For the H2 molecule the equilibrium spacing of the two protons is 0.0740.074 nm. The mass of a hydrogen atom is 1.67×10−271.67\$$times10^{-27} kg. $$Calculate the wavelength of the photon emitted in the rotational transition l=2 l = 2 to l=1l = 1.

Ch 42: Molecules and Condensed Matter

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 42: Molecules and Condensed Matter

Problem 3

Chapter 42, Problem 3

For the H₂ molecule the equilibrium spacing of the two protons is $0.074$ nm. The mass of a hydrogen atom is $1.67 \times 10^{- 27}$ kg. Calculate the wavelength of the photon emitted in the rotational transition $l equals 2$ to $l equals 1$ .

Verified step by step guidance

Step 1: Understand the problem. The question involves calculating the wavelength of a photon emitted during a rotational transition in a hydrogen molecule (H₂). This requires using the energy difference between rotational levels and the relationship between energy and wavelength.

Step 2: Recall the formula for the energy of rotational levels in a diatomic molecule: $ E_l = \frac{l(l+1)h^2}{8\pi^2I} $, where $ l $ is the rotational quantum number, $ h $ is Planck's constant, and $ I $ is the moment of inertia of the molecule. The moment of inertia is given by $ I = \mu r^2 $, where $ \mu $ is the reduced mass and $ r $ is the equilibrium spacing.

Step 3: Calculate the reduced mass $ \mu $ of the H₂ molecule using $ \mu = \frac{m_1 m_2}{m_1 + m_2} $. Since both atoms are hydrogen, $ m_1 = m_2 = 1.67 \times 10^{-27} \, \text{kg} $. Substitute these values into the formula.

Step 4: Substitute the reduced mass $ \mu $ and the equilibrium spacing $ r = 0.074 \times 10^{-9} \, \text{m} $ into the formula for the moment of inertia $ I = \mu r^2 $. Compute $ I $.

Step 5: Use the energy formula $ E_l $ to calculate the energy difference $ \Delta E = E_2 - E_1 $ between the rotational levels $ l = 2 $ and $ l = 1 $. Then, use the relationship $ \lambda = \frac{hc}{\Delta E} $ to express the wavelength $ \lambda $ of the emitted photon, where $ c $ is the speed of light.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Rotational Transitions

Rotational transitions refer to changes in the rotational energy levels of a molecule. In quantum mechanics, molecules can occupy discrete rotational states, characterized by quantum numbers. The transition from one state to another, such as from l = 2 to l = 1, involves the emission or absorption of a photon, with energy corresponding to the difference in rotational energy levels.

Energy of a Photon

The energy of a photon is given by the equation E = hν, where E is energy, h is Planck's constant (6.626 x 10^-34 J·s), and ν (nu) is the frequency of the photon. This relationship shows that the energy of the emitted photon during a transition is directly related to the frequency of the light, which can also be expressed in terms of wavelength using the equation c = λν, where c is the speed of light and λ (lambda) is the wavelength.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. For diatomic molecules like H2, the moment of inertia can be calculated using the formula I = μr², where μ is the reduced mass of the two atoms and r is the equilibrium bond length. This value is crucial for determining the rotational energy levels and, consequently, the energy of the emitted photon during transitions.