Question

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

The separation is approximately 91.0 pm.

Step-by-step solution

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.$$\mu =\frac{1.67\times {10}^{-27}\cdot 2.33\times {10}^{-26}}{1.67\times {10}^{-27}+2.33\times {10}^{-26}}=\frac{3.89\times {10}^{-53}}{2.50\times {10}^{-26}}=1.56\times {10}^{-27}\text{ kg}$$.

Rotational Energy Change

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

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:$$E_{\gamma }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1.780\times {10}^{-9}}=1.12\times {10}^{-17}\text{ J}$$.

Equating Energy and Solving for r

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

Key concepts

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 $\mathrm{\Delta }E$. 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.