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Using the classical equations for momentum and kinetic energy, show that an electron's de Broglie wavelength in nanometers can be written as \(\lambda=1.226 / \sqrt{K}\), in which \(K\) is the electron's kinetic energy in electron-volts.

Short Answer

Expert verified
The de Broglie wavelength \( \lambda = \frac{1.226}{\sqrt{K}} \) comes from substituting known constants into the formula for momentum.

Step by step solution

01

Write down the de Broglie wavelength formula

The de Broglie wavelength of any particle is given by the formula \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle.
02

Express momentum in terms of kinetic energy

The classical momentum \( p \) is given by \( p = \sqrt{2mK} \), where \( m \) is the mass of the electron and \( K \) is its kinetic energy.
03

Substitute momentum in the de Broglie wavelength formula

Substitute \( p \) in the de Broglie formula to get \( \lambda = \frac{h}{\sqrt{2mK}} \).
04

Plug in the values for constants

Use \( h = 6.626 \times 10^{-34} \text{Js} \) and \( m = 9.109 \times 10^{-31} \text{kg} \) and substitute to find \( \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.109 \times 10^{-31} \times K}} \).
05

Convert to nanometers and simplify

When \( K \) is given in electron volts (eV), conversion is required. Use \(1 \text{ eV} = 1.602 \times 10^{-19} \text{J} \). Substitute this in and simplify to get the wavelength in nanometers. Converting yields \( \lambda \approx \frac{1.226}{\sqrt{K}} \) nm by calculating the constants together.

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Key Concepts

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

Electron Kinetic Energy
Kinetic energy is a fascinating and critical concept in physics that explains how much energy a particle possesses due to its motion. For an electron, which is a subatomic particle, its kinetic energy, denoted as \( K \), plays a significant role in its behavior and characteristics.

Kinetic energy can be calculated using the formula \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the electron. However, when we look at electrons on a quantum scale, like when considering their de Broglie wavelength, we often express kinetic energy in electron volts (eV).

An electronvolt is a unit of energy that is particularly handy in quantum and atomic physics. It helps in representing smaller quantities of energy like those encountered with electrons. When we know the kinetic energy of an electron in eV, we can easily explore other properties, such as its de Broglie wavelength, using simplified equations due to predefined constants. This makes the study of quantum behaviors more accessible and comprehensible, allowing more straightforward calculations when assessing particle properties like wavelengths.
Momentum
Momentum, a measure of motion, is a key concept in physics and one that is tightly linked to kinetic energy. For an electron, momentum is essential in determining its behavior, especially at the quantum level.

The classical formula for momentum \( p \) is given by \( p = mv \), where \( m \) is mass and \( v \) is velocity. However, in the scenario of deriving the de Broglie wavelength, momentum can be expressed in terms of electron kinetic energy: \( p = \sqrt{2mK} \).

This expression is particularly useful when working with the de Broglie wavelength, as it allows a clever link between the electron's energy and its wavelength. A key takeaway is that an electron's momentum increases with its kinetic energy, leading to a shorter de Broglie wavelength. This is vital when studying electron behavior at microscopic levels, such as in electron microscopes or particle accelerators, where momentum helps scientists predict an electron's path and interactions.
Planck's Constant
Planck's constant, symbolized as \( h \), is a fundamental constant that appears frequently in quantum physics equations. It connects the wave and particle nature of electrons and is crucial in calculating an electron's de Broglie wavelength.

Defined as \( h = 6.626 \times 10^{-34} \text{Js} \), Planck's constant can be thought of as a scaling factor between energy and frequency. In the context of de Broglie waves, it helps translate momentum into wavelength through the relationship \( \lambda = \frac{h}{p} \).

Understanding Planck's constant allows us to see why electrons and other particles sometimes behave like waves. This dual behavior is a key aspect of quantum theory. When determining the de Broglie wavelength, Planck's constant ensures that even though particles like electrons have mass and momentum, they exhibit wave-like properties that affect how they move and interact in space. Such insights have profound implications in technologies such as semiconductors and lasers, which rely on quantum mechanical properties and calculations.

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Most popular questions from this chapter

An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Satellites must be designed to minimize such charging because it can ruin the sensitive microelectronics. Suppose a satellite is coated with platinum, a metal with a very large work function \((\Phi=5.32 \mathrm{eV}) .\) Find the longest wavelength of incident sunlight that can eject an electron from the platinum.

Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of photographic film and thus recorded on the film. Photon absorption will occur if the photon energy equals or exceeds \(0.6 \mathrm{eV},\) the smallest amount of energy needed to dissociate an AgBr molecule in the film. (a) What is the greatest wavelength of light that can be recorded by the film? (b) In what region of the electromagnetic spectrum is this wavelength located?

What is the maximum kinetic energy of electrons knocked out of a thin copper foil by Compton scattering of an incident beam of \(17.5 \mathrm{keV} \mathrm{x}\) rays? Assume the work function is negligible.

The existence of the atomic nucleus was discovered in 1911 by Ernest Rutherford, who properly interpreted some experiments in which a beam of alpha particles was scattered from a metal foil of atoms such as gold. (a) If the alpha particles had a kinetic energy of \(7.5 \mathrm{MeV},\) what was their de Broglie wavelength? (b) Explain whether the wave nature of the incident alpha particles should have been taken into account in interpreting these experiments. The mass of an alpha particle is \(4.00 \mathrm{u}\) (atomic mass units), and its distance of closest approach to the nuclear center in these experiments was about \(30 \mathrm{fm}\). (The wave nature of matter was not postulated until more than a decade after these crucial experiments were first performed.)

An electron moves through a region of uniform electric potential of \(-200 \mathrm{~V}\) with a (total) energy of \(500 \mathrm{eV}\). What are its (a) kinetic energy (in electron-volts), (b) momentum, (c) speed, (d) de Broglie wavelength, and (e) angular wave number?

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