The reason as to why phosphorescence spectra are better for identifying useful electronic transitions for possible laser transitions than are fluorescence spectra is to be stated. Concept introduction: Fluorescence is a process in which light is emitted from a substance after it absorbs a light. The process involves excitation of electrons and then the relaxation of that electron. The light emitted from the substance has lower energy than the light absorbed by the substance as some radiationless decay takes place.

Question

Want to see the full answer?

Answer 1

Question

Chapter 15, Problem 15.58E

Interpretation Introduction

Interpretation:

The reason as to why phosphorescence spectra are better for identifying useful electronic transitions for possible laser transitions than are fluorescence spectra is to be stated.

Concept introduction:

Fluorescence is a process in which light is emitted from a substance after it absorbs a light. The process involves excitation of electrons and then the relaxation of that electron. The light emitted from the substance has lower energy than the light absorbed by the substance as some radiationless decay takes place.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B Bond A Bond C a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest Bond Strongest Bond b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! Google

Print Last Name, First Name Initial Statifically more chances to abstract one of these 6H 11. (10pts total) Consider the radical chlorination of 1,3-diethylcyclohexane depicted below. 4 4th total • 6H total 래 • 4H total 21 total ZH 2H Statistical H < 3° C-H weakest - product abstraction here bund leads to thermo favored a) (6pts) How many unique mono-chlorinated products can be formed and what are the structures for the thermodynamically and statistically favored products? Product 6 Number of Unique Mono-Chlorinated Products Thermodynamically Favored Product Statistically Favored Product b) (4pts) Draw the arrow pushing mechanism for the FIRST propagation step (p-1) for the formation of the thermodynamically favored product. Only draw the p-1 step. You do not need to include lone pairs of electrons. No enthalpy calculation necessary H H-Cl Waterfox

10. (5pts) Provide the complete arrow pushing mechanism for the chemical transformation → depicted below Use proper curved arrow notation that explicitly illustrates all bonds being broken, and all bonds formed in the transformation. Also, be sure to include all lone pairs and formal charges on all atoms involved in the flow of electrons. CH3O II HA H CH3O-H H ①