/*! 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 36 When a diatomic molecule undergo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 42: Problem 36

When a diatomic molecule undergoes a transition from the \(l=2\) to the \(I=1\) rotational state, a photon with wavelength 63.8\(\mu \mathrm{m}\) is emitted. What is the moment of inertia of the molecule for an axis through its center of mass and perpendicular to the line connecting the nuclei?

Short Answer

Expert verified
The moment of inertia of the molecule is approximately 2.24 x 10^{-46} kg鈰卪虏.

Step by step solution

01

Understanding the Problem

We need to find the moment of inertia, \(I\), of a diatomic molecule given the wavelengths emitted during its rotational transition from the \(l=2\) to the \(l=1\) state. The energy of the emitted photon is given as having a wavelength of 63.8 \(\mu m\).
02

Calculating Energy of Photon

First, we should calculate the energy of the emitted photon using the wavelength \(\lambda = 63.8 \mu m\). We can use the formula \(E = \frac{hc}{\lambda}\) to find the energy, where \(h = 6.626 \times 10^{-34}\, \text{J}\cdot\text{s}\) and \(c = 3 \times 10^{8}\, \text{m/s}\). Convert \(\lambda\) to meters before substituting in. \(\lambda = 63.8 \times 10^{-6}\, \text{m}\).
03

Substitute and Solve for Energy

Substitute the values into the energy formula: \[E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{63.8 \times 10^{-6}}\]Calculating this gives us the energy in joules.
04

Finding Rotational Energy Levels

The energy difference between rotational levels in a diatomic molecule is given by \(\Delta E = E_{l=2} - E_{l=1} = \frac{2 \hbar^2}{I} \cdot \Delta l(l+1)\) with \(\Delta l = 1\) since the transition is from \(l=2\) to \(l=1\). \(\Delta l(l+1) = 2(2+1) - 1(1+1) = 6-2 = 4\).
05

Calculating Moment of Inertia

We now equate the computed energy difference to the photon energy: \[E_{photon} = \frac{2 \hbar^2}{I} \cdot 4\]Solve for \(I\): \[I = \frac{8 \hbar^2}{E_{photon}}\]Substitute the energy value found earlier and numerical value \(\hbar = 1.055 \times 10^{-34} \text{J}\cdot\text{s}\).
06

Substitute and Solve for I

Using the value of \(E_{photon}\) from earlier, substitute into \[I = \frac{8 (1.055 \times 10^{-34})^2}{E_{photon}}\]Calculate to find \(I\) in \(\text{kg}\cdot\text{m}^{2}\).

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.

Diatomic Molecule
Diatomic molecules are molecules that consist of two atoms, which could either be the same or different elements. A common example of diatomic molecules include oxygen (\( O_2 \)), hydrogen (\( H_2 \)), and nitrogen (\( N_2 \)).
These molecules are significant in physics particularly when you explore their rotational motion. The physical structure of a diatomic molecule allows for rotational movement around its center of mass, and this is crucial when studying concepts like rotational transitions and energy levels.
This makes diatomic molecules an excellent subject in the study of molecular physics, where understanding their rotational transitions can provide insight into the properties of these molecules, such as their moment of inertia.
Photon Energy
Photon energy refers to the energy carried by a photon, which is a particle of light. The energy of a photon is directly related to its wavelength. To determine the energy of a photon, the formula \(E = \frac{hc}{\lambda}\) is used, where \(h\) represents Planck鈥檚 constant \((6.626 \times 10^{-34} \, \text{J}\cdot\text{s})\), \(c\) stands for the speed of light \((3 \times 10^{8} \, \text{m/s})\), and \(\lambda\) represents the wavelength of the photon.
This relationship is essential in understanding the transitions of diatomic molecules which emit or absorb light at specific wavelengths. Knowing how to compute photon energy helps in dealing with problems associated with molecular transitions and allows for direct calculation of other related physical quantities, such as the moment of inertia.
Rotational Transition
A rotational transition occurs when a molecule changes its rotational state, often emitting or absorbing a photon in the process. For diatomic molecules, these transitions happen between quantized rotational energy levels.
It happens due to changes in rotational quantum numbers, typically denoted by \(l\), where the molecule moves from one quantized energy state to another. For example, a transition from \(l=2\) to \(l=1\) involves a decrease in the energy state, with the difference being emitted as photon energy.
  • Rotational transitions give insight into the rotational constants of molecules.
  • These transitions are always associated with an angular momentum change.
Understanding these transitions allows for calculated predictions of molecular behavior and their inter-level energy differences.
Rotational Energy Levels
The rotational energy levels of a diatomic molecule are quantized, meaning that the molecule can possess only certain discrete energy levels. These levels are a result of the molecule's rotation about its center of mass and are defined by the quantum number \(l\).
  • Each state is associated with specific energy \(E_l = \frac{l(l+1) \hbar^2}{2I}\), where \(\hbar\) is the reduced Planck鈥檚 constant and \(I\) is the moment of inertia.
  • The differing energy between any two rotational states determines the energy of photons that can be absorbed or emitted as the molecule transitions between these levels.
This quantization plays a pivotal role in spectroscopy as it determines what wavelengths will be emitted or absorbed during molecular transitions.
Understanding rotational energy levels provides the foundation for calculating moments of inertia and analyzing molecular structures and behaviors.

