/*! 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 23 The allowed energies of a simple... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 39: Problem 23

The allowed energies of a simple atom are \(0.00 \mathrm{eV}\) \(4.00 \mathrm{eV},\) and 6.00 eV. An electron traveling with a speed of \(1.30 \times 10^{6} \mathrm{m} / \mathrm{s}\) collides with the atom. Can the electron excite the atom to the \(n=2\) stationary state? The \(n=3\) stationary state? Explain.

Short Answer

Expert verified
The final answer will depend on the calculation in Step 1. For example, if the kinetic energy of the electron is found to be 5 eV, it means that the electron can excite the atom to the n=2 state but not to the n=3 state.

Step by step solution

01

Calculate the kinetic energy of the electron

The electron's kinetic energy can be calculated using the formula \(KE = \frac{1}{2} mv^2\). Here, the mass (m) of an electron is \(9.11 脳 10^{-31} kg\) and the given speed (v) is \(1.30 脳 10^6 m/s\). Substituting these values in the formula, we can find out the kinetic energy in Joules. To convert it into electron-volts (eV), we then divide the result by \(1.6 脳 10^{-19}\).
02

Compare the kinetic energy of the electron with the energies of the atom's stationary states

We know from the problem that the n=2 stationary state of the atom is at 4.00 eV and the n=3 state is at 6.00 eV. By comparing the kinetic energy of the electron with these energies, we can determine if the electron can excite the atom to these states.
03

Determine the excitation condition

If the kinetic energy of the electron is greater than or equal to the energy level at n=2 (4.00 eV) or n=3 (6.00 eV), then the electron is able to excite the atom to that state. If not, it means that the electron does not carry enough energy to excite the atom to that state.

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 91影视!

Key Concepts

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

Electron Excitation
When we talk about electron excitation, we're diving into the world of quantum physics, where electrons change their energy levels inside an atom. An atom consists of electrons orbiting a nucleus, each occupying a specific energy level. To excite an electron means to push it from a lower energy level, or ground state, to a higher energy level, called an excited state.

This requires a very precise amount of energy, equivalent to the difference between the two energy levels. If an incoming electron has sufficient kinetic energy, it can transfer some of that energy to an electron in the atom, thereby exciting it.

It's much like climbing stairs. You need enough energy to jump from one step to another. If you don't have enough energy for a leap to the next step, you're unable to make the jump.
Kinetic Energy Calculation
The kinetic energy of an electron can tell us a lot about what it can do when it interacts with other atoms. Kinetic energy is the energy an object has due to its motion. For an electron moving at a certain speed, it's given by the formula: \[KE = \frac{1}{2} mv^2\]Here, \(m\) is the mass of the electron, and \(v\) is its velocity.

By inserting the values for an electron's mass, \(9.11 \times 10^{-31} \text{kg}\), and its speed, such as \(1.30 \times 10^6 \text{m/s}\) as in our problem, you can calculate its kinetic energy in joules.

Once you have this number, you can convert it to electron-volts to make comparisons easier with the energy levels of atoms.
Energy Conversion to Electron-Volts
In physics, energy is often expressed in electron-volts (eV) when dealing with atomic and subatomic processes. One electron-volt is the amount of kinetic energy gained or lost by an electron when it moves through an electric field with a potential difference of one volt.

To convert kinetic energy from joules to electron-volts, you use a simple conversion factor: \[1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules} \]You divide the energy value in joules by this conversion number to get the energy in electron-volts.

This conversion is crucial because when comparing the kinetic energy of electrons to the energy needed to excite an atom, using the same units鈥攅lectron-volts鈥攎akes the analysis straightforward.

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 confined in a one-dimensional box emits a \(2.0 \mathrm{MeV}\) gamma- ray photon in a quantum jump from \(n=2\) to \(n=1\) What is the length of the box?

At what speed is an electron's de Broglie wavelength (a) \(1.0 \mathrm{pm},\) (b) \(1.0 \mathrm{nm},\) (c) \(1.0 \mu \mathrm{m},\) and (d) \(1.0 \mathrm{mm} ?\)

Is a spectral line with wavelength \(656.5 \mathrm{nm}\) seen in the absorption spectrum of hydrogen atoms? Why or why not?

The diameter of the nucleus is about 10 fm. What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of \(10 \mathrm{fm} ?\)

The moon is a subatomic particle with the same charge as an electron but with a mass that is 207 times greater: \(m_{\mu}=207 m_{e}\) Physicists think of moons as "heavy electrons." However, the moon is not a stable particle; it decays with a half-life of \(1.5 \mu \mathrm{s}\) into an electron plus two neutrinos. Moons from cosmic rays are sometimes "captured" by the nuclei of the atoms in a solid. A captured moon orbits this nucleus, like an electron, until it decays. Because the moon is often captured into an excited orbit \((n>1),\) its presence can be detected by observing the photons emitted in transitions such as \(2 \rightarrow 1\) and \(3 \rightarrow 1\) Consider a moon captured by a carbon nucleus \((Z=6)\) Because of its large mass, the moon orbits well inside the electron cloud and is not affected by the electrons. Thus the moon "sees" the full nuclear charge Ze and acts like the electron in a hydrogen-like ion. a. What are the orbital radius and speed of a moon in the \(n=1\) ground state? Note that the mass of a moon differs from the mass of an electron. b. What is the wavelength of the \(2 \rightarrow 1\) moon transition? c. Is the photon emitted in the \(2 \rightarrow 1\) transition infrared, visible, ultraviolet, or x ray? d. How many orbits will the moon complete during \(1.5 \mu s ?\) Is this a sufficiently large number that the Bohr model "makes sense," even though the moon is not stable?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Work Energy and Power

Read Explanation

Fluids

Read Explanation

Kinematics Physics

Read Explanation
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.