/*! 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 6 (a) A molecule decreases its vib... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 42: Problem 6

(a) A molecule decreases its vibrational energy by 0.250 \(\mathrm{eV}\) by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie? (b) An atom decreases its energy by 8.50 eV by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie? (c) A molecule decreases its rotational energy by \(3.20 \times 10^{-3} \mathrm{eV}\) by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie?

Short Answer

Expert verified
(a) 4960 nm, infrared; (b) 91.3 nm, ultraviolet; (c) 3880 nm, infrared.

Step by step solution

01

Formula for Calculating Wavelength

To find the wavelength of light, use the formula \( \lambda = \frac{hc}{E} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J路s} \), \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \), and \( E \) is the energy in joules. First, convert the energy from electronvolts (eV) to joules using the conversion factor: \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \).
02

Calculate Wavelength for (a)

Convert \( 0.250 \text{ eV} \) to joules: \( E_a = 0.250 \times 1.602 \times 10^{-19} = 4.005 \times 10^{-20} \text{ J} \). Using the wavelength formula, \( \lambda_a = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{4.005 \times 10^{-20}} \approx 4.96 \times 10^{-6} \text{ m} \) or 4960 nm. This places it in the infrared part of the spectrum.
03

Calculate Wavelength for (b)

Convert \( 8.50 \text{ eV} \) to joules: \( E_b = 8.50 \times 1.602 \times 10^{-19} = 1.3617 \times 10^{-18} \text{ J} \). Using the wavelength formula, \( \lambda_b = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1.3617 \times 10^{-18}} \approx 9.13 \times 10^{-8} \text{ m} \) or 91.3 nm. This places it in the ultraviolet part of the spectrum.
04

Calculate Wavelength for (c)

Convert \( 3.20 \times 10^{-3} \text{ eV} \) to joules: \( E_c = 3.20 \times 10^{-3} \times 1.602 \times 10^{-19} = 5.1264 \times 10^{-22} \text{ J} \). Using the wavelength formula, \( \lambda_c = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{5.1264 \times 10^{-22}} \approx 3.88 \times 10^{-3} \text{ m} \) or 3880 nm. This places it in the infrared part of the spectrum.

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.

Vibrational Energy Transition
When a molecule decreases its vibrational energy, it releases a photon. Vibrational energy transitions occur when molecules vibrate less energetically, transitioning from a higher to a lower vibrational state. This energy change is quite significant and involves interactions at the molecular level.
  • Molecules are made up of atoms that are in constant motion. These movements include vibrations - think of bonds stretching and compressing like springs.
  • A vibrational transition involves the release or absorption of a photon as the molecule switches between vibrational states.
Given an energy change of 0.250 eV, the decrease relates directly to the vibrational energy level change. By using the formula to convert this energy release to a wavelength, it becomes possible to identify where this light falls on the electromagnetic spectrum.
The resulting wavelength of 4960 nm indicates the release of a photon in the infrared region, typical for vibrational transitions. This is because infrared light corresponds precisely to the energy levels involved in molecular vibrations.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation. This spectrum ranges from very short wavelengths (gamma rays) to very long wavelengths (radio waves). Each type of radiation is characterized by its unique wavelength and energy.
  • Visible light is just a small part of the spectrum, ranging from about 400 nm (violet) to 700 nm (red).
  • Infrared light is on the spectrum just below visible light, from about 700 nm up to 1 mm. This range involves lower energy than visible light.
  • Ultraviolet light has a shorter wavelength than visible light and can be divided into UV-A, UV-B, and UV-C, with increasing energy levels.
Understanding where a given wavelength fits in the electromagnetic spectrum allows us to deduce the energy changes occurring within photons. In various scientific applications, identifying and categorizing these wavelengths can provide insights into the processes happening at an atomic or molecular scale.
Rotational Energy Transition
Rotational transitions occur when molecules change their rotational energy levels. This energy change is typically much smaller than vibrational or electronic energy changes.
  • Molecules have the ability to rotate around their bond axes, and changes in this rotation involve the release or absorption of very low energy photons.
  • When a molecule gives up 3.20 x 10^{-3} eV to transition to a lower rotational state, the associated photon is of low energy, producing a longer wavelength.
With an energy change resulting in a wavelength of 3880 nm, the photon falls within the infrared region. Rotational transitions, like vibrational ones, often fall within the longer wavelengths of the electromagnetic spectrum due to their comparatively low energy requirements. This highlights the usefulness of infrared spectroscopy in studying the rotational transitions of molecules.

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

The maximum wavelength of light that a certain silicon photocell can detect is 1.11\(\mu \mathrm{m}\) . (a) What is the energy gap (in electron volts) between the valence and conduction bands for this photocell? (b) Explain why pure silicon is opaque.

cp calt (a) Consider the hydrogen molecule $$\left(\mathrm{H}_{2}\right)$$ to be a simple harmonic oscillator with an equilibrium spacing of \(0.074 \mathrm{nm},\) and estimate the vibrational energy-level spacing for H. The mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) . Hint: Estimate the force constant by equating the change in Coulomb repulsion of the protons, when the atoms move slightly closer together than \(r_{0}\) to the "spring" force. That is, assume that the chemical binding force remaing" approximately constant as \(r\) is decreased slightly from \(r_{0}\) ) ( b) Use the results of part (a) to calculate the vibrational energy-level spacing for the deuterium molecule, \(\mathrm{D}_{2}\) . Assume that the spring constant is the same for \(D_{2}\) as for \(\mathrm{H}_{2}\) . The mass of a deuterium atom is \(3.34 \times 10^{-27} \mathrm{kg}\)

Metallic lithium has a bce crystal structure. Each unit cell is a cube of side length \(a=0.35 \mathrm{nm}\) . (a) For a bce lattice, what is the number of atoms per unit volume? Give your answer in terms of \(a\) . (Hint: How many are there per unit cell?) (b) Use the result of part (a) to calculate the zero- temperature Fermi energy \(E_{\mathrm{PO}}\) for metallic lithium. Assume there is one free electron per atom.

For a solid metal having a Fermi energy of 8.500 eV, what is the probability, at room temperature, that a state having an energy of 8.520 eV is occupied by an electron?

CP (a) The equilibrium separation of the two nuclei in an NaCl molecule is 0.24 \(\mathrm{nm}\) . If the molecule is modeled as charges \(+e\) and \(-e\) separated by 0.24 \(\mathrm{nm}\) , what is the electric dipole moment of the molecule (see Section 21.7\() ?\) (b) The measured electric dipole moment arises from point charges \(+q\) and \(-q\) separated by 0.24 \(\mathrm{nm}\) , what is \(q ?\) (c) A definition of the fractional ionic character of the bond is \(q / e\) . If the sodium atom has charge \(+e\) and the chlorine atom has charge \(-e\) the fractional ionic character would be equal to 1. What is the actual fractional ionic character for the bond in NaCl? (d) The equilibrium distance between nuclei in the hydrogen iodide (HI) molecule is \(0.16 \mathrm{nm},\) and the measured electric dipole moment of the molecule is \(1.5 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\) . What is the fractional ionic character for the bond in \(\mathrm{HI}\) ? How does your answer compare to that for NaCl calculated in part (c)? Discuss reasons for the difference in these results.
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Wave Optics

Read Explanation

Classical Mechanics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electrostatics

Read Explanation

Linear Momentum

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.