/*! 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 26 The energy from radiation can be... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 26

The energy from radiation can be used to cause the rupture of chemical bonds. A minimum energy of \(941 \mathrm{~kJ} / \mathrm{mol}\) is required to break the nitrogen-nitrogen bond in \(\mathrm{N}_{2}\). What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?

Short Answer

Expert verified
The longest wavelength of radiation that can break the N-N bond in N2 molecule is \( 1273 \mathrm{nm} \), and it is Infrared radiation.

Step by step solution

01

Convert bond energy to joules per photon

As the given energy is in kJ/mol, we need to convert it into joules per photon. Since we know there are Avogadro's number (6.022 x 10^23) of photons in one mole, we'll divide the bond energy by Avogadro's number. Energy per mol = 941 kJ/mol = 941 x 10^3 J/mol Now, divide this by Avogadro's number to get the energy per photon: Energy per photon = \( \frac{941 \times 10^3 \mathrm{J/mol} }{ 6.022 \times 10^{23} \mathrm{mol^{-1}}} = 1.563 \times 10^{-19} \mathrm{J} \)
02

Use Planck's equation to find frequency

Planck's equation is \( E = h\nu \), where E is the energy, h is Planck's constant (6.626 x 10^-34 J s), and \( \nu \) is the frequency. To find the frequency, divide the energy per photon by Planck's constant : \( \nu = \frac{E}{h} \) \( \nu = \frac{1.563 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}} = 2.358 \times 10^{14} \mathrm{s}^{-1} \)
03

Use the speed of light equation to find the wavelength

The speed of light equation is \( c = \lambda\nu \), where c is the speed of light (3 x 10^8 m/s), \( \lambda\) is the wavelength, and \( \nu \) is the frequency. To find the wavelength, divide the speed of light by the frequency: \( \lambda = \frac{c}{\nu} \) \( \lambda = \frac{3 \times 10^8 \mathrm{m/s}}{2.358 \times 10^{14} \mathrm{s}^{-1}} = 1.273 \times 10^{-6} \mathrm{m} = 1273 \mathrm{nm} \)
04

Determine the type of electromagnetic radiation

Compare the wavelength to the range of wavelengths for each type of electromagnetic radiation: - Infrared: 700 nm to 1 mm - Visible: 400 nm to 700 nm - Ultraviolet: 10 nm to 400 nm Since the wavelength is within the infrared range, the radiation type is Infrared. Therefore, the longest wavelength of radiation that can break the N-N bond in N2 molecule is 1273 nm, and it is Infrared radiation.

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!

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

(a) Account for formation of the following series of oxides in terms of the electron configurations of the elements and the discussion of ionic compounds in Section 2.7: \(\mathrm{K}_{2} \mathrm{O}, \mathrm{CaO}, \mathrm{Sc}_{2} \mathrm{O}_{3}, \mathrm{TiO}_{2}, \mathrm{~V}_{2} \mathrm{O}_{5}, \mathrm{CrO}_{3}\) (b) Name these oxides. (c) Consider the metal oxides whose enthalpies of formation (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) are listed here. $$\begin{array}{lllll} \text { Oxide } & \mathrm{K}_{2} \mathrm{O}(s) & \mathrm{CaO}(s) & \mathrm{TiO}_{2}(s) & \mathrm{V}_{2} \mathrm{O}_{5}(s) \\ \hline \Delta H_{f}^{\circ} & -363.2 & -635.1 & -938.7 & -1550.6 \\ \hline \end{array}$$ Calculate the enthalpy changes in the following general reaction for each case: $$\mathrm{M}_{n} \mathrm{O}_{m}(s)+\mathrm{H}_{2}(g) \longrightarrow n \mathrm{M}(s)+m \mathrm{H}_{2} \mathrm{O}(g)$$ (You will need to write the balanced equation for each case, then compute \(\Delta H^{\circ} .\) ) (d) Based on the data given, estimate a value of \(\Delta H_{f}^{\circ}\) for \(\mathrm{Sc}_{2} \mathrm{O}_{3}(s)\).

Calculate the uncertainty in the position of (a) an electron moving at a speed of \((3.00 \pm 0.01) \times 10^{5} \mathrm{~m} / \mathrm{s},(\mathrm{b})\) a neutron moving at this same speed. (The masses of an electron and a neutron are given in the table of fundamental constants in the inside cover of the text.) (c) What are the implications of these calculations to our model of the atom?

Use the de Broglie relationship to determine the wavelengths of the following objects: (a) an 85 -kg person skiing at \(50 \mathrm{~km} / \mathrm{hr}\), (b) a \(10.0-\mathrm{g}\) bullet fired at \(250 \mathrm{~m} / \mathrm{s},(\mathrm{c})\) a lithium atom moving at \(2.5 \times 10^{5} \mathrm{~m} / \mathrm{s}\), (d) an ozone \(\left(\mathrm{O}_{3}\right)\) molecule in the upper atmosphere moving at \(550 \mathrm{~m} / \mathrm{s}\).

(a) What are "valence electrons"? (b) What are "core electrons"?(c) What does each box in an orbital diagram represent? (d) What quantity is represented by the direction (up or down) of the half-arrows in an orbital diagram?

Sketch the shape and orientation of the following types of orbitals: (a) \(s,(b) p_{z},(c) d_{x y}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

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.