Chapter 39: Problem 29
(a) An atom initially in an energy level with \(E\) = -6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with \(E\) = -2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon?
Short Answer
Step by step solution
Calculate the energy of the absorbed photon
Calculate the internal energy after photon absorption
Calculate the energy of the emitted photon
Calculate the internal energy after photon emission
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Energy Levels
Visualizing energy levels as steps on a staircase can help; an electron can jump from one step to another but cannot stay in between. During such transitions, energy levels change based on whether photons are absorbed (increasing energy) or emitted (decreasing energy).
For example, an atom with an initial energy of \(-6.52 \text{ eV}\) may absorb a photon, which raises it to a higher energy level. Conversely, if the atom's initial energy is \(-2.68 \text{ eV}\), and it emits a photon, its energy is reduced further as it moves to a lower energy level. The energy difference between levels corresponds to the energy of the absorbed or emitted photon.
Photon Energy Calculation
- \( E_{photon} \) is the photon's energy
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \text{ J s}\))
- \( c \) is the speed of light (\(3 \times 10^8 \text{ m/s}\))
- \( \lambda \) represents the wavelength in meters
For instance, the photon absorbed at 860 nm has an energy calculated roughly as\(2.31 \text{ eV}\), demonstrating the formula's practical application. This calculation involves converting the nanometer to meter by multiplying by \(10^{-9}\), ensuring that the final energy is expressed in commonly used units like electron volts (eV).
Electron Volt (eV)
In numbers, 1 eV equals \(1.602 imes 10^{-19} \text{ Joules}\). Experts prefer the eV unit when dealing with subatomic particles because it highlights energy changes on an easily manageable scale. For example, the transition of an atom through different energy levels often involves shifts within ranges of electron volts rather than joules, simplifying comparisons and calculations.
In practical terms, this unit provides a convenient framework for discussing photon energies like those calculated in the earlier sections, such as the \(2.31 \text{ eV}\) (absorption) or \(2.95 \text{ eV}\) (emission).
Wavelength Conversion in Physics
Commonly, light wavelengths are given in nanometers (nm). To convert nm to meters, multiply the given wavelength by \(10^{-9}\). For example, 860 nm converts to \(860 \times 10^{-9}\) meters. This standardization into meters allows physicists to input values directly into the photon energy formula \(E = \frac{hc}{\lambda}\), harmonizing with constants defined in the metric system.
Ensuring precision in unit conversion is vital to achieving accurate and meaningful results, as demonstrated in our absorption and emission calculations, where meticulous conversion led to precise energy values.