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A hypothetical NH molecule makes a rotational-level transition from \(l\) = 3 to \(l\) = 1 and gives off a photon of wavelength 1.780 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\(^-$$^2$$^7\) kg, and the mass of nitrogen is 2.33 \(\times\) 10\(^-$$^2$$^6\) kg.

Short Answer

Expert verified
The separation is approximately 91.0 pm.

Step by step solution

01

Calculating Reduced Mass

The reduced mass \( \mu \) of the diatomic molecule is calculated using the formula:\[ \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \]where \(m_1 = 1.67 \times 10^{-27}\) kg is the mass of hydrogen and \(m_2 = 2.33 \times 10^{-26}\) kg is the mass of nitrogen.\[ \mu = \frac{1.67 \times 10^{-27} \cdot 2.33 \times 10^{-26}}{1.67 \times 10^{-27} + 2.33 \times 10^{-26}} = \frac{3.89 \times 10^{-53}}{2.50 \times 10^{-26}} = 1.56 \times 10^{-27} \text{ kg} \].
02

Rotational Energy Change

The change in rotational energy when the molecule transitions from \( l = 3 \) to \( l = 1 \) is calculated using the formula:\[ \Delta E = E_3 - E_1 = \frac{h^2}{8\pi^2 I} \left(l(l+1)\right) - \frac{h^2}{8\pi^2 I} \left(l'(l'+1)\right) \]where \( l = 3 \) and \( l' = 1 \), and \( I = \mu r^2 \) is the moment of inertia.
03

Photon Energy Calculation

The energy of the emitted photon can also be calculated from its wavelength:\[ E_{\gamma} = \frac{hc}{\lambda} \] Substituting for \( h = 6.626 \times 10^{-34} \) J\(\cdot\)s and \( c = 3 \times 10^8 \) m/s and \( \lambda = 1.780 \times 10^{-9} \) m:\[ E_{\gamma} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.780 \times 10^{-9}} = 1.12 \times 10^{-17} \text{ J} \].
04

Equating Energy and Solving for r

Set the change in rotational energy equal to the photon's energy:\[ \Delta E = E_{\gamma} \Rightarrow \frac{h^2}{8\pi^2 \mu r^2} \Big( l(l+1) - l'(l'+1) \Big) = 1.12 \times 10^{-17} \]Substituting \( \mu \) from Step 1 and solving for \( r \):\[ r = \sqrt{\frac{h^2 \left(12 - 2\right)}{8\pi^2 \mu \cdot 1.12 \times 10^{-17}}} \approx 9.10 \times 10^{-11} \text{ m} \].

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Key Concepts

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

Reduced Mass Calculation
In molecular physics, the reduced mass is a crucial concept often used to simplify the analysis of two-body systems like diatomic molecules. It allows us to treat the motion of two interacting bodies as if it were one single body. The reduced mass \( \mu \) is defined by the formula:\[\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}\]Where \( m_1 \) and \( m_2 \) are the masses of the two atoms. This method focuses on the center of mass of the system, permitting us to eliminate the individual motions of the atoms. In simpler terms, the reduced mass helps us to express the system's behavior by combining both masses into a single effective mass. This makes the mathematics behind calculating physical properties such as rotational energy or vibrational frequency much easier to deal with. In this specific exercise, we calculated the reduced mass for the NH diatomic molecule as \( 1.56 \times 10^{-27} \) kg.
Rotational Energy Change
All molecules exhibit rotational energy due to their rotational motion around a bond. The quantized levels of this energy are given by:\[E_l = \frac{h^2}{8\pi^2 I} l(l+1)\]where \( l \) is the rotational quantum number, \( h \) is Planck’s constant, and \( I \) is the moment of inertia. When a molecule transitions between different rotational levels, there is a change in rotational energy \( \Delta E \). In our problem, the transition occurs from \( l = 3 \) to \( l = 1 \), and the change in rotational energy can be computed by subtracting the energies of these levels. By calculating the difference \( (E_3 - E_1) \), we get the energy emitted or absorbed during the transition. Understanding these energy changes aids in interpreting spectral lines we observe in spectroscopy, providing insights into molecular structure and behavior.
Photon Energy Calculation
Photon energy is directly related to the wavelength of light, via the equation:\[E_{\gamma} = \frac{hc}{\lambda}\]Here, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon. This equation underpins why instruments like spectrometers can measure light energy by evaluating its wavelength. In the given problem, the NH molecule emits a photon with a wavelength of 1.780 nm during its rotational transition. Plugging in the values for \( h \, (6.626 \times 10^{-34} \text{J}\cdot\text{s}) \), \( c \, (3 \times 10^8 \text{ m/s}) \), and the wavelength, we obtain that the emitted photon's energy is \( 1.12 \times 10^{-17} \) J. Understanding this concept helps connect the dots between quantum mechanical phenomena and observable quantities in laboratories.
Moment of Inertia
The moment of inertia \( I \) is a pivotal concept in rotational dynamics for molecular systems, much like mass is for linear dynamics. In diatomic molecules, it reflects how the mass is distributed around the rotational axis and is given by:\[I = \mu r^2\]Where \( \mu \) is the reduced mass, and \( r \) is the distance between the two atoms. This "rotational equivalent of mass" helps determine the resistance of the molecule to changes in its rotation. A larger moment of inertia implies a molecule is harder to spin. In rotational transitions, \( I \) directly influences the energy levels, as seen in the formula for rotational energy. In the original problem, we used the moment of inertia to help link the change in rotational energy to the observed photon energy. Solving for \( r \), we deduced the separation between the nitrogen and hydrogen atoms, giving insights into their molecular structure.

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Most popular questions from this chapter

The maximum wavelength of light that a certain silicon photocell can detect is 1.11 \(\mu\)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.

The gap between valence and conduction bands in silicon is 1.12 eV. A nickel nucleus in an excited state emits a gammaray photon with wavelength 9.31 \(\times\) 10\(^-$$^4\) nm. How many electrons can be excited from the top of the valence band to the bottom of the conduction band by the absorption of this gamma ray?

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 \(2.21 \times 10^{-25} \mathrm{~kg}\).

(a) The equilibrium separation of the two nuclei in an NaCl molecule is 0.24 nm. If the molecule is modeled as charges \(+e\) and \(-e\) separated by 0.24 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 \(3.0 \times 10^{-29}\) \(C \cdot m\). If this dipole moment arises from point charges \(+q\) and \(-q\) separated by 0.24 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) Theequilibrium distance between nuclei in the hydrogen iodide (HI) molecule is 0.16 nm, and the measured electric dipole moment of the molecule is \(1.5 \times 10^{-30}\) \(C \cdot m\). 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.

If the energy of the \(H_{2}\) covalent bond is -4.48 eV, what wavelength of light is needed to break that molecule apart? In what part of the electromagnetic spectrum does this light lie?

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