Checking the possibility for a fluorescent material (one that absorbs and then reemits light) to emit radiation in the ultraviolet region after absorbing visible light should be analyzed. Concept Introduction: The electrons are excited thermally when the light is used by an object. As a result, an emission spectrum comes. The emission spectrum of a substance is seen by energizing a sample of material with either thermal energy or some other form of energy (such as a high-voltage electrical discharge if the substance is a gas). A “red-hot” or “white-hot” iron bar freshly removed from a fire produces a characteristic glow. The glow is the visible portion of its emission spectrum. The heat given off by the same iron bar is another portion of its emission spectrum called the infrared region. A feature common to the emission spectrum of the sun and that of a heated solid is that both are continuous. Hence, all wavelengths of visible light are present in each spectrum.

Question

Want to see the full answer?

Answer 1

Question

Chapter 7, Problem 7.27QP

Interpretation Introduction

Interpretation:

Checking the possibility for a fluorescent material (one that absorbs and then reemits light) to emit radiation in the ultraviolet region after absorbing visible light should be analyzed.

Concept Introduction:

The electrons are excited thermally when the light is used by an object. As a result, an emission spectrum comes. The emission spectrum of a substance is seen by energizing a sample of material with either thermal energy or some other form of energy (such as a high-voltage electrical discharge if the substance is a gas). A “red-hot” or “white-hot” iron bar freshly removed from a fire produces a characteristic glow. The glow is the visible portion of its emission spectrum. The heat given off by the same iron bar is another portion of its emission spectrum called the infrared region. A feature common to the emission spectrum of the sun and that of a heated solid is that both are continuous. Hence, all wavelengths of visible light are present in each spectrum.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

LTS Solid: AT=Te-Ti Trial 1 Trial 2 Trial 3 Average ΔΗ Mass water, g 24.096 23.976 23.975 Moles of solid, mol 0.01763 001767 0101781 Temp. change, °C 2.9°C 11700 2.0°C Heat of reaction, J -292.37J -170.473 -193.26J AH, kJ/mole 16.58K 9.647 kJ 10.85 kr 16.58K59.64701 KJ mol 12.35k Minimum AS, J/mol K 41.582 mol-k Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn & Show your calculations for: AH in J and then in kJ/mole for Trial 1: qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J qsin = qrxn = 292.35 292.37J AH in J = 292.375 0.2923kJ 0.01763m01 =1.65×107 AH in kJ/mol = = 16.58K 0.01763mol mol qrx Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con Kelvin.) AS=AHIT (1.65×10(9.64×103) + (1.0 Jimai

For the compound: C8H17NO2 Use the following information to come up with a plausible structure: 8 This compound has "carboxylic acid amide" and ether functional groups. The peaks at 1.2ppm are two signals that are overlapping one another. One of the two signals is a doublet that represents 6 hydrogens; the other signal is a quartet that represents 3 hydrogens.

Vnk the elements or compounds in the table below in decreasing order of their boiling points. That is, choose 1 next to the substance with the highest bolling point, choose 2 next to the substance with the next highest boiling point, and so on. substance C D chemical symbol, chemical formula or Lewis structure. CH,-N-CH, CH, H H 10: H C-C-H H H H Cale H 10: H-C-C-N-CH, Bri CH, boiling point (C) Сен (C) B (Choose