ORGANIC CHEMISTRY (LL)+ SAPLING ACC >BI
ORGANIC CHEMISTRY (LL)+ SAPLING ACC >BI
6th Edition
ISBN: 9781319306946
Author: LOUDON
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 15.6P
Interpretation Introduction

(a)

Interpretation:

The energy of light with a wavelength of 450nm is to be calculated.

Concept introduction:

The Plank-Einstein equation gives the measure of the energy of an electromagnetic wave. The expression for Plank-Einstein equation is given below.

E=hc/λ

Expert Solution
Check Mark

Answer to Problem 15.6P

The energy of light with a wavelength 450nm is 2.66×102kJmol1.

Explanation of Solution

The expression for the energy by Plank-Einstein equation is shown below.

E=hc/λ…(1)

Where,

E is the energy of the electromagnetic wave.

h is the Planks constant having value 6.626×1034Js.

c is the velocity of light having value 3×108ms1.

λ is the wavelength of light.

The wavelength is given as λ=450nm.

The relation between nm and m is given below.

1nm=109m

The conversion of 450nm into m is done as shown below.

450nm=450nm×109m1nm=450×109m

Now, substitute the values of h, c and λ in the equation (1) and solve for the energy E.

E=(6.626×1034Js)×(3×108ms1)(450×109m)=0.044173×1017J×103kJ1J=0.044173×1017×103kJ=44.173×1023kJ

The energy in kJmol1, is obtained by multiplying the energy calculated above (in kJ) by the total number of photons present in one mole of photon which is 6.022×1023mol1.

E=(44.173×1023kJ)×(6.022×1023mol1)=2.66×102kJmol1

Therefore, the energy of light for the given wavelength (450nm) is 2.66×102kJmol1.

Conclusion

The energy of light is 2.66×102kJmol1 that has wavelength 450nm.

Interpretation Introduction

(b)

Interpretation:

The energy of light with a wavelength 250nm is to be stated.

Concept introduction:

The Plank-Einstein equation gives the measure of the energy of an electromagnetic wave. The expression for Plank-Einstein equation is given below.

E=hc/λ

Expert Solution
Check Mark

Answer to Problem 15.6P

The energy of light with a wavelength 250nm is 4.79×102kJmol1.

Explanation of Solution

The expression for the energy by Plank-Einstein equation is shown below.

E=hc/λ…(1)

Where,

E is the energy of the electromagnetic wave.

h is the Planks constant having value 6.626×1034Js.

c is the velocity of light having value 3×108ms1.

λ is the wavelength of light.

The wavelength is given as λ=250nm.

The relation between nm and m is given below.

1nm=109m

The conversion of 250nm into m is done as shown below.

250nm=250nm×109m1nm=250×109m

Now, substitute the values of h, c and λ in the equation (1) and solve for the energy E.

E=(6.626×1034Js)×(3×108ms1)(250×109m)=0.0795×1017J×103kJ1J=0.0795×1017×103kJ=79.5×1023kJ

The energy in kJmol-1, is obtained by multiplying the energy calculated above (in kJ) by the total number of photons present in one mole of photon which is 6.022×1023mol-1.

E=(79.5×1023kJ)×(6.022×1023mol1)=4.79×102kJmol1

Therefore, the energy of light for the given wavelength (250nm) is 4.79×102kJmol1.

Conclusion

The energy of light is 4.79×102kJmol1 that has wavelength 250nm.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Identify the two geometric isomers of stilbene, C6H5CH=CHC6H5 from their λmax values, 294 nm and 278 nm
an unknown has the formula C₆H₁₂O. The IR spectrum of this substance has a strong vibration near 1700 cm⁻¹ . Which structure below is consistent with these data?
Given the fact that O2, N2, and O in the upper atmosphere absorb most of the light with wavelengths shorter than 240 nm, would you expect thephotodissociation of C ¬ F bonds to be significant in the lower atmosphere?

Chapter 15 Solutions

ORGANIC CHEMISTRY (LL)+ SAPLING ACC >BI

Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42APCh. 15 - Prob. 15.43APCh. 15 - Prob. 15.44APCh. 15 - Prob. 15.45APCh. 15 - Prob. 15.46APCh. 15 - Prob. 15.47APCh. 15 - Prob. 15.48APCh. 15 - Prob. 15.49APCh. 15 - Prob. 15.50APCh. 15 - Prob. 15.51APCh. 15 - Prob. 15.52APCh. 15 - Prob. 15.53APCh. 15 - Prob. 15.54APCh. 15 - Prob. 15.55APCh. 15 - Prob. 15.56APCh. 15 - Prob. 15.57APCh. 15 - Prob. 15.58APCh. 15 - Prob. 15.59APCh. 15 - Prob. 15.60APCh. 15 - Prob. 15.61APCh. 15 - Prob. 15.62APCh. 15 - Prob. 15.63APCh. 15 - Prob. 15.64APCh. 15 - Prob. 15.65APCh. 15 - Prob. 15.66APCh. 15 - Prob. 15.67APCh. 15 - Prob. 15.68APCh. 15 - Prob. 15.69APCh. 15 - Prob. 15.70APCh. 15 - Prob. 15.71APCh. 15 - Prob. 15.72APCh. 15 - Prob. 15.73APCh. 15 - Prob. 15.74APCh. 15 - Prob. 15.75APCh. 15 - Prob. 15.76APCh. 15 - Prob. 15.77APCh. 15 - Prob. 15.78APCh. 15 - Prob. 15.79APCh. 15 - Prob. 15.80APCh. 15 - Prob. 15.81APCh. 15 - Prob. 15.82APCh. 15 - Prob. 15.83APCh. 15 - Prob. 15.84APCh. 15 - Prob. 15.85APCh. 15 - Prob. 15.86AP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Quantum Mechanics - Part 1: Crash Course Physics #43; Author: CrashCourse;https://www.youtube.com/watch?v=7kb1VT0J3DE;License: Standard YouTube License, CC-BY