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 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

All textbooks

Young & Freedman Calc 15th Edition

Ch 42: Molecules and Condensed Matter

Problem 3

Chapter 41, 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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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.

Recommended video:

Guided course

11:47

Intro to Moment of Inertia

Related Practice

Textbook Question

The rotational energy levels of CO are calculated in Example $42.2$ . If the energy of the rotating molecule is described by the classical expression $K=($\frac$12)I$\omega$^2$ , for the $l = 1$ level, what is the angular speed of the rotating molecule?

views

Textbook Question

During each of these processes, a photon of light is given up. In each process, what wavelength of light is given up, and in what part of the electromagnetic spectrum is that wavelength? A molecule decreases its vibrational energy by $0.198$ eV.

views

Textbook Question

The H₂ molecule has a moment of inertia of $4.6 \times 10^{- 48}$ kg-m². What is the wavelength $l$ of the photon absorbed when H₂ makes a transition from the $l equals 3$ to the $l equals 4$ rotational level?

views

Textbook Question

Two atoms of cesium (Cs) can form a $C s sub 2$ molecule. The equilibrium distance between the nuclei in a $C s sub 2$ molecule is $0.447$ 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 $2.21 \times 10^{- 25}$ kg.

views

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\(\times\)10^{-27} kg. Calculate the wavelength of the photon emitted in the rotational transition l=2 l = 2 to l=1l = 1.

Verified video answer for a similar problem:

Key Concepts

Rotational Transitions

Energy of a Photon

Moment of Inertia

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 $l equals 1$ .