Question

(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

(a) -4.21 eV; (b) -5.63 eV.

Step-by-step solution

Calculate the energy of the absorbed photon

The energy of a photon can be calculated using the formula: \( E_{photon} = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \), \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \), and \( \lambda \) is the wavelength in meters. First, convert 860 nm to meters: \( 860 \text{ nm} = 860 \times 10^{-9} \text{ m} \). Calculate the photon energy: \[ E_{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{860 \times 10^{-9}} \approx 2.31 \text{ eV}. \]

Calculate the internal energy after photon absorption

The internal energy of the atom after absorbing the photon is the sum of its initial energy and the energy of the absorbed photon. Initial energy \( E = -6.52 \text{ eV} \). Total energy after absorption: \[ E_{total} = E + E_{photon} = -6.52 + 2.31 = -4.21 \text{ eV}. \]

Calculate the energy of the emitted photon

Use the photon energy formula to find the energy of the emitted photon: \( \lambda = 420 \text{ nm} = 420 \times 10^{-9} \text{ m} \). Calculate the photon energy: \[ E_{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{420 \times 10^{-9}} \approx 2.95 \text{ eV}. \]

Calculate the internal energy after photon emission

The internal energy of the atom after emitting the photon is its initial energy minus the energy of the emitted photon. Initial energy \( E = -2.68 \text{ eV} \). Total energy after emission: \[ E_{total} = E - E_{photon} = -2.68 - 2.95 = -5.63 \text{ eV}. \]

Key concepts

Energy Levels

Atoms can occupy different energy levels, which are specific discrete values of energy. The chemical properties of an atom and how it interacts with other atoms depend on this energy arrangement. When an atom absorbs or emits a photon, it transitions between these energy levels. The energy levels are quantized, meaning electrons within the atom can only exist at specific energies.
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

Calculating the energy of a photon involves understanding the relationship between energy, wavelength, and fundamental constants. A photon's energy is inversely related to its wavelength. The equation \( E_{photon} = \frac{hc}{\lambda} \) shows this relationship, where:

• \( E_{photon} \) 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

By using these constants, you can determine the energy of a photon associated with any specific wavelength.
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)

The electron volt (eV) is a unit of energy that is commonly used in the context of atomic and particle physics. It simplifies calculations involving tiny energy amounts. Particularly, an electron volt is the amount of kinetic energy gained by an electron when it is accelerated through an electric potential difference of 1 volt.
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

Converting wavelength measurements to different units is essential for calculations in physics, specifically when applying formulas involving light. Since the energy of a photon \(E\) is determined using its wavelength \(\lambda\), it's crucial to have wavelength in meters.
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.