Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Q22P (page 488)

The end of Box 19-2 states that “the 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.” 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Infrared spectra are customarily recorded on a transmittance scale so that weak and strong bands can be displayed on the same scale. The region near 2000cm-1in the infrared spectra of compounds A and B is shown in the figure. Note that absorption corresponds to a downward peak on this scale. The spectra were recorded from a 0.0100M solution of each, in cells with 0.00500 - cm path lengths. A mixture of A and B in a 0.00500 - cm cell gave a transmittance 34.0 % of at2022cm and 383% at 1093 cm. Find the concentrations of A and B.

The indicator xylenol orange (Table 12-3) forms a complex with Zr(IV)in HCIsolution. Prepare a Job plot from the data in the table and suggest the stoichiometry of the complex (xylenol orange)Zrzx.

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

The figure shows spectra of1.00×10-4MMnO4-,1.00×10-4and an unknown mixture of both, all in1.000‐cm‐path length cells. Absorbances are given in the table. Use the least squares procedure in Figure 19-3 to find the concentration of each species in the mixture.

Visible spectrum ofMnO4-,Cr2O72-, and an unknown mixture containing both ions.

Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for CO2in the ocean was designed to operate without the need for calibration.33


The sensor compartment is separated from seawater by a silicone membrane through which CO2, but not dissolved ions, can diffuse. Inside the sensor, CO2equilibrates with HCO3and CO32. For each

measurement, the sensor is flushed with fresh solution containingbromothymol blue indicator. All indicator is in the formnear neutral pH, so we can

write two mass balances:

[HIn−]+[ln2−]=FIn=50.0μMand[Na+]=FNα=50.0μM+42.0μM=92.0μM

has an absorbance maximum at 434 nm andhas a maximum at 620 nm. The sensor measures the absorbance ratio RA=A620/A434reproducibly without need for calibration. From this ratio, we can findin the seawater as outlined here:

(a).From Beer’s law for the mixture, write equations forin terms of the absorbance at 620 and 434 nmThen show that

[ln2−][Hln−]=RAε434HHn−−ε6,20Hln−ε620ln2−−RAε434ln2−=Rln (A)

(b) From the mass balance (1) and the acid dissociation constant

, show that

[Hln−]=F1nRln+1 (B)

[ln2−]=KlnFln[H+](Rln+1) (C)

(c) Show that H+=Kln/Rln (D)

(d) From the carbonic acid dissociation equilibria, show that

[HCO3−]=K1[CO(aq)]E[H+][CO32−]=K1K2[CO(aq)]F[H+]2

(e) Write the charge balance for the solution in the sensor compartment. Substitute in expressions B, C, E, and F forHln,In2-,[HCO3], and[CO32−]

(f) Suppose that the various constants have the following values:

ε4344HHn−=8.00×103M−1cm−1    K1=3.0×10−7ε6620Hn−=0    K2=3.3×10−11ε434ln2=1.90×103M−1cm−1    Kln=2.0×10−7ε620ln2−=1.70×104M−1cm−1    Kw=6.7×10−15

From the measured absorbance ratio=2.84, findin the seawater.

(g) Approximately what is the ionic strength inside the sensor compartment? Were we justified in neglecting activity coefficients in working this problem?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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.

Sign-up for free