Question

The ionization energy of a certain element is \(412 \mathrm{~kJ} / \mathrm{mol}\). When the atoms of this element are in the first excited state, however, the ionization energy is only \(126 \mathrm{~kJ} / \mathrm{mol}\). Based on this information, calculate the wavelength of light emitted in a transition from the first excited state to the ground state.

Short answer

The wavelength of the emitted light is approximately 418 nm.

Step-by-step solution

Understand the Problem

We need to find the wavelength of light emitted when this element transitions from its first excited state to its ground state. The ionization energy difference between these states will give us the energy of the emitted photon.

Calculate Energy Difference

The energy difference between the ground state and the first excited state is given by their ionization energies. Calculate it using: \(E = E_{ground} - E_{excited}\) where \(E_{ground} = 412 \, \mathrm{kJ/mol}\) and \(E_{excited} = 126 \, \mathrm{kJ/mol}\).

Convert Energy to Joules

Convert the energy difference from \(\mathrm{kJ/mol}\) to \(\mathrm{J/molecule}\). First, convert to joules: \(E = (412 - 126) \, \mathrm{kJ/mol} = 286 \, \mathrm{kJ/mol}\). In joules, this is \(286,000 \, \mathrm{J/mol}\). Then divide by Avogadro's number \(N_A = 6.022 \times 10^{23} \, \mathrm{mol}^{-1}\): \[ E = \frac{286,000 \, \mathrm{J}}{6.022 \times 10^{23}} = 4.75 \times 10^{-19} \, \mathrm{J} \]

Use Energy-Wavelength Relationship

Use the relationship between energy and wavelength: \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \, \mathrm{J}\cdot\mathrm{s}\) is Planck's constant and \(c = 3.00 \times 10^8 \, \mathrm{m/s}\) is the speed of light. Solve for \(\lambda\):\[ \lambda = \frac{hc}{E} \]

Substitute the Values

Substitute the calculated energy and constants into the formula to find the wavelength:\[ \lambda = \frac{6.626 \times 10^{-34} \, \mathrm{J\cdot s} \times 3.00 \times 10^8 \, \mathrm{m/s}}{4.75 \times 10^{-19} \, \mathrm{J}} \]

Calculate Wavelength

Perform the calculation: \[ \lambda = \frac{1.9878 \times 10^{-25}}{4.75 \times 10^{-19}} = 4.18 \times 10^{-7} \, \mathrm{m} \] Convert \(m\) to \(nm\) (nanometers) by multiplying by \(10^9\): \(\lambda = 418 \, \mathrm{nm}\).

Key concepts

Wavelength Calculation

To determine the wavelength of light emitted during a transition between atomic states, we use the energy-wavelength relationship. This is crucial when evaluating how energy, in terms of photons, translates into visible or other forms of light. From physics, we know that this relationship is defined by the equation:\[ E = \frac{hc}{\lambda} \]Where:

• E is the energy in joules (J).
• h is Planck's constant, approximately \(6.626 \times 10^{-34} \, \mathrm{J\cdot s}\).
• c is the speed of light, about \(3.00 \times 10^8 \, \mathrm{m/s}\).
• \lambda (lambda) is the wavelength in meters (converted to nanometers when needed).

In the context of an ionization energy problem, you will first determine the energy of the light emitted due to the transition, which corresponds directly to the energy difference between two states. From there, solving for \lambda requires dividing the product of h and c by E. This provides the wavelength in meters, which can then be converted to nanometers (\times 10^9). A practical understanding of this conversion and calculation is impeccably useful in fields like chemistry and physics.

First Excited State

The first excited state of an atom is a key concept in understanding energy transitions. When an atom absorbs energy, its electrons jump to a higher energy level above the ground state, which is the lowest possible energy level. This new level is called the first excited state.In our example, the atom in question requires less energy to be removed (ionized) from this excited state compared to the ground state. Specifically, less energy is needed because the electron is already about halfway out of its atomic "orbit." This is why the ionization energy of the first excited state (\(126 \, \mathrm{kJ/mol}\)) is less than that of the ground state (\(412 \, \mathrm{kJ/mol}\)). The difference between these energies is what's released as a photon when the electron returns to the ground state.Understanding the states and energies allows scientists to calculate the changes an atom undergoes during reactions and when transitioning between states. Such calculations provide insights into the structures and behaviors of molecules.

Photon Emission

Photon emission is a fundamental concept in quantum mechanics and atomic physics. It occurs when an electron transitions from a higher energy level to a lower one, typically from an excited state back to the ground state. During this transition, the electron emits energy in the form of a photon. The amount of energy released is precisely the difference between the two energy levels involved. For instance, if an electron drops back to the ground state from the first excited state, it releases this specific energy difference as a photon. The emitted photon's energy determines its wavelength, thanks to the energy-wavelength relationship discussed earlier. Hence, understanding photon emission is not only essential for explaining the observable colors in substances but also plays a critical role in technologies such as lasers, LEDs, and stellar spectroscopy, where transitions between states lead to the emission of distinct wavelengths of light.