Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 67

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{~s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Short Answer

Expert verified
The final principal quantum level for the electronic transition in this excited hydrogen atom is approximately \(n_f = 3\).

Step by step solution

01

Recall the Rydberg Formula for Hydrogen Atom

The Rydberg formula gives the frequency of the photon emitted or absorbed during electronic transition in the hydrogen atom. The formula is given by: \[\nu = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right),\] where \(\nu\) is the frequency of the emitted light, \(n_i\) and \(n_f\) are the initial and final principal quantum numbers, and \(R_H\) is the Rydberg constant for the hydrogen atom (\(R_H = 3.29 \times 10^{15} \mathrm{Hz}\)).
02

Rearrange the Rydberg formula to solve for \(n_f\)

In order to find the principal quantum number of the final state, we need to rearrange the Rydberg formula to solve for \(n_f\). This gives: \[n_f = \frac{1}{\sqrt{\frac{1}{n_i^2} - \frac{\nu}{R_H}}}\]
03

Substitute the given values into the formula

Now we will substitute the given values into the formula and get the value of the \(n_f\). We have \(n_i = 5\), \(\nu = 6.90 \times 10^{14} \mathrm{Hz}\), and \(R_H = 3.29 \times 10^{15} \mathrm{Hz}\). Substitute these values into the formula: \[n_f = \frac{1}{\sqrt{\frac{1}{5^2} - \frac{6.90 \times 10^{14}}{3.29 \times 10^{15}}}}\]
04

Calculate the value of \(n_f\)

Perform the calculation to get the value of the final quantum number: \[n_f = \frac{1}{\sqrt{\frac{1}{25} - \frac{6.90 \times 10^{14}}{3.29 \times 10^{15}}}} \approx 2.84\]
05

Round off the value of \(n_f\)

Since \(n_f\) must be an integer, we round off the calculated value of \(n_f\) to the nearest integer: \[n_f \approx 3\]
06

Report the result

The final principal quantum level for the electronic transition in this excited hydrogen atom is approximately \(n_f = 3\).

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

Using the Heisenberg uncertainty principle, calculate \(\Delta x\) for each of the following. a. an electron with \(\Delta v=0.100 \mathrm{~m} / \mathrm{s}\) b. a baseball (mass \(=145 \mathrm{~g}\) ) with \(\Delta v=0.100 \mathrm{~m} / \mathrm{s}\) c. How does the answer in part a compare with the size of a hydrogen atom? d. How does the answer in part \(b\) correspond to the size of a baseball?

Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

Predict the atomic number of the next alkali metal after francium and give its ground-state electron configuration.

Calculate, to four significant figures, the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the \(n=5\) state and then fall to states with smaller values of \(n\).

Which of elements \(1-36\) have one unpaired electron in the ground state?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

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