Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 28

Calculate the increase in (a) distance from the nucleus and (b) energy when an electron is excited from the first to the third Bohr orbit.

Short Answer

Expert verified
The increase in distance from the nucleus when an electron is excited from the first to the third Bohr orbit is 3.176 Å, and the increase in energy is + 10.2 eV.

Step by step solution

01

Formula for the Radius of Bohr Orbit

The formula for the radius \(r_n\) of the nth Bohr orbit is given by \(r_n = 0.529n^2\) Å, where n denotes the number of the orbit. First calculate the radius of the first Bohr orbit \(r_1\) by substituting \(n = 1\) into the formula, and then calculate the radius of the third Bohr orbit \(r_3\) by substituting \(n = 3\). The increase in distance from the nucleus can be found by subtracting the radius of the first orbit \(r_1\) from the third \(r_3\).
02

Formula for the Energy of Bohr Orbit

The energy \(E_n\) of an electron in the nth Bohr orbit is given by \(E_n = -13.6/n^2\) eV. Find the energy of the first orbit \(E_1\) by substituting \(n = 1\), and then find the energy of the third orbit \(E_3\) by substituting \(n = 3\). The increase in energy can be found by subtracting the energy of the first orbit \(E_1\) from the third \(E_3\). Note that the energy is negative, indicating a bound system, and the magnitude of the energy increases (becomes more negative) as the electron gets closer to the nucleus.
03

Calculation

Now, we can find the values of the increase in distance and the increase in energy. After performing the calculations, we find that the increase in distance from the nucleus is \(r_3 - r_1 = 0.529*(3^2 - 1^2)\) Å = 3.176 Å, and the increase in energy is \(E_3 - E_1 = -13.6 * (1/3^2 - 1/1^2)\) eV = + 10.2 eV.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

For electromagnetic radiation transmitted through a vacuum, state whether each of the following properties is directly proportional to, inversely proportional to, or independent of the frequency: (a) velocity; (b) wavelength; (c) energy per mole. Explain.

What is the expected ground-state electron configuration for each of the following elements? (a) tellurium; (b) cesium; (c) selenium; (d) platinum; (e) osmium; (f) chromium.

The lowest-frequency light that will produce the photoelectric effect is called the threshold frequency. (a) The threshold frequency for indium is \(9.96 \times\) \(10^{14} \mathrm{s}^{-1} .\) What is the energy, in joules, of a photon of this radiation? (b) Will indium display the photoelectric effect with UV light? With infrared light? Explain.

The following electron configurations correspond to the ground states of certain elements. Name each element. (a) \([\mathrm{Rn}] 7 s^{2} 6 d^{2} ;\) (b) \([\mathrm{He}] 2 s^{2} 2 p^{2} ;\) (c) \([\mathrm{Ar}] 3 d^{3} 4 s^{2}\) (d) \([\mathrm{Kr}] 4 d^{10} 5 s^{2} 5 p^{4} ;\) (e) \([\mathrm{Xe}] 4 f^{2} 6 s^{2} 6 p^{1}\)

Which must possess a greater velocity to produce matter waves of the same wavelength (such as \(1 \mathrm{nm}\) ), protons or electrons? Explain your reasoning.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

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.

Sign-up for free