Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 82

If a sample of calcium chloride is introduced into a nonluminous flame, the color of the flame turns to orange ("flame test"). The light is emitted because calcium atoms become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is $422.7 \mathrm{nm} .\( Calculate its frequency. (b) What is the energy of \)1.00 \mathrm{~mol}$ of these photons (a mole of photons is called an Einstein)? (c) Calculate the energy gap between the excited and ground states for the calcium atom.

Short Answer

Expert verified
The frequency of the emitted light is \(7.10 * 10^{14} Hz\). The energy of 1 mole of photons is \(2.84 * 10^{5} J/mol\). The energy gap between the excited and ground states for the calcium atom is \(4.71 * 10^{-19} J\).

Step by step solution

01

Convert wavelength to meters

We are given the wavelength in nanometers (nm), so we need to convert it to meters (m) for our calculations. Use the conversion factor 1 nm = 10⁻⁹ m: λ = 422.7 nm * (1 * 10⁻⁹ m / 1 nm) = 422.7 * 10⁻⁹ m
02

Calculate the frequency

Now that we have the wavelength in meters, we can use the speed of light equation to calculate the frequency: c = λ * ν Rearrange the equation to solve for frequency: ν = c / λ ν = (3 * 10⁸ m/s) / (422.7 * 10⁻⁹ m) = 7.10 * 10¹⁴ Hz The frequency of the emitted light is 7.10 * 10¹⁴ Hz. #b) Calculating the energy of 1 mole of photons#
03

Calculate the energy of one photon

Now that we have the frequency, we can use Planck's equation to calculate the energy of one photon: E = h * ν E = (6.626 * 10⁻³⁴ Js) * (7.10 * 10¹⁴ Hz) = 4.71 * 10⁻¹⁹ J The energy of one photon is 4.71 * 10⁻¹⁹ J.
04

Calculate the energy of 1 mole of photons

To find the energy of 1 mole of photons, multiply the energy of one photon by Avogadro's number (Nₐ): E₁ (1 mole of photons) = Nₐ * E E₁ = (6.022 * 10²³ mol⁻¹) * (4.71 * 10⁻¹⁹ J) = 2.84 * 10⁵ J/mol The energy of 1 mole of photons is 2.84 * 10⁵ J/mol. #c) Calculating the energy gap between the excited and ground states of the calcium atom# Since the energy of a single photon corresponds to the energy gap between the excited and ground states of the calcium atom, we have already found the required energy gap in step 3. Thus, the energy gap between the excited and ground states for the calcium atom is 4.71 * 10⁻¹⁹ J.

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

Scientists have speculated that element 126 might have a moderate stability, allowing it to be synthesized and characterized. Predict what the condensed electron configuration of this element might be.

It is possible to convert radiant energy into electrical energy using photovoltaic cells. Assuming equal efficiency of conversion, would infrared or ultraviolet radiation yield more electrical energy on a per-photon basis?

The first 25 years of the twentieth century were momentous for the rapid pace of change in scientists' understanding of the nature of matter. (a) How did Rutherford's experiments on the scattering of \(\alpha\) particles by a gold foil set the stage for Bohr's theory of the hydrogen atom? (b) In what ways is de Broglie's hypothesis, as it applies to electrons, consistent with J. J. Thomson's conclusion that the electron has mass? In what sense is it consistent with proposals preceding Thomson's work that the cathode rays are a wave phenomenon?

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(2.94 \times 10^{14} \mathrm{~s}^{-1}\). (b) Calculate the energy of a photon of radiation whose wavelength is 413 \(\mathrm{nm} .\) (c) What wavelength of radiation has photons of energy $6.06 \times 10^{-19} \mathrm{~J} ?$

The familiar phenomenon of a rainbow results from the diffraction of sunlight through raindrops. (a) Does the wavelength of light increase or decrease as we proceed outward from the innermost band of the rainbow? (b) Does the frequency of light increase or decrease as we proceed outward? [Section 6.3]
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

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