Problem 20 - College Physics

Question

(a) A molecule decreases its vibrational energy by 0.250 eV by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie? (b) An atom decreases its energy by 8.50 eV by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electromagnetic spectrum does that wavelength of light lie? (c) A molecule decreases its rotational energy by \(3.20 \times\) \(10 ^ { - 3 }\) eV by giving up a photon of light. What wavelength of light does it give up during this process, and in what part of the electro- magnetic spectrum does that wavelength of light lie?

Step-by-Step Solution

Answer

(a) 4960 nm, infrared; (b) 146 nm, ultraviolet; (c) 388,000 nm, far infrared/microwave.

Step 1: Understanding Energy Loss and Photon Emission

When a molecule or an atom loses energy by emitting a photon, the energy of the photon equals the energy lost. The energy of a photon is related to its wavelength by the equation \( E = \frac{hc}{\lambda} \) where \( E \) is the energy in electronvolts (eV), \( h \) is Planck’s constant \( (4.135667696 \times 10^{-15} \, \text{eV·s}) \), \( c \) is the speed of light \( (3.00 \times 10^{8} \, \text{m/s}) \), and \( \lambda \) is the wavelength in meters.

Step 2: Convert Energy to Wavelength for Vibrational Transition

For the vibrational energy decrease of 0.250 eV, use the equation \( \lambda = \frac{hc}{E} \). Substitute the values: \( h = 4.135667696 \times 10^{-15} \, \text{eV·s} \), \( c = 3.00 \times 10^{8} \, \text{m/s} \), and \( E = 0.250 \, \text{eV} \). Calculate: \( \lambda = \frac{4.135667696 \times 10^{-15} \, \text{eV·s} \times 3.00 \times 10^{8} \, \text{m/s}}{0.250 \, \text{eV}} \approx 4.96 \times 10^{-6} \, \text{m} \) or 4960 nm. This wavelength is in the infrared region of the electromagnetic spectrum.

Step 3: Convert Energy to Wavelength for Atomic Transition

For the atomic energy decrease of 8.50 eV, use the same formula: \( \lambda = \frac{hc}{E} \). Substitute \( E = 8.50 \, \text{eV} \) into the equation: \( \lambda = \frac{4.135667696 \times 10^{-15} \, \text{eV·s} \times 3.00 \times 10^{8} \, \text{m/s}}{8.50 \, \text{eV}} \approx 1.46 \times 10^{-7} \, \text{m} \) or 146 nm. This wavelength lies in the ultraviolet region of the spectrum.

Step 4: Convert Energy to Wavelength for Rotational Transition

For the rotational energy decrease of \(3.20 \times 10^{-3} \, \text{eV}\), again use \( \lambda = \frac{hc}{E} \). Substituting \( E = 3.20 \times 10^{-3} \, \text{eV} \) into the formula: \( \lambda = \frac{4.135667696 \times 10^{-15} \, \text{eV·s} \times 3.00 \times 10^{8} \, \text{m/s}}{3.20 \times 10^{-3} \, \text{eV}} \approx 3.88 \times 10^{-4} \, \text{m} \) or 388,000 nm. This wavelength is in the far infrared or microwave region of the spectrum.

Key Concepts

Photon Emission

Photon emission refers to the process where an atom or molecule loses energy and releases a photon—a particle of light. Think of photons as tiny packets of energy. When an atom or molecule transitions from a higher energy state to a lower one, it emits a photon to shed the excess energy.
This emitted photon carries away energy equal to the difference between the high and low energy states.

• Photon energy is directly linked to its wavelength: As energy increases, the wavelength becomes shorter. Conversely, as the energy level decreases, the wavelength gets longer.
• This relationship allows scientists to determine the wavelength or color of the light emitted during photon emission by using the formula: \( E = \frac{hc}{\lambda} \).

By understanding photon emission, we can uncover useful information about the energy transitions within atoms and molecules.

Energy Transitions

Energy transitions occur when atoms or molecules move between different states of energy. Imagine climbing or descending energy 'stairs'—each step represents a shift in energy levels. Every move, whether a jump up or down the steps, involves a change in energy.

• An atom absorbs energy to move to a higher energy level (excitation).
• When it falls back to a lower energy level (emission), it releases energy, often in the form of a photon.

The energy difference in these transitions determines the wavelength and type of photon emitted or absorbed. This knowledge helps us categorize photon emissions in various segments of the electromagnetic spectrum, such as infrared or ultraviolet.

Electromagnetic Spectrum

The electromagnetic spectrum is a full range of wavelengths and frequencies of electromagnetic radiation, from the longest and lowest energy radio waves to the shortest and highest energy gamma-rays.

The spectrum categorizes this range into several types:

• Radio waves have the longest wavelength and lowest energy.
• Microwaves are used in communication technology and have longer wavelengths than infrared light.
• Infrared light, not visible to the human eye, is experienced as heat.
• Visible light is the small part of the spectrum we can see, consisting of colors from red to violet.
• Ultraviolet light comes after visible light and has shorter wavelengths with higher energies.
• X-rays and gamma-rays possess the shortest wavelengths and highest energies, often used in medical imaging.

This spectrum helps us understand where specific light falls when a photon is emitted.

Planck's Constant

Planck's constant, denoted as \( h \), is a crucial element in the quantum world, bridging the gap between energy and frequency. It is expressed in Joule-seconds or electronvolt-seconds, with an approximate value of \( 4.135667696 \times 10^{-15} \) eV·s.

Planck’s constant appears in the energy-wavelength relationship formula \( E = \frac{hc}{\lambda} \), linking a photon's energy to its wavelength and the speed of light.

• It provides a consistent method for calculating energy transition wavelengths across various parts of the electromagnetic spectrum.
• Planck’s constant is fundamental in understanding the quantum behavior of particles—illuminating the interaction of matter and energy at the smallest scales.

By applying Planck’s constant, we can solve problems involving photon emission and energy transitions with ease.