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Chapter 17: Problem 2421

Through what potential difference should an electron be accelerated so it's de-Broglie wavelength is \(0.3 \AA\). (A) \(1812 \mathrm{~V}\) (B) \(167.2 \mathrm{~V}\) (C) \(1516 \mathrm{~V}\) (D) \(1672.8 \mathrm{~V}\)

Short Answer

Expert verified
To find the potential difference through which an electron must be accelerated for its de Broglie wavelength to be 0.3 脜, we can use the combined formula \(V = \frac{h^2}{2m_eq位^2}\). Plugging in the values for Planck's constant (h), electron's mass (\(m_e\)), charge (q), and the given wavelength (位), we get \(V \approx 167.2 \, \text{V}\). Therefore, the correct answer is (B) \(167.2 \mathrm{~V}\).

Step by step solution

01

Understand the de Broglie wavelength formula

For any particle, the de Broglie wavelength (位) is given by the formula: \(位 = \frac{h}{p}\) Where h is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)), and p is the particle's momentum. For an electron, its momentum can be calculated using the formula: \(p = \sqrt{2m_eE}\) Where \(m_e\) is the electron's mass (\(9.11 \times 10^{-31} \, \text{kg}\)), and E is its energy.
02

Relating energy to potential difference

To find the energy of the electron when it is accelerated through a potential difference V, we can use the formula: \(E = qV\) Where E is the energy, q is the charge of the electron (which is \(1.6 \times 10^{-19} \, \text{C}\)), and V is the potential difference.
03

Combine formulas to calculate potential difference

We want to find the potential difference V, so we will combine the formulas for the de Broglie wavelength, momentum, and energy as follows: \(位 = \frac{h}{\sqrt{2m_eE}} = \frac{h}{\sqrt{2m_eqV}}\) Now, we can solve for V: \(V = \frac{h^2}{2m_eq位^2}\)
04

Calculate the potential difference V

Now that we have the formula for the potential difference V, we can plug in the given values and solve for V: \(V = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})^2}{2(9.11 \times 10^{-31} \, \text{kg})(1.6 \times 10^{-19} \, \text{C})(0.3 \times 10^{-10} \, \text{m})^2}\) After calculating, we get: \(V \approx 167.2 \, \text{V}\) So, the correct answer is: (B) \(167.2 \mathrm{~V}\)

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Key Concepts

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

Electron Acceleration
When an electron is accelerated, it gains kinetic energy which directly influences its motion. This is a crucial concept in quantum mechanics, as we often need to manipulate electrons to achieve desired outcomes in experiments and technology. When we talk about accelerating an electron through a "potential difference," we're discussing the work done on the electron by an electric field.
  • This process increases the electron's kinetic energy, turning it into a speeding particle.
  • The energy gained by the electron is directly related to the potential difference it is accelerated through.
  • This acceleration impacts how we calculate other properties of the electron, like its momentum and wavelength.
In summary, understanding how electrons are accelerated helps us connect the dots between electrical forces and particle movement, which is vital in many areas of physics.
Potential Difference
A potential difference, often referred to as voltage, plays a significant role in the dynamics of electrons and their behavior. This is particularly important when considering the de Broglie wavelength of particles, such as electrons.
The potential difference is essentially the "push" that moves electrons from one place to another, energizing them in the process.
  • Electrons gain energy when subjected to a difference in electric potential between two points.
  • This energy gain is key to altering their speed and, consequently, other properties like wavelength and momentum.
  • The measurement of this voltage is critical for determining how electrons behave in electric fields and circuits.
Understanding potential difference is fundamental because it allows us to predict and manipulate the energy levels of electrons, crucial for designing electronic devices and conducting experiments.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior of very small particles like electrons. It introduces unique concepts such as wave-particle duality, where particles like electrons exhibit both particle-like and wave-like characteristics.
One of the intriguing aspects of quantum mechanics is the use of the de Broglie wavelength, which helps us understand experimental outcomes concerning particles.
  • The de Broglie hypothesis states that every moving particle, including electrons, can be associated with a wave.
  • This wave property is essential for understanding phenomena at microscopic levels, such as interference and diffraction.
  • Quantum mechanics provides the theoretical framework to analyze particle behavior, which classical physics cannot explain.
Ultimately, quantum mechanics opens the door to understanding the deeper principles of nature, offering insights that are foundational to modern technology and scientific advancements.
Particle Momentum
Momentum is a key concept in physics that describes the quantity of motion an object possesses. In the realm of quantum mechanics, particle momentum becomes even more intriguing due to its relationship with the de Broglie wavelength.
Learn how momentum interconnects with other quantum physics concepts:
  • An electron's momentum is calculated as the product of its mass and velocity, but can also be determined from its wave-like properties.
  • According to the de Broglie relationship, momentum is inversely proportional to wavelength: as momentum increases, wavelength decreases.
  • Understanding an electron's momentum is crucial for calculating its behavior in quantum experiments and for technologies like electron microscopes.
By grasping particle momentum, we gain a deeper appreciation for the dual nature of matter and its applications in various scientific and technological domains.

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Most popular questions from this chapter

Which of the following phenomenon can not be explained by quantum theory of light? (A) Emission of radiation from black body (B) Photo electric effect (C) Polarization (D) Crompton effect

Particle \(\mathrm{A}\) and \(\mathrm{B}\) have electric charge \(+\mathrm{q}\) and \(+4 \mathrm{q} .\) Both have mass \(\mathrm{m}\). If both are allowed to fall under the same p.d., ratio of velocities \(\left(\mathrm{V}_{\mathrm{A}} / \mathrm{V}_{\mathrm{B}}\right)=\ldots \ldots \ldots \ldots \ldots\) (A) \(2: 1\) (B) \(1: 2\) (C) \(1: 4\) (D) \(4: 1\)

Photoelectric effect on surface is found for frequencies \(5.5 \times 10^{8} \mathrm{MHz}\) and \(4.5 \times 10^{8} \mathrm{MHz}\) If ratio of maximum kinetic energies of emitted photo electrons is \(1: 5\), threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) \(7.55 \times 10^{8} \mathrm{MHz}\) (B) \(4.57 \times 10^{8} \mathrm{MHz}\) (C) \(9.35 \times 10^{8} \mathrm{MHz}\) (D) \(5.75 \times 10^{8} \mathrm{MHz}\)

A proton and electron are lying in a box having impenetrable walls, the ratio of uncertainty in their velocities are \(\ldots \ldots\) \(\left(\mathrm{m}_{\mathrm{e}}=\right.\) mass of electron and \(\mathrm{m}_{\mathrm{p}}=\) mass of proton. (A) \(\left(\mathrm{m}_{\mathrm{e}} / \mathrm{m}_{\mathrm{p}}\right)\) (B) \(\mathrm{m}_{\mathrm{e}} \cdot \mathrm{m}_{\mathrm{p}}\) (C) \(\left.\sqrt{\left(m_{e}\right.} \cdot m_{p}\right)\) (D) \(\sqrt{\left(m_{e} / m_{p}\right)}\)

Direction Read the following question choose if: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) Assertion is true but the Reason is false. (d) Both Assertion and Reason is false. Assertion: Metals like Na or \(\mathrm{K}\), emit electrons even when visible lights fall on them. Reason: This is because their work function is low. (A) a (B) \(\mathrm{b}\) (C) \(\mathrm{c}\) (D) d
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