kinetic energy: \[ KE = eV = (1.6 \times 10^{-19} \, \mathrm{C})(25,000 \, \mathrm{V}) = 4.0 \times 10^{-15} \, \mathrm{J} \]
03

Relate Kinetic Energy to Momentum

The kinetic energy for non-relativistic particles is given by \( KE = \frac{p^2}{2m} \), where \( p \) is the momentum and \( m \) is the mass of an electron \( (9.11 \times 10^{-31} \, \mathrm{kg}) \). Rearrange the equation to solve for momentum: \[ p = \sqrt{2mKE} \]
04

Calculate the Momentum of the Electron

Substitute the known values into the momentum equation: \[ p = \sqrt{2 \times 9.11 \times 10^{-31} \, \mathrm{kg} \times 4.0 \times 10^{-15} \, \mathrm{J}} \approx 6.02 \times 10^{-24} \, \mathrm{kg \, m/s} \]
05

Use de Broglie's Equation

The de Broglie wavelength is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( (6.63 \times 10^{-34} \, \mathrm{J \, s}) \).
06

Calculate de Broglie Wavelength

Substitute the calculated momentum and Planck's constant into the de Broglie wavelength equation: \[ \lambda = \frac{6.63 \times 10^{-34} \, \mathrm{J \, s}}{6.02 \times 10^{-24} \, \mathrm{kg \, m/s}} \approx 1.10 \times 10^{-11} \, \mathrm{m} \]

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Key Concepts

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

Electron Acceleration
When electrons move in a television set, they are sped up or accelerated using a potential difference. Imagine potential difference as a means of giving energy to the electrons, which propels them forward. This is similar to giving a ball a push to roll it down a hill. The potential difference provides the energy needed to increase the speed of the electrons. In this exercise, the given potential difference is 25.0 kV, which is equivalent to 25,000 volts.
The potential difference creates an electric field that pushes the electrons, allowing them to gain speed as they travel through its influence. Understanding electron acceleration in this way helps to conceptualize how electrons acquire the necessary speed to reach the television screen and create an image.
Kinetic Energy and Potential Difference
The potential difference is crucial in calculating the kinetic energy gained by the electrons. Kinetic energy is determined using the formula: \[ KE = eV \] where \( e \) is the electron charge \( (1.6 \times 10^{-19} \, \text{C}) \) and \( V \) is the potential difference (25,000 V in this scenario). This equation shows that kinetic energy is directly related to the amount of work done on electrons by the potential difference.
  • High potential difference = More kinetic energy
  • Low potential difference = Less kinetic energy
For our exercise, multiplying the electron charge by the potential difference gives a kinetic energy of \( 4.0 \times 10^{-15} \, \text{J} \). This value is essential for further calculations in determining the de Broglie wavelength.
Momentum Calculation
Understanding momentum is crucial for computing the de Broglie wavelength. Momentum, denoted as \( p \), connects with kinetic energy through the equation: \[ KE = \frac{p^2}{2m} \] Here, \( m \) is the mass of an electron \( (9.11 \times 10^{-31} \, \text{kg}) \). Rearranging this equation allows us to solve for momentum: \[ p = \sqrt{2mKE} \]
By substituting the known values for electron mass and kinetic energy, we can calculate the momentum of the electron. In this case, the result is \( 6.02 \times 10^{-24} \, \text{kg} \, \text{m/s} \). Calculating momentum accurately is necessary because it feeds directly into the de Broglie wavelength equation.
Planck's Constant
Planck's constant \( h \) is a fundamental value in quantum mechanics, linking a particle's wavelength with its momentum. In the de Broglie equation, a particle's wavelength is defined by: \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the de Broglie wavelength and \( p \) is momentum.
For electron wavelength calculation, Planck's constant is typically \( 6.63 \times 10^{-34} \, \text{J} \, \text{s} \). By substituting the previously calculated momentum and this constant, we find the de Broglie wavelength of an electron. The final computed wavelength for our example is approximately \( 1.10 \times 10^{-11} \, \text{m} \). This wavelength represents the wave-like properties of particles, showcasing the intriguing dual nature of matter.

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Most popular questions from this chapter

A special kind of lightbulb emits monochromatic light of wavelength \(630 \mathrm{~nm}\). Electrical energy is supplied to it at the rate of \(60 \mathrm{~W}\), and the bulb is \(93 \%\) efficient at converting that energy to light energy. How many photons are emitted by the bulb during its lifetime of \(730 \mathrm{~h}\) ?

In a photoelectric experiment using a sodium surface, you find a stopping potential of \(1.85 \mathrm{~V}\) for a wavelength of \(300 \mathrm{~nm}\) and a stopping potential of \(0.820 \mathrm{~V}\) for a wavelength of \(400 \mathrm{~nm}\). From these data find (a) a value for the Planck constant, (b) the work function \(\Phi\) for sodium, and (c) the cutoff wavelength \(\lambda_{0}\) for sodium.

Consider a collision between an x-ray photon of initial energy \(50.0 \mathrm{keV}\) and an electron at rest, in which the photon is scattered backward and the electron is knocked forward. (a) What is the energy of the backscattered photon? (b) What is the kinetic energy of the electron?

(a) The smallest amount of energy needed to eject an electron from metallic sodium is \(2.28 \mathrm{eV}\). Does sodium show a photoelectric effect for red light, with \(\lambda=680 \mathrm{~nm} ?\) (That is, does the light cause electron emission?) (b) What is the cutoff wavelength for photoelectric emission from sodium? (c) To what color does that wavelength correspond?

Find the maximum kinetic energy of electrons ejected from a certain material if the material's work function is \(2.3 \mathrm{eV}\) and the frequency of the incident radiation is \(3.0 \times 10^{15} \mathrm{~Hz}\).

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