/*! 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 34 An electron is confined to a o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron is confined to a one-dimensional box. When the electron makes a transition from its first excited state to the ground state, it emits a photon of energy 1.2 eV. (a) What is the ground-state energy (in electronvolts) of the electron? (b) List all energies (in electronvolts) of photons that could be emitted when the electron starts in its second excited state and makes transitions downward to the ground state either directly or through intervening states. Show all these transitions on an energy level diagram. (c) What is the length of the box (in nanometers)?

Short Answer

Expert verified
The ground-state energy of the electron is 0.4 eV. 2. What are the photon energies emitted when the electron transitions from the second excited state to the ground state? There are two possible transitions: a. Direct transition from n=3 to n=1 (E_3 -> E_1): The energy of the photon emitted in this transition will be [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 1^2) / (8 * m * L^2)]. b. Transition from n=3 to n=2 (E_3 -> E_2) and then from n=2 to n=1 (E_2 -> E_1): The energy of the photon emitted in this transition will be [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 2^2) / (8 * m * L^2)]. 3. How can the length of the box be calculated? The length of the box (L) can be calculated using the formula: L = sqrt((h^2) / (8 * m * E_1)) where h is Planck's constant, m is the mass of the electron, and E_1 is the ground-state energy.

Step by step solution

01

Calculate the energy difference between the first excited state and the ground state

We are given that the energy difference between the first excited state and the ground state (photon energy) is 1.2 eV. We can use the formula `E_n = (h^2 * n^2) / (8 * m * L^2)` to find this energy difference. For the first excited state (n = 2) and for the ground state (n = 1), the energy difference can be written as: `E_2 - E_1 = (h^2 * 2^2) / (8 * m * L^2) - (h^2 * 1^2) / (8 * m * L^2)` This energy difference is equal to 1.2 eV.
02

Calculate the ground-state energy

Now that we have an equation for the energy difference, we can solve it for the ground-state energy `E_1`. `E_1 = (h^2 * 1^2) / (8 * m * L^2) = (1/3) * (E_2 - E_1) = (1/3) * 1.2 eV` After the calculation, we get `E_1 = 0.4 eV`.
03

Calculate the photon energies emitted when the electron starts in its second excited state

We need to find the energies of the photons emitted when the electron transitions from the second excited state (n=3) to the ground state (n=1) either directly or through the first excited state (n=2). The possible transitions are: 1. Direct transition from n=3 to n=1 (E_3 -> E_1) 2. Transition from n=3 to n=2 (E_3 -> E_2) and then from n=2 to n=1 (E_2 -> E_1) We can use the energy formula to calculate the energies of these transitions: 1. `E_3 - E_1`: `E_3 = (h^2 * 3^2) / (8 * m * L^2)` The energy of the photon emitted in this transition will be: `E_3 - E_1 = [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 1^2) / (8 * m * L^2)]` 2. `E_3 - E_2`: `E_3 = (h^2 * 3^2) / (8 * m * L^2)` `E_2 = (h^2 * 2^2) / (8 * m * L^2)` The energy of the photon emitted in this transition will be: `E_3 - E_2 = [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 2^2) / (8 * m * L^2)]` The emissions for each transition can be represented on an energy level diagram.
04

Calculate the length of the box

We know the ground-state energy, E_1, and we have the expression for energy in terms of the length L: `E_1 = (h^2 * 1^2) / (8 * m * L^2)` We can solve for L: `L = sqrt((h^2) / (8 * m * E_1))` Given E_1 = 0.4 eV = 0.4 * 1.6 * 10^(-19) J and the mass of an electron, m = 9.11 * 10^(-31) kg, and Planck's constant, h = 6.63 * 10^(-34) J*s, plug in the values: `L = sqrt((6.63 * 10^(-34))^2 / (8 * 9.11 * 10^(-31) * 0.4 * 1.6 * 10^(-19)))` The length L will be in meters. To get the length in nanometers, multiply the result by 10^9. `L (nm) = L (m) * 10^9`

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!

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 beam of neutrons is used to study molecular structure through a series of diffraction experiments. A beam of neutrons with a wide range of de Broglie wavelengths comes from the core of a nuclear reactor. In a time-offlight technique, used to select neutrons with a small range of de Broglie wavelengths, a pulse of neutrons is allowed to escape from the reactor by opening a shutter very briefly. At a distance of \(16.4 \mathrm{m}\) downstream, a second shutter is opened very briefly 13.0 ms after the first shutter. (a) What is the speed of the neutrons selected? (b) What is the de Broglie wavelength of the neutrons? (c) If each shutter is open for 0.45 ms, estimate the range of de Broglie wavelengths selected.
What is the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus \((1.0 \mathrm{fm}) ?\)

A nickel crystal is used as a diffraction grating for x-rays. Then the same crystal is used to diffract electrons. If the two diffraction patterns are identical, and the energy of each \(\mathrm{x}\) -ray photon is \(E=20.0 \mathrm{keV},\) what is the kinetic energy of each electron?

The phenomenon of Brownian motion is the random motion of microscopically small particles as they are buffeted by the still smaller molecules of a fluid in which they are suspended. For a particle of mass $1.0 \times 10^{-16} \mathrm{kg},\( the fluctuations in velocity are of the order of \)0.010 \mathrm{m} / \mathrm{s} .$ For comparison, how large is the change in this particle's velocity when the particle absorbs a photon of light with a wavelength of \(660 \mathrm{nm}\) such as might be used in observing its motion under a microscope?
The energy-time uncertainty principle allows for the creation of virtual particles, that appear from a vacuum for a very brief period of time $\Delta t,\( then disappear again. This can happen as long as \)\Delta E \Delta t=\hbar / 2,\( where \)\Delta E$ is the rest energy of the particle. (a) How long could an electron created from the vacuum exist according to the uncertainty principle? (b) How long could a shot put with a mass of \(7 \mathrm{kg}\) created from the vacuum exist according to the uncertainty principle?
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.