/*! 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 19 An electron is in a box of width... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron is in a box of width 3.0 \(\times\) 10\(^{-10}\) m. What are the de Broglie wavelength and the magnitude of the momentum of the electron if it is in (a) the \(n\) = 1 level; (b) the \(n\) = 2 level; (c) the \(n\) = 3 level? In each case how does the wavelength compare to the width of the box?

Short Answer

Expert verified
The electron's de Broglie wavelength decreases with increasing energy levels (higher n), and each wavelength becomes comparable to or smaller than the box width.

Step by step solution

01

Understand the Concept of de Broglie Wavelength

The de Broglie wavelength \( \lambda \) of a particle is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) Js) and \( p \) is the momentum of the particle. For particles in a box, the momentum \( p \) can also be determined using quantum mechanical principles.
02

Calculate the Momentum

The momentum \( p \) of an electron in a box is given by \( p = \frac{n h}{2L} \), where \( n \) is the principal quantum number, and \( L \) is the width of the box. Substituting \( L = 3.0 \times 10^{-10} \) m, we find \( p = \frac{n \times 6.63 \times 10^{-34} }{2 \times 3.0 \times 10^{-10} }\).
03

Calculate the de Broglie Wavelength for n = 1

Substituting \( n = 1 \) into the equation for \( p \) from Step 2 gives \( p = \frac{1 \times 6.63 \times 10^{-34}}{6.0 \times 10^{-10}} \). Calculate \( p \) and then use it to find \( \lambda = \frac{h}{p} \).
04

Calculate for n = 2

Substitute \( n = 2 \) into the momentum equation from Step 2 and find \( p \). Use this \( p \) to calculate \( \lambda = \frac{h}{p} \).
05

Calculate for n = 3

Use \( n = 3 \) in the momentum formula to solve for \( p \) and then find \( \lambda \) with \( \lambda = \frac{h}{p} \).
06

Compare Wavelengths with Box Width

For each \( n \), compute the ratio of the de Broglie wavelength \( \lambda \) to the width of the box \( L \) to see how they compare. Use the calculated \( \lambda \) from each previous step to find \( \frac{\lambda}{L} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

quantum mechanics
Quantum mechanics is a fundamental theory in physics that describes the physical properties of nature at the smallest scales. It provides a framework for understanding the behavior of particles such as electrons, photons, and atoms, which do not always adhere to classical laws of physics. Quantum mechanics introduces concepts like wave-particle duality, which tells us that particles exhibit both wave-like and particle-like properties.

One of the key insights of quantum mechanics is that certain physical quantities, such as energy and angular momentum, are quantized. This means they can only take on discrete values—much like how money can only be divided into whole cents. In the context of an electron in a box, quantum mechanics allows us to calculate properties like momentum and wavelength using these quantized values.

Thus, quantum mechanics forms the backbone for understanding why electrons behave the way they do in confined spaces, like a "box," leading to practical applications such as quantum computing and the precise calculation of an electron’s de Broglie wavelength.
principle quantum number
The principal quantum number, often denoted by the symbol \(n\), represents the quantized energy levels of an electron within an atom or a confined system, such as the electron in a box. It determines the electron's energy and distance from the nucleus when considering atomic structures.

In the case of an electron in a box, the principal quantum number \(n\) helps define the electron's quantized states. The number can typically take positive integer values: 1, 2, 3, and so forth. Each value of \(n\) corresponds to a specific energy level and thus, a unique momentum.
  • For \(n = 1\), the electron is in the lowest energy state and momentum is the least, leading to a longer wavelength.
  • For \(n = 2\), the electron is higher in energy, momentum increases, and wavelength decreases.
  • For \(n = 3\) and higher, we see even further quantized states with incrementally increasing momentum and decreasing wavelength.
The principal quantum number is crucial for the mathematical calculations that predict the behavior of electrons and leads to the de Broglie wavelength calculation in the "electron in a box" model.
electron in a box
The concept of an "electron in a box" is a simple yet powerful model used in quantum mechanics to help understand how electrons behave under certain constraints. Imagine an electron trapped in a narrow one-dimensional space where it can move freely but is unable to escape due to the boundaries formed by very high, impenetrable walls.

