Ch 38: Quantization

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 38: Quantization

Problem 17

Chapter 38, Problem 17

INT Through what potential difference must an electron be accelerated from rest to have a de Broglie wavelength of 500 nm?

Verified step by step guidance

Step 1: Recall the formula for the de Broglie wavelength, which is \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle.

Step 2: Relate the momentum \( p \) to the kinetic energy \( K \) of the electron using \( p = \sqrt{2m_e K} \), where \( m_e \) is the mass of the electron and \( K \) is the kinetic energy.

Step 3: The kinetic energy \( K \) of the electron is equal to the work done on it by the electric field, which is \( K = e \Delta V \), where \( e \) is the charge of the electron and \( \Delta V \) is the potential difference.

Step 4: Combine the equations \( \lambda = \frac{h}{\sqrt{2m_e e \Delta V}} \) to express \( \Delta V \) in terms of \( \lambda \), \( h \), \( m_e \), and \( e \). Rearrange to solve for \( \Delta V \): \( \Delta V = \frac{h^2}{2m_e e \lambda^2} \).

Step 5: Substitute the known values: \( h = 6.626 \times 10^{-34} \, \text{J·s} \), \( m_e = 9.109 \times 10^{-31} \, \text{kg} \), \( e = 1.602 \times 10^{-19} \, \text{C} \), and \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \) into the formula to calculate \( \Delta V \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

de Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics that relates the wavelength of a particle to its momentum. It is given by the formula λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. For an electron, this wavelength can be calculated when its velocity is known, linking wave-like and particle-like behavior.

Kinetic Energy and Potential Difference

When an electron is accelerated through a potential difference (V), it gains kinetic energy equal to the work done on it by the electric field. This relationship is expressed as KE = eV, where KE is the kinetic energy, e is the charge of the electron, and V is the potential difference. This principle is crucial for determining the energy required to achieve a specific velocity, which in turn affects the de Broglie wavelength.

Momentum of an Electron

The momentum of an electron is defined as the product of its mass and velocity (p = mv). In the context of quantum mechanics, the momentum is also related to the de Broglie wavelength. As the electron is accelerated, its velocity increases, leading to a change in momentum, which directly influences the wavelength according to the de Broglie relation.

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