/*! 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 11 Use Planck's equation (8.3) to d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use Planck's equation (8.3) to determine (a) the energy, in joules per photon, of radiation of frequency \(7.39 \times 10^{15} \mathrm{s}^{-1}\) (b) the energy, in kilojoules per mole, of radiation of frequency \(1.97 \times 10^{14} \mathrm{s}^{-1}\)

Short Answer

Expert verified
a) The energy of radiation per photon is around \(4.9 \times 10^{-18}\) Joules. b) The energy of radiation per mole is around \(78.1\) Kilojoules.

Step by step solution

01

Calculating photon energy (part a)

Using the Planck's equation \(E = h\times v\), and substituting Planck's constant \(h\) as \(6.626 \times10^{-34}\) Js and frequency \(v\) as \(7.39 \times 10^{15}\) s^(-1), we can calculate the energy \(E\).
02

Convert joules to kilojoules (part b)

Still using the Planck's equation, we substitute Planck's constant and frequency \(v\) as \(1.97 \times 10^{14}\) s^(-1) to find the energy per photon. Then we multiply this value by Avogadro's number, \(6.022 \times 10^{23}\) to find the energy per mole, and divide by \(10^{3}\) to convert from Joules to Kilojoules.

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.

Photon Energy
Photon energy refers to the energy carried by a single photon, which is the particle-like unit of light. An essential equation used in calculating photon energy is Planck's equation, given by: \[E = h \times v\] Here, \(E\) is the energy of the photon, \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{Js})\), and \(v\) is the frequency of the photon in hertz (Hz).
Planck's equation is crucial in understanding the relationship between the energy of light and its frequency. It shows that photon energy directly depends on frequency: the higher the frequency, the greater the energy of the photon.
By applying this formula, you can convert the frequency of radiation, given in hertz, to energy in joules per photon. Understanding this concept helps in various fields like quantum physics, chemistry, and even in applications such as photovoltaic energy systems, where the energy from light is used to generate electricity.
Avogadro's Number
Avogadro's number is a fundamental constant that represents the number of constituent particles (usually atoms or molecules) in one mole of a substance. Its value is: \(6.022 \times 10^{23}\) entities per mole. This large number is significant because it allows chemists to count particles by weighing macroscopic amounts of material, which is easier to measure in the lab. When talking about electron or photon energy in chemistry, Avogadro's number is crucial. It helps convert the energy of a single photon (in joules) to the energy contained per mole, which is more useful for practical chemical and physical calculations.
In part b of our exercise, to determine the energy in kilojoules per mole instead of joules per photon, we multiply the energy for one photon by Avogadro's number. This calculation shows how much energy a mole of these photons would contain and is typically expressed in kilojoules when dealing with larger values of energy.
Frequency
Frequency is a term commonly used in physics to describe the number of cycles of a wave that pass a point in one second. The unit of frequency is hertz (Hz), where 1 Hz means one cycle per second. In the context of light and electromagnetic radiation, frequency refers to how often the wave oscillates as it travels through space.
The relationship between frequency, wavelength, and the speed of light is given by the equation: \[c = \lambda \times v\] where \(c\) is the speed of light \((3.00 \times 10^8 \text{ m/s})\), \(\lambda\) is the wavelength, and \(v\) is the frequency.
Understanding frequency is important in the calculation of photon energy using Planck's equation. A change in frequency will directly affect the energy of photons because, as mentioned, higher frequency means higher energy. This concept is fundamental when studying the spectral characteristics of light and analyzing phenomena like the photoelectric effect, where light frequency determines the energy needed to eject electrons from a material.

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

What is the wavelength, in nanometers, associated with a \(1000 \mathrm{kg}\) automobile traveling at a speed of \(25 \mathrm{m} \mathrm{s}^{-1},\) that is, considering the automobile to be a matter wave? Comment on the feasibility of an experimental measurement of this wavelength.

The angular momentum of an electron in the Bohr hydrogen atom is mur , where \(m\) is the mass of the electron, \(u,\) its velocity, and \(r,\) the radius of the Bohr orbit. The angular momentum can have only the values nh/2 \(\pi\), where \(n\) is an integer (the number of the Bohr orbit). Show that the circum frences of the various Bohr orbits are integral multiples of the de Broglie wavelengths of the electron treated as a matter wave.

In your own words, define the following terms or symbols: (a) \(\lambda ;\) (b) \(\nu ;\) (c) \(h ;\) (d) \(\psi ;\) (e) principal quantum number, \(n\).

An atom in which just one of the outer-shell electrons is excited to a very high quantum level \(n\) is called a "high Rydberg" atom. In some ways, all these atoms resemble a Bohr hydrogen atom with its electron in a high-numbered orbit. Explain why you might expect this to be the case.

A molecule of chlorine can be dissociated into atoms by absorbing a photon of sufficiently high energy. Any excess energy is translated into kinetic energy as the atoms recoil from one another. If a molecule of chlorine at rest absorbs a photon of 300 nm wavelength, what will be the velocity of the two recoiling atoms? Assume that the excess energy is equally divided between the two atoms. The bond energy of \(\mathrm{Cl}_{2}\) is \(242.6 \mathrm{kJ} \mathrm{mol}^{-1}\)

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.