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In a TV picture tube the accelerating voltage is 15.0 \(\mathrm{kV}\) , and the electron beam passes through an aperture 0.50 \(\mathrm{mm}\) in diameter to a screen 0.300 \(\mathrm{m}\) away. (a) Calculate the uncertainty in the component of the electron's velocity perpendicular to the line between aperture and screen. (b) What is the uncertainty in position of the point where the electrons strike the screen? (c) Does this uncertainty affect the clarity of the picture significantly? (Use nonrelativistic expressions for the motion of the electrons. This is fairly accurate and is certainly adequate for obtaining an estimate of uncertainty effects.)

Short Answer

Expert verified
The uncertainty in electron's velocity is \( \Delta v \approx 5.79 \times 10^{3} \mathrm{m/s} \). Uncertainty in position on screen \( \Delta y \approx 1 \mathrm{nm} \), not significantly affecting clarity.

Step by step solution

01

Understand the Problem

We need to calculate uncertainty in velocity and position of electrons in a TV picture tube under the influence of an accelerating voltage. We'll need the aperture dimensions and distance to the screen to find these uncertainties.
02

(a): Calculate Electron Velocity

Given voltage is 15.0 \( \mathrm{kV} \). Use \( v = \sqrt{\frac{2 e V}{m_e}} \) to find the velocity of an electron. Here \( e = 1.6 \times 10^{-19} \mathrm{C} \), \( V = 15000 \mathrm{V} \), and \( m_e = 9.11 \times 10^{-31} \mathrm{kg} \). Calculate \( v \).
03

Apply Heisenberg Uncertainty Principle

Heisenberg's principle states \( \Delta x \Delta p \geq \frac{h}{4\pi} \). In this context, \( \Delta x = 0.5 \times 10^{-3} \mathrm{m} \) (aperture diameter), and \( \Delta p = m_e \Delta v \). Substitute \( \Delta x \) and solve for \( \Delta v \).
04

(a): Solve for Uncertainty in Velocity

Substitute known values into the equality \( \Delta v = \frac{h}{4\pi m_e \Delta x} \), with \( h = 6.63 \times 10^{-34} \mathrm{m^2 kg/s} \). Compute \( \Delta v \).
05

(b): Calculate Uncertainty in Position on Screen

Use \( \Delta y = \frac{\Delta v \cdot L}{v} \) to find positional uncertainty. Here, \( L = 0.300 \mathrm{m} \) is the distance to the screen. Substitute for \( \Delta v \) and initial velocity \( v \).
06

Calculate and Interpret Result

Calculate \( \Delta y \). If \( \Delta y \) is smaller than typical screen pixel size, it does not affect the clarity significantly. If larger, discuss impact.

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Key Concepts

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

Electron Velocity
Electrons within a TV picture tube are accelerated between two points, speeding them up to high velocities. In this scenario, we are given an accelerating voltage of 15.0 kV. This voltage is crucial as it determines just how fast the electrons will travel.
To compute their velocity, use the formula:
  • \( v = \sqrt{\frac{2 e V}{m_e}} \)
where:
  • \( e = 1.6 \times 10^{-19} \mathrm{C} \) is the elementary charge,
  • \( V = 15000 \mathrm{V} \) is the given accelerating voltage,
  • \( m_e = 9.11 \times 10^{-31} \mathrm{kg} \) is the mass of an electron.
By substituting the known values, this results in a precise calculation of how fast the electrons move after being accelerated, allowing them to travel through the TV picture tube to the screen. Fast-moving electrons play a vital part in the functioning of TV picture tubes, helping them display clear images.
Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle is fundamental in quantum mechanics and describes a limit to how precisely we can measure certain pairs of properties simultaneously. For an electron traveling in a TV picture tube, it addresses our uncertainty about both its position and momentum (related closely to velocity).
This principle is expressed mathematically as:
  • \( \Delta x \Delta p \geq \frac{h}{4\pi} \)
where:
  • \( \Delta x \) represents the uncertainty in position,
  • \( \Delta p \) relates to momentum which is \( m_e \Delta v \),
  • \( h = 6.63 \times 10^{-34} \mathrm{m^2 kg/s} \) is Planck's constant.
With the given aperture's diameter being \( \Delta x = 0.5 \times 10^{-3} \mathrm{m} \), we aim to solve for the velocity uncertainty, \( \Delta v \), using the relationship:
  • \( \Delta v = \frac{h}{4\pi m_e \Delta x} \)
This calculation gives insight into how much the speed of the electron might deviate when confined to such a small aperture.
TV Picture Tube
A TV picture tube, or cathode-ray tube (CRT), uses high-speed electrons to create images on a screen. The basic operation involves firing electrons through a tiny aperture towards the phosphorescent screen situated at a fixed distance away (in this case, 0.300 m).
The precision with which these electrons are directed is essential for producing a clear picture on the screen. However, due to the principles of quantum mechanics, as noted by the uncertainty principle, the electrons' exact positions can never be perfectly determined.
The position uncertainty where electrons strike the screen is computed using:
  • \( \Delta y = \frac{\Delta v \cdot L}{v} \)
where \( L = 0.300 \mathrm{m} \), \( \Delta v \) is the uncertainty in velocity derived previously, and \( v \) is the calculated velocity.
If the calculated \( \Delta y \) is less than the pixel dimension of the screen, the clarity of the picture remains unaffected. However, significant uncertainty may blur these pixels, thus impacting display quality. A careful balance in designing CRTs has allowed for this principle to not visibly affect picture resolution at the typical viewing distances.

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Most popular questions from this chapter

(a) What is the energy of a photon that has wavelength 0.10\(\mu \mathrm{m} ?\) (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10\(\mu \mathrm{m}\) in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

A hydrogen atom is in a state with energy \(-1.51 \mathrm{eV}\) . In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

Suppose that the uncertainty of position of an electron is equal to the radius of the \(n=1\) Bohr orbit for hydrogen. Calculate the simultaneous minimum uncertainty of the corresponding momentum component, and compare this with the magnitude of the momentum of the electron in the \(n=1\) Bohr orbit. Discuss your results.

Electrons go through a single slit 150 \(\mathrm{nm}\) wide and strike a screen 24.0 \(\mathrm{cm}\) away. You find that at angles of \(\pm 20.0^{\circ}\) from the center of the diffraction pattern, no electrons hit the screen but electrons hit at all points closer to the center, (a) How fast were these electrons moving when they went through the slit? (b) What will be the next larger angles at which no electrons hit the screen?

A \(10.0-\mathrm{g}\) marble is gently placed on a horizontal tabletop that is 1.75 \(\mathrm{m}\) wide. (a) What is the maximum uncertainty in the horizontal position of the marble? (b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately 14 billion years. (Hint: Can you know that the horizontal velocity of the marble is exactly zero?)

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