Question

The following values are the only energy levels of a hypothetical one-electron atom: $$\begin{array}{ll} E_{6}=-2 \times 10^{-19} \mathrm{~J} & E_{5}=-7 \times 10^{-19} \mathrm{~J} \\\ E_{4}=-11 \times 10^{-19} \mathrm{~J} & E_{3}=-15 \times 10^{-19} \mathrm{~J} \\\ E_{2}=-17 \times 10^{-19} \mathrm{~J} & E_{1}=-20 \times 10^{-19} \mathrm{~J} \end{array}$$(a) If the electron were in the \(n=3\) level, what would be the highest frequency (and minimum wavelength) of radiation that could be emitted? (b) What is the ionization energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) of the atom in its ground state? (c) If the electron were in the \(n=4\) level, what would be the shortest wavelength (in \(\mathrm{nm}\) ) of radiation that could be absorbed without causing ionization?

Short answer

For part (a), the highest frequency is 7.55 × 10^14 Hz and minimum wavelength is 398 nm. For part (b), the ionization energy is 1.204 × 10^3 kJ/mol. For part (c), the shortest wavelength is 181 nm.

Step-by-step solution

Find the Highest Frequency for Part (a)

When the electron is in the n = 3 level (d = E_nd = -15 × 10^{-19} J), the highest frequency occurs when the electron drops to the E_1 energy level (d = -20 × 10^{-19} J). Calculate the energy difference ΔE: d = |d_3 - d_1|d = | -15 × 10^{-19} J -(-20 × 10^{-19} J) |d = | 5 × 10^{-19} J |d = 5 × 10^{-19} JNext, use the energy-frequency relationship, d = hu, where h = 6.626 × 10^{-34} J·s.Solve for frequency: u = d ÷ hu = (5 × 10^{-19} J) ÷ (6.626 × 10^{-34} J·s)u ≈ 7.55 × 10^{14} Hz

Find the Minimum Wavelength for Part (a)

From the previous step, the energy change ΔE = 5 × 10^{-19} J.Using the relationship u = c ÷ λ, where c = 3.00 × 10^8 m/s, first find ν.From Step 1, ν ≈ 7.55 × 10^{14} Hz.Now, solve for the wavelength λ:λ = c ÷ νλ = (3.00 × 10^8 m/s) ÷ (7.55 × 10^{14} Hz)λ ≈ 3.98 × 10^{-7} mConvert meters to nanometers:λ ≈ 398 nm

Calculate Ionization Energy for Part (b)

The ionization energy is the energy required to remove the electron from the ground state (n = 1, E_1 = -20 × 10^{-19} J) to E_∞ (0 J). ΔE = |d_1 - d_∞|d = | -20 × 10^{-19} J - 0 J |d = 20 × 10^{-19} JConvert this energy to kJ/mol using Avogadro's number (N_A = 6.022 × 10^{23} mol^{-1}):20 × 10^{-19} J × (1 kJ/10^3 J) × (6.022 × 10^{23} mol^{-1})d_1 ≈ 1.204 × 10^3 kJ/mol

Find the Shortest Wavelength for Part (c)

For an electron in the n = 4 level (d_4 = -11 × 10^{-19} J), the shortest wavelength absorbed before ionization is for E_∞ (0 J). Calculate the ΔE:ΔE = |d_4 - d_∞|nd = | -11 × 10^{-19} J - 0 J |nd = | 11 × 10^{-19} J |Convert this energy to wavelength using λ = h*c ÷ nd:λ = h*c ÷ ΔEλ ≈ (6.626 × 10^{-34} J·s) × (3.00 × 10^8 m/s) ÷ (11 × 10^{-19} J)λ ≈ 1.81 × 10^{-7} mConvert to nanometers:λ ≈ 181 nm

Key concepts

Quantum Mechanics

Quantum mechanics is the branch of physics that explains the behavior of particles at the atomic and subatomic levels. It introduces the concept of quantized energy levels in atoms, where electrons can only occupy specific energy states.
In our exercise, we see energy levels such as \(E_{1}=-20 \times 10^{-19} \text{ J}\) and \(E_{6}=-2 \times 10^{-19} \text{ J}\). Each of these levels corresponds to a different amount of energy an electron can hold. When an electron jumps from a higher energy level to a lower one, it emits energy in the form of radiation. Conversely, when it absorbs energy, it moves to a higher energy level.
Understanding these principles helps us calculate quantities such as the frequency and wavelength of the emitted or absorbed radiation, which are essential for solving the given problems.

Ionization Energy

Ionization energy is the energy required to remove an electron from an atom entirely, making it ionized. This value changes depending on the electron’s initial energy level.
In our exercise, we calculate the ionization energy for an electron in its ground state (lowest energy level) using the provided energy values. For the hypothetical atom given, the ground state energy level is \(E_{1}=-20 \times 10^{-19} \text{ J}\). The energy required to ionize this electron is obtained by transitioning it from this ground state to \(E_{\text{∞}} = 0 \text{ J}\), which represents the electron being completely removed from the atom.
Using the formula:◾ \(\text{Ionization Energy} = |E_{1} - E_{\text{∞}}|\),
we find \(\text{ΔE} = 20 \times 10^{-19} \text{ J}\). This value is then converted to kJ/mol to give a practical unit of measurement.

Wavelength and Frequency

Wavelength and frequency are two important properties of waves, including those of electromagnetic radiation emitted or absorbed by atoms during electron transitions between energy levels. The relationship between energy (\(ΔE\)), frequency (\(ν\)) and wavelength (\(λ\)) is given by the equations:

• \(ΔE = hν\)
• \(ν = \frac{c}{λ}\)
• \(λ = \frac{c}{ν}\)

Where \(h\) is Planck’s constant (\(6.626 \times 10^{-34} \text{ J·s}\)) and \(c\) is the speed of light (\(3.00 \times 10^{8} \text{ m/s}\)).
In our problems, we use these formulas to calculate the wavelength and frequency of radiation emitted or absorbed when an electron transitions through various energy levels. For example, when determining the highest frequency radiation emitted from \(n=3\) to \(n=1\), we use:◾ \(ν = \frac{ΔE}{h}\) and◾ \(λ = \frac{c}{ν}\)
These calculations help us better understand the light spectral properties associated with electron transitions in atoms.