/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q22P The end of Box 19-2 states that ... [FREE SOLUTION] | 91影视

91影视

The end of Box 19-2 states that 鈥渢he advantage of up conversion for biomedical probes is that low energy near-infrared (800 to 1000 nm) incident radiation stimulates little background emission from the complex biological matrix that can be highly fluorescent under visible radiation.鈥 Suggest why near-infrared radiation stimulates less emission than visible radiation and why this behavior is useful.

Short Answer

Expert verified

The reason for the promotion of infra-red radiation shows less emissions than visible radiation and the significance of these behaviors was explained.

Step by step solution

01

Reason for less stimulation:

The reason for the stimulation of infra-red radiation shows less emissions than visible radiation because fewer biological molecules draw near infrared radiation at a distance of 800-1000nm..

02

STEP-2:  Reason for its useful behavior:

The importance of this behavior is due to the fact that a small matrix fluorescence is promoted

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Explain what is done in flow injection analysis and sequential injection. What is the principal difference between the two techniques? Which one is called 鈥渓ab-on-a-valve鈥?

The graph shows the effect of pH on quenching of luminescence of tris(2,2'-bipyridine) Ru(II) by 2,6-dimethylphenol. The ordinate, KSV, is the collection of constants, kq /(ke + kd), in the Stern-Volmer equation. The greater KSV, the greater the quenching. Suggest an explanation for the shape of the graph and

estimate pKa for 2,6-dimethylphenol.

Now we use Solver to find Kfor the previous problem. The only absorbing species at 332 nmis the complex, so, from Beer鈥檚 law [complex]=A/(becausepathlength=1.000cm).I2is either free or bound in the complex,so [I2]=[I2]tot-[complex].There is a huge excess of mesitylene, so[mesitylene][mesitylene]tot

K=[complex][l2][mesitylene]=A/(l2tot-A/)[mestitylene]tot

The spreadsheet shows some of the data. You will need to use all the data. Column A contains [mesitylene] and column B contains [l2]tot. Column C lists the measured absorbance. Guessa value of the molar absorptivity of the complex,,incellA7.Then compute the concentration of the complex (=A/)in column D. The equilibrium constant in column EisgivenbyE2=[complex]/([I2][mesitylene])=(D2)/((B2-D2)*A2).

should we minimize with Solver? We want to vary蔚in cell A7 until the values of Kin column E are as constant as possible. We would like to minimize a function like (Ki-Kaverage)2, where Kiis the value in each line of the table and Kaverage is the average of all computed values. The problem with (Ki-Kaverage)2is that we can minimize this function simply by making Kivery small, but not necessarily constant. What we really want is for all the Kito be clustered around the mean value. A good way to do this is to minimize the relative standard deviationof the K, which is (standard deviation)/average. In cell E5we compute the average value of Kand in cell E6the standard deviation. Cell E7contains the relative standard deviation. Use Solver to minimize cell E7by varying cell A7. Compare your answer with that of Problem 19-13.

The spreadsheet lists molar absorptivities of three dyes and the absorbance of a mixture of the dyes in a 1.000-cm cell. Use the least-squares procedure in Figure 19-3 to find the concentration of each dye in the mixture.

Challenging your acid-base prowess. A solution was prepared by mixing 25.00mL of 0.800Maniline, 25.00mLsulfanilic acid, andand then diluting to 100.0mL. (stands for protonated indicator.)


The absorbance measured at550nmin 5.00 - cmwas 0.110.Find the concentrations ofHIn and In andpafor HIn

See all solutions

Recommended explanations on Chemistry Textbooks

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.