/*! 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 4 The water molecule has an \(l=1\... [FREE SOLUTION] | 91Ó°ÊÓ

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The water molecule has an \(l=1\) rotational level \(1.01 \times 10^{-5} \mathrm{eV}\) above the \(l=0\) ground level. Calculate the wave length and frequency of the photon absorbed by water when it undergoes a rotational-level transition from \(l=0\) to \(l=1\) . The magnetron oscillator in a microwave oven generates microwaves with a frequency of 2450 \(\mathrm{MHz}\) Does this make sense, in view of the frequency you calculated in this problem? Explain.

Short Answer

Expert verified
The photon's frequency is 2440 MHz, and its wavelength is 12.3 cm, which is close to the microwave oven frequency of 2450 MHz.

Step by step solution

01

Calculate the Energy Difference

The energy difference between two rotational levels can be given directly, but here it is explicitly stated as \(1.01 \times 10^{-5}\, \text{eV}\). This is the energy of the photon absorbed during the rotational transition from \(l=0\) to \(l=1\).
02

Convert Energy from eV to Joules

Use the conversion factor \(1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J}\) to convert the energy from electron volts to joules:\[E = 1.01 \times 10^{-5}\, \text{eV} \times 1.602 \times 10^{-19}\, \text{J/eV} = 1.617 \times 10^{-24}\, \text{J}\]
03

Calculate the Frequency of the Photon

Use Planck’s equation \(E = h \cdot f\), where \(E\) is energy, \(h = 6.626 \times 10^{-34}\, \text{J}\cdot\text{s}\) is Planck’s constant, and \(f\) is frequency:\[f = \frac{E}{h} = \frac{1.617 \times 10^{-24}\, \text{J}}{6.626 \times 10^{-34}\, \text{J}\cdot\text{s}} \approx 2.44 \times 10^{9}\, \text{Hz}\]This is equivalent to a frequency of 2440 MHz.
04

Calculate the Wavelength of the Photon

Use the speed of light equation \(c = \lambda \cdot f\), where \(c = 3 \times 10^{8}\, \text{m/s}\), to find the wavelength \(\lambda\):\[\lambda = \frac{c}{f} = \frac{3 \times 10^{8}\, \text{m/s}}{2.44 \times 10^{9}\, \text{Hz}} \approx 0.123\, \text{m} = 12.3\, \text{cm}\]
05

Compare with Microwave Oven Frequency

The calculated frequency for the rotational transition is approximately 2440 MHz. A typical microwave oven operates at 2450 MHz. This is quite close, indicating that the rotational transitions of water molecules match the frequency at which microwave ovens operate, thereby absorbing energy effectively.

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Key Concepts

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

Molecular Energy Levels
When we talk about rotational spectroscopy in molecules, we refer to energy levels associated with the rotation of molecules. These energy levels are quantized, meaning that a molecule can only rotate at specific, discrete energy levels rather than a continuous range. For a molecule like water (\(H_2O\)), these levels are characterized by the rotational quantum number \(l\).

  • \(l=0\) corresponds to the lowest rotational energy level, often referred to as the ground state.
  • \(l=1\) is the next energy level, positioned slightly higher in energy.
To transition between these levels, the molecule must absorb a specific amount of energy. For water, moving from \(l=0\) to \(l=1\) requires absorbing \(1.01 \times 10^{-5} \, \text{eV}\). This absorbed energy allows the molecule to spin faster, moving it to the higher energy state.
Photon Absorption
Photon absorption is the process by which a molecule takes in energy in the form of a photon. In rotational spectroscopy, this absorption causes the molecule to jump from one rotational energy level to a higher one.

When water absorbs a photon with energy \(1.01 \times 10^{-5} \, \text{eV}\), it undergoes a transition from the \(l=0\) (ground) rotational level to the \(l=1\) level. This photon absorption aligns with the concept that energy is absorbed in discrete packets, or quanta. This quantum nature of photon absorption is a fundamental principle of quantum mechanics that helps explain how molecular transitions occur.

This absorption process is crucial for technologies like microwave ovens, which leverage this concept to heat water by aligning the frequency of the microwaves with the rotational transition frequencies of water molecules.
Frequency Calculation
Calculating the frequency of a photon involved in rotational transitions is a key part of understanding spectroscopy. Frequency \(f\) in this context is directly related to the energy \(E\) absorbed by the molecule using the formula from Planck's theory: \[E = h \cdot f\]Here:
  • \(E\) is the energy in joules;
  • \(h\) is Planck’s constant \(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\);
  • \(f\) is the frequency in hertz (Hz).
Given a photon energy of \(1.617 \times 10^{-24} \, \text{J}\), the frequency is:\[f = \frac{1.617 \times 10^{-24} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} \approx 2.44 \times 10^{9} \, \text{Hz}\]This calculation reveals that the photon has a frequency of approximately 2440 MHz, close to the operational frequency of a microwave oven, confirming its effectiveness in heating through molecular rotations.
Wavelength Calculation
Wavelength \(\lambda\) is another critical property of a photon that is inversely related to its frequency, defined by the speed of light \(c\). The relationship is expressed by:\[c = \lambda \cdot f\]To find the wavelength, rearrange the formula:\[\lambda = \frac{c}{f}\]Plugging in known values where \(c = 3 \times 10^{8} \, \text{m/s}\) and \(f = 2.44 \times 10^{9} \, \text{Hz}\), we have:\[\lambda \approx \frac{3 \times 10^{8} \, \text{m/s}}{2.44 \times 10^{9} \, \text{Hz}} \approx 12.3 \, \text{cm}\] This wavelength is characteristic of microwave radiation, falling within the regions where water molecules absorb energy efficiently. Understanding wavelength is crucial for designing devices like microwave ovens, as it determines how effectively they generate the desired rotational molecular transitions.

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Most popular questions from this chapter

(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 of an NaCl molecule is \(3.0 \times 10^{-29} \mathrm{C} \cdot \mathrm{m} .\) If this dipole moment arises from point charges \(+q\) and \(-q\) separated by 0.24 \(\mathrm{nm}\) , what is \(q ?(\mathrm{c})\) 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 HI? How does your answer compare to that for NaCl calculated in part (c)? Discuss reasons for the difference in these results.

When a diatomic molecule undergoes a transition from the \(l=2\) to the \(l=1\) 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?

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 \(6.49 \times 10^{-26} \mathrm{kg}\) .

The force constant for the internuclear force in a hydrogen molecule \(\left(\mathrm{H}_{2}\right)\) is \(k^{\prime}=576 \mathrm{N} / \mathrm{m} .\) A hydrogen atom has mass \(1.67 \times 10^{-27} \mathrm{kg}\) . Calculate the zero-point vibrational energy for \(\mathrm{H}_{2}\) (that is, the vibrational energy the molecule has in the \(n=0\) ground vibrational level). How does this energy compare in magnitude with the \(\mathrm{H}_{2}\) bond energy of \(-4.48 \mathrm{eV} ?\)

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 clectron volts) between the valcnce and conduction bands for this photocell? (b) Explain why pure silicon is opaque.

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