/*! 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 76 Calculate the energy in electron... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the energy in electron volts of (a) an electron that has de Broglie wavelength 400 \(\mathrm{nm}\) and (b) a photon that has wavelength 400 \(\mathrm{nm} .\)

Short Answer

Expert verified
Electron energy: 0.941 eV; photon energy: 3.10 eV.

Step by step solution

01

Understand the Problem

We need to calculate the energy of both an electron and a photon, each with a wavelength of 400 nm. We'll use different formulas for each since electrons and photons have different behaviors.
02

Convert Wavelength to Standard Unit

Convert 400 nm to meters for calculations since most constants are in SI units. 400 nm = 400 x 10^{-9} m
03

Calculate de Broglie Wavelength Energy for Electron

The energy of an electron can be calculated using the de Broglie wavelength formula: \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the wavelength, \( h \) is Planck’s constant \(6.626 \times 10^{-34} \ \text{m}^2 \text{kg/s}\), and \( p \) is the momentum.First determine momentum: \[ p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{400 \times 10^{-9}} = 1.6565 \times 10^{-24} \ \text{kg m/s} \]
04

Electron Kinetic Energy Calculation

The kinetic energy \( K \) of an electron is given by \[ K = \frac{p^2}{2m} \]where \( m \) is the mass of the electron \(9.109 \times 10^{-31} \text{ kg} \). So,\[ K = \frac{(1.6565 \times 10^{-24})^2}{2 \times 9.109 \times 10^{-31}} = 1.508 \times 10^{-19} \text{ J} \] Convert \( K \) from joules to electron volts (1 eV = \(1.602 \times 10^{-19} \text{ J} \)):\[ E_{e^-} = \frac{1.508 \times 10^{-19}}{1.602 \times 10^{-19}} = 0.941 \text{ eV} \]
05

Calculate Energy for Photon

For a photon, use the Planck-Einstein relation:\[ E = \frac{hc}{\lambda} \]where \( c \) is the speed of light \(3 \times 10^8 \ \text{m/s}\).\[ E_{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \text{ J} \]Convert \( E_{photon} \) to electron volts: \[ E_{photon} = \frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} = 3.10 \text{ eV} \]
06

Final Results

For the electron, the energy is approximately 0.941 eV. For the photon, the energy is approximately 3.10 eV.

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Key Concepts

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

de Broglie Wavelength
The de Broglie wavelength is a fundamental concept in quantum mechanics, suggesting that all particles exhibit both particle and wave characteristics. Introduced by Louis de Broglie, it relates a particle's momentum to its wavelength. This notion helps bridge the gap between classical and quantum physics, explaining phenomena that classical physics cannot. In simple terms, the de Broglie wavelength (\[ \lambda = \frac{h}{p} \] where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is momentum) gives us a way to calculate the wave-like nature of matter.
  • **Planck's constant (\( h \)):** A crucial constant in quantum mechanics, it has the value \(6.626 \times 10^{-34} \text{m}^2\text{kg/s}\).
  • **Momentum (\( p \)):** The product of mass and velocity for an object, reflecting its motion.
At a subatomic level, particles like electrons have significant wave properties as depicted by their de Broglie wavelength. This equation allows us to understand how particles can behave like waves, influencing their energy and interactions at a quantum scale.
Planck-Einstein Relation
The Planck-Einstein relation is a cornerstone of quantum physics, emphasizing the dual nature of light and defining a direct relationship between the energy of a photon and its frequency. This vital equation \[ E = hf \] provides a method for calculating the energy (\( E \)) of photons by using Planck's constant (\( h \)) and the frequency (\( f \)) of the light wave. When dealing with wavelengths instead of frequency, the relation expands: \[ E = \frac{hc}{\lambda} \] where \( c \) is the speed of light and \( \lambda \) is the wavelength.
  • **Speed of light (\( c \)):** The constant speed at which light travels in a vacuum, approximately \(3 \times 10^8 \text{m/s}\).
  • **Wavelength (\( \lambda \)):** The distance between successive peaks of the wave, affecting the energy level.
This relationship is pivotal for understanding light's properties and computing photon energy in exercises, such as determining the energy conversion from joules to electron volts for different wavelengths.
Kinetic Energy Calculation
Kinetic energy is a fundamental concept in physics describing the energy an object possesses due to its motion. In quantum mechanics, calculating the kinetic energy of particles such as electrons requires an understanding of their wave properties and momentum. We utilize the formula:\[ K = \frac{p^2}{2m} \] where \( K \) is the kinetic energy, \( p \) is the momentum, and \( m \) is the mass of the particle.
  • **Momentum (\( p \)):** Derived from the de Broglie relation, expressing the wave-like motion of the electron.
  • **Mass (\( m \)):** The intrinsic property of the particle contributing to its kinetic energy.
This formula helps in converting the momentum, an abstraction of motion, into kinetic energy, a measurable quantity. For electrons, typically measured in joules or converted to electron volts. Through these calculations, we discern how an electron's motion contributes to its energy state in quantum systems. Understanding this concept allows students to appreciate how velocity and mass interplay to define a particle's kinetic dynamics.

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Most popular questions from this chapter

A beam of protons and a beam of alpha particles (of mass \(6.64 \times 10^{-27} \mathrm{kg}\) and charge \(+2 e )\) are accelerated from rest through the same potential difference and pass through identical circular holes in a very thin, opaque film. When viewed far from the hole, the diffracted proton beam forms its first dark ring at \(15^{\circ}\) with respect to its original direction. When viewed similarly, at what angle will the alpha particle form its first dark ring?

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