In this model, the width of the box determines the spatial constraint on the electron, affecting the possible wavelengths the electron can have. The model aligns with the quantum mechanical principle that particles exhibit wave-like characteristics, and these characteristics are dictated by the dimensions of their confinement.
  • The box has a specified width, which in this exercise is \(3.0 \times 10^{-10}\) m.
  • The electron cannot exist just anywhere; its wavelength must fit "neatly" within the boundaries, much like notes on a guitar string fit between the ends of the string.
This concept highlights the quantization of the system and allows for the application of the de Broglie hypothesis to determine specific properties, such as momentum, at given quantum states.
momentum calculation
Calculating momentum in quantum systems like the electron in a box involves using the relationship established by quantum mechanics. The momentum \(p\) of a particle in this context is linked to its motion and wavelength, quantified as it corresponds to the particle's energy state numbered by the principal quantum number \(n\).

In the given problem, the momentum is calculated using the formula:\[ p = \frac{n h}{2L} \]where:
  • \(n\) is the principal quantum number.
  • \(h\) is Planck's constant, \(6.63 \times 10^{-34}\) J·s.
  • \(L\) is the width of the box, \(3.0 \times 10^{-10}\) m.
This relationship shows that momentum is directly proportional to the principal quantum number, signifying that higher energy states correspond to greater momentum.

Once momentum is determined, it can be used to calculate the de Broglie wavelength of the electron using:\[ \lambda = \frac{h}{p} \]This formula further elucidates how we see wave-particle duality in action through the quantum state analysis provided by the electron's computed momentum and resulting wavelength.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A proton with initial kinetic energy 50.0 eV encounters a barrier of height 70.0 eV. What is the width of the barrier if the probability of tunneling is 8.0 \(\times\) 10\(^{-3}\)? How does this compare with the barrier width for an electron with the same energy tunneling through a barrier of the same height with the same probability?

An electron in a one-dimensional box has ground-state energy 2.00 eV. What is the wavelength of the photon absorbed when the electron makes a transition to the second excited state?

When low-energy electrons pass through an ionized gas, electrons of certain energies pass through the gas as if the gas atoms weren't there and thus have transmission coefficients (tunneling probabilities) \(T\) equal to unity. The gas ions can be modeled approximately as a rectangular barrier. The value of \(T\) = 1 occurs when an integral or half-integral number of de Broglie wavelengths of the electron as it passes over the barrier equal the width \(L\) of the barrier. You are planning an experiment to measure this effect. To assist you in designing the necessary apparatus, you estimate the electron energies \(E\) that will result in \(T\) = 1. You assume a barrier height of 10 eV and a width of 1.8 \(\times\) 10\(^{-10}\) m. Calculate the three lowest values of \(E\) for which \(T\) = 1.

(a) Show by direct substitution in the Schr\(\ddot{o}\)dinger equation for the one-dimensional harmonic oscillator that the wave function \(\psi_1(x) = A_1xe^{-a^2x^2/2}\), where \(\alpha^2 = m\omega/\hslash\), is a solution with energy corresponding to \(n\) = 1 in Eq. (40.46). (b) Find the normalization constant A1. (c) Show that the probability density has a minimum at \(x\) = 0 and maxima at \(x = \pm1/\alpha\), corresponding to the classical turning points for the ground state \(n\) = 0.

A free particle moving in one dimension has wave function $$\Psi(x, t) = A[e^{i(kx-vt)} - e^{i(2kx-4vt)}]$$ where \(k\) and \(\omega\) are positive real constants. (a) At \(t\) = 0 what are the two smallest positive values of \(x\) for which the probability function \(\mid \Psi(x, t) \mid ^2\) is a maximum? (b) Repeat part (a) for time \(t = 2\pi/\omega\). (c) Calculate \(v_{av}\) as the distance the maxima have moved divided by the elapsed time. Compare your result to the expression \(v_{av} = (\omega_2 - \omega_1)/(k_2 - k_1)\) from Example 40.1.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.