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Chapter 42: Problem 3

hypothetical NH molecule makes a rotational-level transition from \(l=3\) to \(l=1\) and gives off a photon of wavelength 1.780 \(\mathrm{nm}\) in doing so. What is the separation between the two atoms in this molecule if we model them as point masses? The mass of hydrogen is \(1.67 \times 10^{-27} \mathrm{kg}\) , and the mass of mitrogen is \(2.33 \times 10^{-26} \mathrm{kg}\) .

Short Answer

Expert verified
The separation between the nitrogen and hydrogen atoms is approximately 1.035 pm.

Step by step solution

01

Understand Rotational Transitions and Energies

In a diatomic molecule, rotational energy levels are quantized and described by\[ E = \frac{l(l+1)\hbar^2}{2I} \]where \( l \) is the rotational quantum number, \( \hbar \) is the reduced Planck constant (\( 1.054 \times 10^{-34} \mathrm{J}\cdot\mathrm{s} \)), and \( I \) is the moment of inertia. The energy of a photon emitted during a transition between two energy levels is equal to the difference in these energies.
02

Calculate Photon Energy

The energy of the emitted photon is given by \[ E_{\text{photon}} = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \mathrm{J}\cdot\mathrm{s} \) is Planck's constant, \( c = 3.00 \times 10^8 \mathrm{m/s} \) is the speed of light, and \( \lambda = 1.780 \times 10^{-9} \mathrm{m} \) is the wavelength of the photon. Calculate this energy.
03

Calculate Moment of Inertia

The change in energy due to rotational transition is given by\[ \Delta E = E_{l=3} - E_{l=1} = \frac{6\hbar^2}{I} = \frac{12\hbar^2}{2I} \].From step 2, \( \Delta E = E_{\text{photon}} \). Use \( \Delta E \) calculated previously to solve for \( I \).
04

Calculate Bond Length

The moment of inertia \( I \) for a diatomic molecule modeled as point masses is given by\[ I = \mu r^2 \]where \( \mu \) is the reduced mass:\[ \mu = \frac{m_{\text{H}} m_{\text{N}}}{m_{\text{H}} + m_{\text{N}}} \].Use the known masses \( m_{\text{H}} = 1.67 \times 10^{-27} \mathrm{kg} \) and \( m_{\text{N}} = 2.33 \times 10^{-26} \mathrm{kg} \) to find \( \mu \) and solve for \( r \).
05

Solve for Separation Distance

Rearrange the equation \( I = \mu r^2 \) to solve for \( r \):\[ r = \sqrt{\frac{I}{\mu}} \].Use the previously calculated values of \( I \) and \( \mu \) to find the separation distance \( r \) between the nitrogen and hydrogen atoms.

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Key Concepts

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

Diatomic Molecule
A diatomic molecule is a molecule composed of two atoms, which can be the same or different elements. In our exercise, we're looking at a diatomic molecule made of nitrogen (N) and hydrogen (H), known as the NH molecule. These molecules are often modeled as rigid rotors to simplify the understanding of their rotational transitions.

The rotational energy levels of a diatomic molecule are quantized. This means that they can only take on specific energy values. The energy levels are described by the formula: \[ E = \frac{l(l+1)\hbar^2}{2I} \]where \( l \) is the quantum number associated with rotation, \( \hbar \) is the reduced Planck constant, and \( I \) is the moment of inertia of the molecule.

When a diatomic molecule undergoes a transition between these energy levels, it can emit or absorb photons. Understanding these quantized energy levels is essential to determine the behavior of diatomic molecules.
Moment of Inertia
The moment of inertia \( I \) is a crucial concept in understanding rotational transitions of diatomic molecules. It's a measure of how much resistance a body has to change its rotation. For a diatomic molecule treated as two point masses, it describes how the mass is distributed around the axis of rotation.

The formula to calculate the moment of inertia for a diatomic molecule is: \[ I = \mu r^2 \],where \( \mu \) is the reduced mass and \( r \) is the separation distance between the two atoms. The greater the moment of inertia, the lower the rotational energy levels for a given rotational quantum number \( l \).

Calculating the moment of inertia allows us to find the energy difference between rotational states, helping us understand phenomena like photon emissions during transitions.
Reduced Mass
The reduced mass \( \mu \) is a simplified representation of the mass in a diatomic system, which helps in treating the two-body problem as a one-body problem. When dealing with diatomic molecules, like nitrogen and hydrogen in the NH system, calculating the reduced mass is crucial for understanding their rotational dynamics.

The formula to determine the reduced mass is:\[ \mu = \frac{m_{\text{H}} m_{\text{N}}}{m_{\text{H}} + m_{\text{N}}} \],where \( m_{\text{H}} \) and \( m_{\text{N}} \) are the masses of hydrogen and nitrogen, respectively.

By calculating the reduced mass, we can better understand the physical properties of the diatomic molecule, as it's used in the moment of inertia and, therefore, directly impacts the energy levels and transitions.
Photons Emission
Photon emission is a process where an atom or molecule, after transitioning from a higher energy level to a lower energy level, releases a photon. This phenomenon is upmost important for understanding rotational transitions in diatomic molecules. In our scenario, the NH molecule emits a photon as it transitions from a rotational level \( l=3 \) to \( l=1 \).

The energy of this photon, \( E_{\text{photon}} \), is calculated using:\[ E_{\text{photon}} = \frac{hc}{\lambda} \],where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the emitted photon. This energy corresponds to the difference in the rotational energy levels of the molecule.

Understanding photon emission helps in the calculation of relevant properties like the moment of inertia and the distance between atoms in a diatomic molecule.

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Most popular questions from this chapter

Silver has a Fermi energy of 5.48 eV. Calculate the electron contribution to the molar heat capacity at constant volume of silver, \(C_{V},\) at 300 \(\mathrm{K}\) . Express your result \((\mathrm{a})\) as a multiple of \(R\) and (b) as a fraction of the actual value for silver, \(C_{V}=25.3 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) . (c) Is the value of \(C_{V}\) due principally to the electrons? If not, to what is it due? (Hint: See Section \(18.4 . )\)

42.27. 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?

An Ionic Bond. (a) Calculate the electric potential energy for \(a K^{+}\) ion and \(a B r^{-}\) ion separated by a distance of \(0.29 \mathrm{nm},\) the equilibrium separation in the KBr molecule. Treat the ions as point charges. (b) The ionization energy of the potassium atom is 4.3 \(\mathrm{eV}\) . Atomic bromine has an electron affinity of 3.5 \(\mathrm{eV}\) . Use these data and the results of part (a) to estimate the binding energy of the KBr molecule. Do you expect the actual binding energy to be higher or lower than your estimate? Explain your reasoning.

Hydrogen is found in two naturally occurring isotopes; normal hydrogen (containing a single proton in its nucleus) and deuterium (having a proton and a neutron). Assuming that both molecules are the same size and that the proton and neutron have the same mass (which is almost the case), find the ratio of (a) the energy of any given rotational state in a diatomic hydrogen molecule to the energy of the same state in a diatomic deuterium molecule and (b) the energy of any given vibrational state in hydrogen to the same state in deuterium (assuming that the force constant is the same for both molecules). Why is it physically reasonable that the force constant would be the same for hydrogen and deuterium molecules?

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.
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