/*! 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 21 (II) A certain type of film is s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 37: Problem 21

(II) A certain type of film is sensitive only to light whose wavelength is less than 630 \(\mathrm{nm}\) . What is the energy \((\mathrm{eV}\) and \(\mathrm{kcal} / \mathrm{mol} )\) needed for the chemical reaction to occur which causes the film to change?

Short Answer

Expert verified
Energy needed: 1.97 eV or 45.3 kcal/mol.

Step by step solution

01

Convert Wavelength to Energy in Joules

Use the energy-wavelength relationship, given by the equation: \[E = \frac{hc}{\lambda}\]where:- \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\),- \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\),- \(\lambda\) is the wavelength \(630 \text{ nm} = 630 \times 10^{-9} \text{ m}\).Substituting these values, we calculate:\[E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{630 \times 10^{-9}}\]\[E \approx 3.15 \times 10^{-19} \text{ J}\]
02

Convert Energy from Joules to Electronvolts (eV)

To convert energy from joules to electronvolts, use the conversion factor: \[1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\].Thus, \[E \text{ (in eV)} = \frac{3.15 \times 10^{-19}}{1.602 \times 10^{-19}}\]\[E \approx 1.97 \text{ eV}\]
03

Convert Energy from Joules to kcal/mol

To find the energy in kcal/mol:- First convert joules to kilojoules per mole using Avogadro's number \(6.022 \times 10^{23}\text{ mol}^{-1}\).- Then convert kJ to kcal using the conversion \[1 \text{ kcal} = 4.184 \text{ kJ}\].\[E \text{ (in kJ/mol)} = \left( 3.15 \times 10^{-19} \text{ J} \right) \times \left( \frac{1 \text{ kJ}}{1000 \text{ J}} \right) \times (6.022 \times 10^{23} \text{ mol}^{-1})\]\[E \approx 189.7 \text{ kJ/mol}\]\[E \text{ (in kcal/mol)} = \frac{189.7}{4.184}\]\[E \approx 45.3 \text{ kcal/mol}\]

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!

Key Concepts

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

Photon Energy
Photon energy is the amount of energy carried by a single photon. Photons are the smallest units of light and electromagnetic radiation. The energy of a photon is directly proportional to its frequency, which means that higher frequency light has more energy. However, we often find it easier to relate photon energy to its wavelength using an important equation involving Planck's constant and the speed of light. Photon energy can be measured in various units including joules (J), electronvolts (eV), and calories per mole (kcal/mol). Different contexts use different units based on convenience. It's crucial to learn how to convert between these units to apply photon energy in different scenarios such as photoelectric effects and other applications in physics.
Wavelength-Energy Relationship
The wavelength-energy relationship is a fundamental concept in physics, which ties together the characteristics of light. It is expressed using the formula \[E = \frac{hc}{\lambda}\]where \(E\) represents energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the light. This equation shows us that energy is inversely proportional to wavelength. This means shorter wavelengths have higher energies and vice versa. Light's wavelength can range from very short gamma rays to long radio waves, with visible light sitting in between. Understanding this relationship is essential in fields like quantum physics and chemistry. It helps explain behaviors of light and its interaction with matter, such as in reactions sensitive to light like in photography or photosynthesis.
Planck's Constant
Planck's constant is a universal constant that plays a critical role in quantum mechanics. Denoted by the symbol \(h\), its value is approximately \(6.626 \times 10^{-34}\) Joule seconds. This constant is vital as it relates the energy of a photon to its frequency by the equation \[E = h \cdot f\]where \(E\) is energy and \(f\) is frequency. In the context of the photoelectric effect, Planck鈥檚 constant allows us to determine the energy of photons based on their wavelength or frequency. It effectively bridges the gap between the macroscopic classical view of light as a wave and the quantum view of light in terms of photon particles. Planck鈥檚 constant is fundamental to our understanding of the quantum world, serving as a mix of light-as-wave and light-as-particle theories.
Speed of Light
The speed of light in a vacuum is an essential constant in physics, expressed as \(c = 3.00 \times 10^8\) meters per second. This speed is a definitive factor in all formulas related to light, including those connecting wavelength, frequency, and photon energy. Light speed is the ultimate speed limit in the universe, meaning nothing can travel faster than light in a vacuum. It also plays an integral role in the equation \(E = \frac{hc}{\lambda}\), helping to translate wavelength values into energy. Understanding the speed of light is crucial for grasping concepts in relativity as well. In practical terms, the speed of light determines how we communicate over long distances and how we explore the universe. It serves as a foundation for understanding how energy is transmitted and how electromagnetic waves propagate through space.

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

In an \(X\) -ray tube, the high voltage between filament and target is \(V\) . After being accelerated through this voltage, an electron strikes the target where it is decelerated (by positively charged nuclei) and in the process one or more X-ray photons are emitted. (a) Show that the photon of shortest wavelength will have $$ \lambda_{0}=\frac{h c}{e V} $$ $$ \begin{array}{l}{\text { (b) What is the shortest wavelength of X-ray emitted when }} \\ {\text { accelerated electrons strike the face of a } 33-\mathrm{kV} \text { television }} \\ {\text { picture tube? }}\end{array} $$

Light of wavelength \(424 \mathrm{nm}\) falls on a metal which has a work function of \(2.28 \mathrm{eV}\). ( \(a\) ) How much voltage should be applied to bring the current to zero? (b) What is the maximum speed of the emitted electrons? (c) What is the de Broglie wavelength of these electrons?

(II) X-rays of wavelength \(\lambda=0.120 \mathrm{nm}\) are scattered from carbon. What is the expected Compton wavelength shift for photons detected at angles (relative to the incident beam) of exactly \((a) 60^{\circ},(b) 90^{\circ},(c) 180^{\circ} ?\)

(I) Calculate the momentum of a photon of yellow light of wavelength \(6.20 \times 10^{-7} \mathrm{~m}.\)

In a particular photoelectric experiment, a stopping potential of 2.70 \(\mathrm{V}\) is measured when ultraviolet light of wavelength 380 \(\mathrm{nm}\) is incident on the metal. If blue light of wavelength 440 \(\mathrm{nm}\) is used, what is the new stopping potential?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Rotational Dynamics

Read Explanation

Electromagnetism

Read Explanation

Engineering Physics

Read Explanation

Electricity

Read Explanation

Oscillations

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.