   Chapter 6, Problem 16PS

Chapter
Section
Textbook Problem

What are the wavelength and frequency of the radiation involved in the least energetic emission line in the Lyman series? What are the values of ninitial and nfinal?

Interpretation Introduction

Interpretation: The wavelength and frequency of the least energetic line has to be calculated. The initial and the final electronic state the least energetic line in the Lyman series have to be identified.

Concept introduction:

• Electronic transitions that take place in excited H atom is,

Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.

Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.

Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.

Brackett series: electronic transitions take place from n>4 to the n=4 level.

Pfund series: electronic transitions take place from n>5 to the n=5 level

• Electromagnetic radiations are a type of energy surrounding us. They are of different types like radio waves, IR, UV, X-ray etc.
• Visible radiation in the wavelength of 400nm to 700nm
• Planck’s equation,

E==hcλwhere, E=energyh=Planck'sconstantν=frequency

The energy increases as the wavelength of the light decrease. Also the energy increases as the frequency of the light increases.

• The frequency of the light is inversely proportional to its wavelength.

ν=cλwhere, c=speedoflightν=frequencyλ=wavelength

Explanation

In Lyman series, the electronic transitions take place from n>2 to the n=2 level and it is in visible region. The wavelength of lines in are 93.1nm, 93.8nm, 95nm, 97.3nm, 102.6nm and 121.6nm.

• The wavelength of the least energetic line is determined

E==hcλ so the energy increases as the wavelength of the light decrease. So the radiation having lower wavelength has higher energy per photon.

Here the prominent line which found in 121.6nm has the highest wavelength comparing to other lines.

Therefore the prominent line which found in 121.6nm is the least energetic line.

The wavelength of the least energetic line is 121.6nm=1.216 ×107m

• The frequency of the least energetic line is calculated

ν=cλ

Substituting the values

ν=2.998×108m/s1

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

What is meant by the term subunit?

Chemistry for Today: General, Organic, and Biochemistry

The maturation of sperm takes 7080 days. ______ True ______ False

Nutrition Through the Life Cycle (MindTap Course List)

What things were unique about the 2005 Atlantic hurricane season?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 