/*! 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} Problem 36 Is it possible for a fluorescent... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Is it possible for a fluorescent material to emit radiation in the ultraviolet region after absorbing visible light? Explain your answer.

Short Answer

Expert verified
No, a fluorescent material cannot emit ultraviolet light after absorbing visible light due to energy loss during re-emission.

Step by step solution

01

Understanding Fluorescence

Fluorescence is the process by which a material absorbs light at one wavelength and then emits light at a different, typically longer wavelength. This emission usually continues as long as the stimulating light is present.
02

Relation Between Wavelength and Energy

Light wavelengths are inversely related to energy according to the equation \( E = \frac{hc}{\lambda} \), where \( E \) is energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. Thus, shorter wavelengths (like UV) correspond to higher energy.
03

Comparing Visible and Ultraviolet Light

Visible light ranges from about 400 nm to 700 nm, while ultraviolet light ranges from about 10 nm to 400 nm. Ultraviolet light has shorter wavelengths and therefore higher energy than visible light.
04

Analyzing Energy Transition in Fluorescence

Since fluorescence involves absorbing light energy and then re-emitting this energy, the emitted light typically has the same or longer wavelength than the absorbed light due to loss of some energy (often as heat). Thus, it is common for fluorescence to emit light in a lower energy range than the absorbed light.
05

Conclusion on Possibility of Emission

Given that emitting ultraviolet light would require the emitted light to have a higher energy than the absorbed visible light, this would violate the typical process of fluorescence where some energy loss is inevitable during transition. Hence, a fluorescent material cannot emit in the ultraviolet region after absorbing visible light.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength and Energy
The concept of wavelength and energy is fundamental to understanding light and its properties. Light can be thought of as waves, and each wave has a particular wavelength, which is the distance between two consecutive peaks. The energy of these light waves is inversely proportional to their wavelength.

The equation that describes this relationship is:
  • \( E = \frac{hc}{\lambda} \)
- Where \( E \) represents energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the light.
- A shorter wavelength means higher energy.
- Conversely, a longer wavelength corresponds to lower energy.

This equation highlights why ultraviolet light, which has a shorter wavelength than visible light, possesses more energy.
Visible Light
Visible light is the portion of the electromagnetic spectrum that can be detected by the human eye. It consists of wavelengths ranging from approximately 400 nanometers (nm) to 700 nm.
  • The shortest wavelength in this range (approximately 400 nm) is associated with violet light.
  • The longest wavelength (around 700 nm) corresponds to red light.

This range of light encompasses all the colors that we can see in a rainbow.
- Since visible light has longer wavelengths than ultraviolet light, it consequently has lower energy. Understanding this helps in distinguishing how materials interact with different types of light.
Ultraviolet Light
Ultraviolet (UV) light is found just beyond the visible spectrum, with wavelengths from about 10 nm to 400 nm.
  • Because its wavelengths are shorter, ultraviolet light carries higher energy than visible light.
  • UV light is not visible to the human eye, but it can have significant effects, such as causing sunburn.

This high energy means that UV light can cause more energetic reactions. For example, it can change the chemical structure of some molecules, which is why materials sensitive to UV light often need to be protected.
Energy Transitions
In the process of fluorescence, a material absorbs light at one wavelength and emits it at another. Generally, these emissions occur at a longer wavelength than the absorbed light because some energy is lost in the process, often as heat.
  • This means the emitted light has less energy.
  • For example, if a material absorbs visible light, it usually emits light at a visible or even longer wavelength, such as in the infrared range.

Fluorescence cannot cause a material to emit at a shorter wavelength (higher energy), like UV light, after absorbing visible light. This is due to the intrinsic energy loss inherent in the fluorescence process. Hence, it is not possible for a fluorescent material to emit ultraviolet light after absorbing visible light.

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

An electron in a hydrogen atom is excited from the ground state to the \(n=4\) state. Comment on the correctness of the following statements (true or false). (a) \(n=4\) is the first excited state. (b) It takes more energy to ionize (remove) the electron from \(n=4\) than from the ground state. (c) The electron is farther from the nucleus (on average) in \(n=4\) than in the ground state. (d) The wavelength of light emitted when the electron drops from \(n=4\) to \(n=1\) is longer than that from \(n=4\) to \(n=2\) (e) The wavelength the atom absorbs in going from \(n=1\) to \(n=4\) is the same as that emitted as it goes from \(n=4\) to \(n=1\)

When copper is bombarded with high-energy electrons, \(\mathrm{X}\) rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the \(\mathrm{X}\) rays is \(0.154 \mathrm{nm}\).

A 368 -g sample of water absorbs infrared radiation at \(1.06 \times 10^{4} \mathrm{nm}\) from a carbon dioxide laser. Suppose all the absorbed radiation is converted to heat. Calculate the number of photons at this wavelength required to raise the temperature of the water by \(5.00^{\circ} \mathrm{C}\).

Describe the shapes of \(s, p\), and \(d\) orbitals. How are these orbitals related to the quantum numbers \(n, \ell,\) and \(m_{\ell} ?\)

A baseball pitcher's fastballs have been clocked at about \(100 \mathrm{mph}\). (a) Calculate the wavelength of a \(0.141-\mathrm{kg}\) baseball (in \(\mathrm{nm}\) ) at this speed. (b) What is the wavelength of a hydrogen atom at the same speed \((1 \mathrm{mile}=1609 \mathrm{~m})\) ?

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.