Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 64

An excited hydrogen atom emits light with a wavelength of \(397.2 \mathrm{~nm}\) to reach the energy level for which \(n=2\). In which principal quantum level did the electron begin?

Short Answer

Expert verified
The electron initially started in the principal quantum level \(n_{2} = 3\).

Step by step solution

01

Recall the Rydberg formula

The Rydberg formula is given by: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_{1}^2} - \frac{1}{n_{2}^2}\right)\] where \(\lambda\) is the wavelength of emitted light, \(R_H\) is the Rydberg constant for hydrogen \((R_H = 1.097 \times 10^7 \, \text{m}^{-1})\), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.
02

Insert known values into the Rydberg formula

We know the final energy level \(n_1 = 2\), and we are given the wavelength of the emitted light \(\lambda = 397.2\, \text{nm}\). We need to find the initial energy level \(n_2\). Convert the given wavelength from nanometers to meters: \[\lambda = 397.2\, \text{nm} \times \frac{1\, \text{m}}{10^9\, \text{nm}} = 397.2 \times 10^{-9}\, \text{m}\] Now, we insert the known values into the Rydberg formula and solve for \(n_2\): \[\frac{1}{397.2\times10^{-9} \, \text{m}} = R_H \left(\frac{1}{4} - \frac{1}{n_{2}^2}\right)\]
03

Solve for the initial energy level, n2

Rearrange the equation and solve for \(n_{2}^2\): \[n_{2}^2 = \frac{1}{\frac{1}{R_H\lambda}+\frac{1}{4}}\]
04

Calculate n2, the initial energy level

Substitute the given values into the equation: \[n_{2}^2 = \frac{1}{\frac{1}{(1.097\times10^7 \, \text{m}^{-1})(397.2\times10^{-9} \, \text{m})}+\frac{1}{4}}\] Calculate \(n_{2}^2\): \[n_{2}^2 \approx 6.003\] Since the principal quantum level must be a positive integer, we round up to the nearest whole number: \[n_{2} = 3\] So, the electron initially started in the principal quantum level \(n_{2} = 3\).

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

Without looking at data in the text, sketch a qualitative graph of the third ionization energy versus atomic number for the elements Na through Ar, and explain your graph.

How many electrons in an atom can have the designation \(1 p\), \(6 d_{x^{2}-y^{2}}, 4 f, 7 p_{y}, 2 s, n=3 ?\)

Using only the periodic table inside the front cover of the text, write the expected ground-state electron configurations for a. the third element in Group \(5 \mathrm{~A}\). b. element number 116 . c. an element with three unpaired \(5 d\) electrons. d. the halogen with electrons in the \(6 p\) atomic orbitals.

The first-row transition metals from chromium through zinc all have some biologic function in the human body. How many unpaired electrons are present in each of these first-row transition metals in the ground state?

Photogray lenses incorporate small amounts of silver chloride in the glass of the lens. When light hits the AgCl particles, the following reaction occurs: $$ \mathrm{AgCl} \stackrel{h v}{\longrightarrow} \mathrm{Ag}+\mathrm{Cl} $$ The silver metal that is formed causes the lenses to darken. The enthalpy change for this reaction is \(3.10 \times 10^{2} \mathrm{~kJ} / \mathrm{mol}\). Assuming all this energy must be supplied by light, what is the maximum wavelength of light that can cause this reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

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