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

Two atoms of cesium (Cs) can form a Cs \(_{2}\) molecule. The equilibrium distance between the nuclei in a \(\mathrm{Cs}_{2}\) molecule is 0.447 \(\mathrm{nm} .\) Calculate the moment of inertia about an axis through the center of mass of the two nuclei and perpendicular to the line joining them. The mass of a cesium atom is 2.2 \(\mathrm{I} \times 10^{-25} \mathrm{kg}\) .

The gap between valence and conduction bands in diamond is 5.47 eV. (a) What is the maximum wavelength of a photon that can excite an electron from the top of the valence band into the conduction band? In what region of the electromagnetic spectrum does this photon lie? (b) Explain why pure diamond is transparent and colorless. (c) Most gem diamonds have a yellow color. Explain how impurities in the diamond can cause this color.

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.

Compute the Fermi energy of potassium by making the simple approximation that each atom contributes one free electron. The density of potassium is \(851 \mathrm{kg} / \mathrm{m}^{3},\) and the mass of a single potassium atom is \(6.49 \times 10^{-26} \mathrm{kg}\) .

CALC The one-dimensional calculation of Example 42.4 (Section 42.3\()\) can be extended to three dimensions. For the three-dimensional fce NaCl lattice, the result for the potential energy of a pair of \(\mathrm{Na}^{+}\) and \(C^{-}\) ions due to the electrostatic interaction with all of the ions in the crystal is \(U=-\alpha e^{2} / 4 \pi \epsilon_{0} r,\) where \(\alpha=1.75\) is the Madelung constant. Another contribution to the potential energy is a repulsive interaction at small ionic separation \(r\) due to overlap of the electron clouds. This contribution can be represented by \(A / r^{8},\) where \(A\) is a positive constant, so the expression for the total potential energy is $$U_{\mathrm{tot}}=-\frac{\alpha e^{2}}{4 \pi \epsilon_{0} r}+\frac{A}{r^{8}}$$ (a) Let \(r_{0}\) be the value of the ionic separation \(r\) for which \(U_{\text { ot }}\) is a minimum. Use this definition to find an equation that relates \(r_{0}\) and \(A,\) and use this to write \(U_{\text { ot }}\) in terms of \(r_{0 .}\) For \(\mathrm{NaCl}\) , \(r_{0}=0.281 \mathrm{nm} .\) Obtain a numerical value (in electron volts) of \(U_{\mathrm{tot}}\) for NaCl. (b) The quantity \(-U_{\text { tot }}\) is the energy required to remove a \(\mathrm{Na}^{+}\) ion and a \(\mathrm{Cl}^{-}\) ion from the crystal. Forming a pair of neutral atoms from this pair of ions involves the release of 5.14 eV (the ionization energy of \(\mathrm{Na}\) ) and the expenditure of 3.61 \(\mathrm{eV}\) (the electron affinity of Cl). Use the result of part (a) to calculate the energy required to remove a pair of neutral Na and Cl atoms from the crystal. The experimental value for this quantity is \(6.39 \mathrm{eV} ;\) how well does your calculation agree?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Solid State Physics

Read Explanation

Magnetism

Read Explanation

Quantum Physics

Read Explanation

Force

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.