/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 What is the kinetic energy, in e... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the kinetic energy, in eV, of an electron with a de Broglie wayelength of \(1.0 \mathrm{nm}\) ?

Short Answer

Expert verified
The kinetic energy of the electron is approximately 15 eV

Step by step solution

01

Find momentum from de Broglie wavelength

Knowing the de Broglie wavelength \(\lambda = 1.0 \mathrm{nm} = 1.0 \times 10^{-9} \mathrm{m}\) and \(h\) as Planck's constant which is \(6.62607004 \times 10^{-34} \mathrm{m^2 kg / s}\), we can recast the formula to find momentum as \(p = \frac{h}{\lambda} = \frac{6.626\times10^{-34}}{1.0 \times 10^{-9}} = 6.626 \times 10^{-25} \mathrm{kg.m/s}\)
02

Find the kinetic energy

Knowing that the kinetic energy can be found by \(E_k = \frac{p^2}{2m}\), and given the electron mass \(m = 9.11 \times 10^{-31} \mathrm{kg}\), we substitute the values to find \(E_k = \frac{(6.626 \times 10^{-25})^2}{2(9.11 \times 10^{-31})} = 2.403 \times 10^{-17} \mathrm{J}\)
03

Convert the energy to Electronvolts

Finally, we need to convert the energy from Joules to electronvolts. We know that \(1 \mathrm{J} = 6.242 \times 10^{18} \mathrm{eV}\). Thus, the kinetic energy becomes \(E_k = 2.403 \times 10^{-17} \times 6.242 \times 10^{18} = 14.998 \mathrm{eV}\)

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Key Concepts

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

de Broglie Wavelength
One of the key concepts in this exercise is the de Broglie wavelength. This principle is fundamental in quantum mechanics, suggesting that every moving particle or object has an associated wave. The de Broglie wavelength is given by the equation \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) represents momentum.
For electrons, the de Broglie wavelength can help determine how they behave under different conditions. Electrons are considered both as particles and waves, and this concept blurs the line between classical and quantum physics. "Why is this important?" you might ask. Understanding this is crucial because it allows us to predict behaviors of particles like electrons in a precise manner.
It helps not only in the theoretical aspects of physics but also in practical applications, such as electron microscopy, wherein understanding an electron's wavelength enables us to achieve greater resolution.
  • Formula: \( \lambda = \frac{h}{p} \)
  • Relates to: Quantum mechanics and wave-particle duality
  • Applications: Electron microscopy, wave behavior prediction
Planck's Constant
Planck's constant is another essential part of understanding this exercise. It is a fundamental constant linked directly to the quantization of energy, revolutionizing how we understand quantum physics. The constant \( h \), approximately \( 6.62607004 \times 10^{-34} \ \mathrm{m^2 \cdot kg / s} \), acts as the bridge connecting the energy of a photon to its frequency via the relation \( E = hf \), where \( E \) is energy and \( f \) is frequency.
In the context of the de Broglie wavelength, Planck’s constant becomes a pivotal part of the equation \( \lambda = \frac{h}{p} \), thereby evidencing the wave behavior of particles.
"So why does it matter?" Simply put, Planck's constant allows physicists to scale down classical mechanics laws to quantum physics laws.
It's one of those constants that make life more predictable on a microscopic scale by providing a basis from which we can define not just a single system, but universal laws.
  • Value: \( 6.62607004 \times 10^{-34} \ \mathrm{m^2 \cdot kg / s} \)
  • Importance: Links energy and frequency
  • Role in de Broglie Equation: Determines particle wavelength
  • Significance: Fundamental in the formulation of quantum mechanics
Electronvolt Conversion
Finally, converting energy units to electronvolts (eV) is the last essential concept here. Often in physics, particularly in the realm of particle physics and quantum mechanics, energy quantities are more practically expressed in terms of electronvolts rather than joules.
An electronvolt is defined as the amount of kinetic energy gained or lost by an electron when it accelerates through an electric potential difference of one volt. Essentially, it's a more convenient way of expressing small energies.
This exercise involves a conversion from joules to electronvolts using the relation: \( 1 \ \mathrm{J} = 6.242 \times 10^{18} \ \mathrm{eV} \). Understanding how to convert between these units is not only crucial for solving problems but also for interpreting scientific data, as electronvolts are a standard unit of measure in fields like high-energy physics.
  • Definition: Energy unit commonly used in particle physics
  • Relations: \( 1 \ \mathrm{J} = 6.242 \times 10^{18} \ \mathrm{eV} \)
  • Purpose: Describing small energies with easier numerical values

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Most popular questions from this chapter

Dinoflagellates are singlecell creatures that float in theworld's oceans; many types are bioluminescent. When disurbed by motion in the water, a typical bioluminescent dinoflagellate emits 100,000,000 photons in a 0.10 -s-long flash of light of wavelength 460 nm. What is the power of the flash in watts?

II A study of photosynthesis in phytoplankton in the open ocean used short pulses of laser light to trigger photosynthetic reactions. The investigator's system used \(0.10 \mathrm{mW}\) pulses of \(640 \mathrm{nm}\) laser light of length 200 ps. How many photons were contained in each pulse?

As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a channel in a microprocessor. If we know that an electron is somewhere along the \(50 \mathrm{nm}\) length of the channel, what is \(\Delta v_{x} ?\) If we treat the electron as a classical particle moving at a speed at the outer edge of the uncertainty range, how long would it take to traverse the channel?

Exposure to a sufficient quantity of ultraviolet light will redden the skin, producing erythema-a sunburn. The amount of exposure necessary to produce this reddening depends on the wavelength. For a \(1.0 \mathrm{cm}^{2}\) patch of skin, \(3.7 \mathrm{mJ}\) of ultraviolet light at a wavelength of 254 nm will produce reddening; at \(300 \mathrm{nm}\) wavelength, \(13 \mathrm{mJ}\) are required. What is the photon energy corresponding to each of these wavelengths? b. How many total photons does each of these exposures correspond to? c. Explain why there is a difference in the number of photons needed to provoke a response in the two cases.

It is stated in the text that special relativity must be used to calculate the de Broglie wavelength of electrons in an electron microscope. Let us discover how much of an effect relativity has. Consider an electron accelerated through a potential difference of \(1.00 \times 10^{5} \mathrm{V}\) a. Using the Newtonian (nonrelativistic) expressions for kinetic energy and momentum, what is the electron's de Broglie wavelength? b. The de Broglie wavelength is \(\lambda=h / p,\) but the momentum of a relativistic particle is not \(m v .\) Using the relativistic expressions for kinetic energy and momentum, what is the electron's de Broglie wavelength?

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