/*! 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 18 Show that the angular wave numbe... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that the angular wave number \(k\) for a nonrelativistic free particle of mass \(m\) can be written as $$ k=\frac{2 \pi \sqrt{2 m K}}{h} $$ in which \(K\) is the particle's kinetic energy.

Short Answer

Expert verified
The expression for the angular wave number is derived as \( k = \frac{2 \pi \sqrt{2 m K}}{h} \).

Step by step solution

01

Recall the relationship between momentum and wave number

We know that the momentum \( p \) of a particle is related to the wave number \( k \) through the de Broglie relation: \( p = \hbar k \), where \( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant.
02

Express momentum in terms of kinetic energy

For a nonrelativistic particle, the kinetic energy \( K \) can be expressed in terms of momentum \( p \) as: \( K = \frac{p^2}{2m} \). Solving for \( p \), we get \( p = \sqrt{2mK} \).
03

Substitute momentum into the wave number equation

Substitute the expression for momentum \( p = \sqrt{2mK} \) into the de Broglie relation \( p = \hbar k \). This gives us \( \hbar k = \sqrt{2mK} \).
04

Solve for the wave number \( k \)

Isolate \( k \) in the equation \( \hbar k = \sqrt{2mK} \) to get \( k = \frac{\sqrt{2mK}}{\hbar} \). Since \( \hbar = \frac{h}{2\pi} \), substitute this into the expression to get \( k = \frac{2\pi \sqrt{2mK}}{h} \).

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Key Concepts

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

Nonrelativistic Free Particle
In physics, a nonrelativistic free particle refers to a particle that moves freely without any forces acting on it, and its speed is much lower than the speed of light. This implies that we do not need to consider the effects of relativity; thus, calculations become more straightforward. The energy and momentum of the particle can be described using classical mechanics.
In this scenario, the relationship between momentum and kinetic energy is crucial. Since no external forces apply, the particle's energy remains constant, which simplifies the analysis. This assumption allows us to use classical formulas like \(K = \frac{p^2}{2m}\) for kinetic energy, where \(m\) is the particle's mass.
Kinetic Energy
Kinetic energy, represented as \(K\), is the energy possessed by a particle due to its motion. For a nonrelativistic free particle, we express it using the simple classical formula \(K = \frac{1}{2}mv^2\), where \(v\) is the particle's velocity.
Alternatively, since momentum \(p\) is related to velocity by \(p = mv\), another useful expression is \(K = \frac{p^2}{2m}\). By using this equation, we can connect kinetic energy directly to momentum. This connection is fundamental in deriving the wave number \(k\) for a particle.
Understanding kinetic energy in this context is vital because it links the particle's motion to wave phenomena through the relation of its momentum to the wave number.
de Broglie Relation
The de Broglie relation forms a bridge between particle properties and wave behavior. According to de Broglie, every particle has an associated wave characterized by a wavelength \(\lambda\). This relationship is given by \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant and \(p\) is the particle's momentum.
Rearranging this, we find the wave number \(k\), which is defined as \(k = \frac{2\pi}{\lambda}\). Substituting the de Broglie wavelength, we obtain \(k = \frac{2\pi p}{h}\). Connecting this to the step-by-step solution, when we express momentum as \(p = \sqrt{2mK}\), we arrive at a fundamental expression for the wave number of a particle in terms of its kinetic energy.
The de Broglie relation is a cornerstone of wave-particle duality, highlighting that particles can exhibit both particle-like and wave-like properties.
Reduced Planck's Constant
The reduced Planck's constant, denoted as \(\hbar\), is an important constant in quantum mechanics. It is equal to \(\frac{h}{2\pi}\), where \(h\) is Planck's constant. This version of Planck's constant often simplifies equations that involve angular frequencies or wave numbers.
In the context of a nonrelativistic free particle, reduced Planck's constant links the particle's momentum to its wave number through the equation \(p = \hbar k\). This formula is useful because it ties together the concepts of momentum and wave-like behavior in quantum mechanics.
The reduced Planck's constant appears frequently in equations due to its ability to simplify the mathematical relationship between classical and quantum descriptions.

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Most popular questions from this chapter

You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to "observe" such an orbiting electron by using a light microscope to measure the electron's presumed orbital position with a precision of, say, \(10 \mathrm{pm}\) (a typical atom has a radius of about \(100 \mathrm{pm}\). The wavelength of the light used in the microscope must then be about \(10 \mathrm{pm}\). (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of "viewing" an atomic electron at two or more points along its presumed orbital path? (Hint. The outer electrons of atoms are bound to the atom by energies of only a few electron-volts.)

Under ideal conditions, a visual sensation can occur in the human visual system if light of wavelength \(550 \mathrm{~nm}\) is absorbed by the eye's retina at a rate as low as 110 photons per second. What is the corresponding rate at which energy is absorbed by the retina?

What are (a) the energy of a photon corresponding to wavelength \(1.00 \mathrm{~nm}\), (b) the kinetic energy of an electron with de Broglie wavelength \(1.00 \mathrm{~nm}\), (c) the energy of a photon corresponding to wavelength \(1.00 \mathrm{fm}\), and (d) the kinetic energy of an electron with de Broglie wavelength \(1.00 \mathrm{fm}\) ?

The beam emerging from a \(1.5 \mathrm{~W}\) argon laser \((\lambda=515 \mathrm{~nm})\) has a diameter \(d\) of \(3.5 \mathrm{~mm}\). The beam is focused by a lens system with an effective focal length \(f_{\mathrm{L}}\) of \(2.5 \mathrm{~mm}\). The focused beam strikes a totally absorbing screen, where it forms a circular diffraction pattern whose central disk has a radius \(R\) given by \(1.22 f_{\mathrm{L}} \lambda / d\). It can be shown that \(84 \%\) of the incident energy ends up within this central disk. At what rate are photons absorbed by the screen in the central disk of the diffraction pattern?

A 100 W sodium lamp \((\lambda=589 \mathrm{~nm})\) radiates energy uniformly in all directions. (a) At what rate are photons emitted by the lamp? (b) At what distance from the lamp will a totally absorbing screen absorb photons at the rate of \(1.00\) photon/ \(\mathrm{cm}^{2} \cdot \mathrm{s} ?\) (c) What is the photon flux (photons per unit area per unit time) on a small screen \(2.00 \mathrm{~m}\) from the lamp?

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