/*! 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 8 The water molecule has an \(I=1\... [FREE SOLUTION] | 91Ó°ÊÓ

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The water molecule has an \(I=1\) rotational level 1.0 \(\mathrm{I} \times\) \(10^{-5} \mathrm{eV}\) above the \(I=0\) ground level. Calculate the wavelength and frequency of the photon absorbed by water when it undergoes a rotational- level transition from \(I=0\) to \(I=1 .\) The magnetron oscillator in a microwave oven generates microwaves with a frequency of 2450 MHz. Does this make sense, in view of the frequency you calculated in this problem? Explain.

Short Answer

Expert verified
The wavelength is 0.124 m, and the frequency is approximately 2.42 GHz. This frequency is close to the microwave oven frequency of 2.45 GHz, validating its use for exciting rotational transitions in water.

Step by step solution

01

Identify the Energy Difference

The energy difference between the rotational levels is given as \( \Delta E = 1.0 \times 10^{-5} \text{ eV} \). This is the energy of the photon absorbed during the transition from \( I=0 \) to \( I=1 \).
02

Convert Energy to Frequency

To find the frequency \( u \) of the photon absorbed, use the energy-frequency relation \( E = hu \), where \( h = 4.135667696 \times 10^{-15} \text{ eV}\cdot \text{s} \) is Planck's constant. Thus, \( u = \frac{\Delta E}{h} = \frac{1.0 \times 10^{-5} \text{ eV}}{4.135667696 \times 10^{-15} \text{ eV}\cdot \text{s}} \approx 2.42 \times 10^{9} \text{ Hz} \).
03

Convert Frequency to Wavelength

To find the wavelength \( \lambda \), use the relation \( c = \lambda u \), where \( c = 3.00 \times 10^{8} \text{ m/s} \). Thus, \( \lambda = \frac{c}{u} = \frac{3.00 \times 10^{8} \text{ m/s}}{2.42 \times 10^{9} \text{ Hz}} \approx 0.124 \text{ m} \).
04

Compare with Microwave Oven Frequency

The microwave oven operates at \( 2450 \text{ MHz} \) or \( 2.450 \times 10^{9} \text{ Hz} \). The frequency calculated for the water molecule's rotational transition is \( 2.42 \times 10^{9} \text{ Hz} \), which is very close to the microwave oven frequency. This makes sense as microwaves are used to excite rotational transitions in water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy-Frequency Relation
The energy-frequency relation is a fundamental concept in physics that connects the energy of a photon to its frequency. This relationship is given by the equation \( E = h u \), where \( E \) is the energy in electron volts (eV), \( h \) is Planck's constant, and \( u \) is the frequency in hertz (Hz). This simple yet powerful equation allows us to calculate the frequency of a photon if we know its energy and vice versa.

In the original exercise involving the water molecule, the energy difference between rotational levels is given as \( \Delta E = 1.0 \times 10^{-5} \) eV. By using the energy-frequency relation, the photon that's absorbed during the transition between these levels can have its frequency determined.

  • Each energy level transition involves the absorption or emission of a photon with a specific energy corresponding to that transition.
  • Understanding this relation is crucial for studying various phenomena such as rotational transitions used in spectroscopy.
Microwave Spectroscopy
Microwave spectroscopy is a technique used to study molecular rotational transitions. In this technique, microwaves are used to excite molecules from one rotational energy level to another. This makes it an invaluable tool for analyzing molecular structures and shapes. In particular, it helps physicists and chemists determine the moment of inertia and bond lengths in molecules.

Water molecules, when exposed to microwave radiation, absorb energy and undergo rotational transitions. This is exactly what happens in microwave ovens, where the microwave radiation excites the rotational states of water molecules present in the food, leading to heating.

  • Microwave spectroscopy is commonly used to study not only water but also other polar molecules.
  • This method provides detailed information about molecular geometry and internal motions.
  • It is fundamental in applications such as remote sensing and astrophysics missions to observe molecular clouds.
Planck's Constant
Planck's constant is a key fundamental constant in physics that plays a central role in the quantum mechanics domain. Its value is \( 4.135667696 \times 10^{-15} \text{ eV} \cdot \text{s} \), representing the proportionality constant that relates the energy of a photon to its frequency in the energy-frequency relation (\( E = h u \)).

Named after Max Planck, who was a pioneer in quantum theory, this constant has widespread applications across various fields including atomic, molecular, and optical physics. It enables precise calculations in quantum phenomena predictions.

  • Planck's constant is instrumental in understanding black-body radiation and energy quantization.
  • This constant helps bridge classical physics concepts to the quantum level.
  • It serves as the foundation for developing technologies such as lasers and semiconductors.
Photon Absorption
Photon absorption is a process where a photon is taken in by an atom or molecule, causing an excitation to a higher energy level. In the context of rotational transitions within a molecule, photon absorption results in a change of rotational energy state.

When a water molecule transitions from one rotational level to another, it absorbs a photon with a specific energy corresponding to that transition. The energy of this photon is exactly equal to the energy gap between the two rotational levels.

  • Photon absorption is pivotal in spectroscopy techniques, providing insights into molecular structures and dynamics.
  • This process aids in understanding how molecules interact with electromagnetic radiation.
  • In fields such as chemistry and physics, it helps in tracking reaction progress and performing material analysis.

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Most popular questions from this chapter

A hypothetical diatomic molecule of oxygen (mass =\( 2.656 \times 10^{-26} \mathrm{kg} ) \quad \text { and } \quad \text { hydrogen } \quad\left(\text { mass }=1.67 \times 10^{-27} \mathrm{kg}\right)\) emits a photon of wavelength 2.39\(\mu \mathrm{m}\) when it makes a transition from one vibrational state to the next lower state. If we model this molecule as two point masses at opposite ends of a massless spring, (a) what is the force constant of this spring, and (b) how many vibrations per second is the molecule making?

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.

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.

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 \(3.34 \times 10^{-27} \mathrm{kg}\)

Pure germanium has a band gap of 0.67 ev. The Fermienergy is in the middle of the gap. (a) For temperatures of 250 \(\mathrm{K}\) ,\(300 \mathrm{K},\) and 350 \(\mathrm{K}\) , calculate the probability \(f(E)\) that a state at the bottom of the conduction band is occupied. (b) For each temperature in part (a), calculate the probability that a state at the top of the valence band is empty.

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