   Chapter 6, Problem 7PS

Chapter
Section
Textbook Problem

The most prominent line in the emission spectrum of aluminum is at 396.15nm. What is the frequency of this line? What is the energy of one photon with this wavelength? Of 1.00 mol of these photons?

Interpretation Introduction

Interpretation: The frequency, the energy per photon and the energy of one mole of photons of prominent line in emission spectrum of aluminium have to be calculated.

Concept introduction:

• Planck’s equation,

E=where, 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

The frequency, the energy per photon and the energy per mole of photons of prominent line in emission spectrum of aluminium is calculated below.

Given,

The wavelength of prominent line in emission spectrum of aluminum is,

396.15nm=396.15 ×109m

Planck'sconstant,h=6.626×1034J.sThe speed of light,c=2.998×108m/sAvogadro's number,NA=6.022×1023photons/mol

• The frequency of prominent line in emission spectrum of aluminium is calculated by using the equation,

ν=cλ

Substituting the values to the above equation,

ν=2.998×108m/s396.15 ×109m=7.5676×1014s1

The frequency of prominent line in emission spectrum of aluminum is 7.5676×1014s-1

• The energy per photon of prominent line in emission spectrum of aluminium is calculated,

E=

Substituting the values to the above equation,

E==6

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 