Problem 34 - Chemistry The Molecular Science

Question

Spectroscopists have observed \(\mathrm{He}^{+}\) in outer space. This ion is a one-electron species like a neutral hydrogen atom. Calculate the energy of the photon emitted for the transition from the \(n=5\) to the \(n=3\) state in this ion using the equation: \(E_{n}=-Z^{2} / n^{2}\left(2.179 \times 10^{-18} \mathrm{~J}\right) . \mathrm{Z}\) is the positive charge of the nucleus and \(n\) is the principal quantum number. In what part of the electromagnetic spectrum does this radiation lie?

Step-by-Step Solution

Answer

The photon emitted is in the ultraviolet range with a wavelength of 320 nm.

Step 1: Understand the Ion

The ion mentioned is He⁺, which behaves like a hydrogen atom because it is a one-electron system, removing its neutral outer electron. As such, we will use the formula for hydrogen-like atoms to calculate the energy of photon transitions.

Step 2: Identify the Variables

For \mathrm{He}^{+}, the nuclear charge (\mathrm{Z}) is 2. We are examining a transition from \(\textit{n}\) = 5 to \(\textit{n}\) = 3. We'll apply the formula: \[E_{n} = -\frac{Z^{2}}{n^{2}}\left(2.179 \times 10^{-18} \text{ J} \right)\] to both levels.

Step 3: Calculate Energy at n=5

Insert the values for n = 5 and Z = 2 into the energy equation: \[E_{5} = -\frac{2^{2}}{5^{2}} \left(2.179 \times 10^{-18} \text{ J}\right) = -\frac{4}{25} \times 2.179 \times 10^{-18} \text{ J}\] \[E_{5} = -3.4864 \times 10^{-19} \text{ J}\]

Step 4: Calculate Energy at n=3

Insert the values for n = 3 and Z = 2 into the energy equation: \[E_{3} = -\frac{2^{2}}{3^{2}} \left(2.179 \times 10^{-18} \text{ J}\right) = -\frac{4}{9} \times 2.179 \times 10^{-18} \text{ J}\] \[E_{3} = -9.6844 \times 10^{-19} \text{ J}\]

Step 5: Determine Photon Emission Energy

The energy of the emitted photon is the difference in energy between the two energy states:\[\Delta E = E_{3} - E_{5} = -9.6844 \times 10^{-19} \text{ J} + 3.4864 \times 10^{-19} \text{ J} \]\[\Delta E = -6.198 \times 10^{-19} \text{ J}\]This negative sign indicates the release of energy.

Step 6: Determine the Wavelength

Use the energy-wavelength relationship \( E = \frac{hc}{\lambda} \):\[\lambda = \frac{hc}{\Delta E} \]Using \(h = 6.626 \times 10^{-34} \text{ J s}\) and \(c = 3 \times 10^{8} \text{ m/s}\), \[\lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{6.198 \times 10^{-19}} \approx 3.20 \times 10^{-7} \text{ m}\]Convert meters to nanometers: \[3.20 \times 10^{-7} \text{ m} = 320 \text{ nm}\].

Step 7: Determine Part of Electromagnetic Spectrum

Given \( \lambda = 320 \text{ nm} \), this wavelength is in the ultraviolet (UV) range of the electromagnetic spectrum.

Key Concepts

Photon Emission

When an electron in an atom transitions from a higher energy state to a lower one, it releases energy in the form of a photon. This phenomenon is known as photon emission. The energy of the emitted photon is equivalent to the difference in energy between the two states, often calculated using the formula \[\Delta E = E_{ ext{lower}} - E_{ ext{higher}}\]In many atomic systems like the hydrogen atom or hydrogen-like ions, the energy levels can be determined precisely using the principal quantum number \(n\). By knowing both the initial and final states, we can calculate the specific energy of the photon that will be emitted during such a transition.

Understanding this process is essential in spectroscopy, which is the study of how atoms absorb and emit light. This concept is widely applied in astronomy where spectroscopists observe emissions from celestial objects to comprehend their composition.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from low-energy radio waves to high-energy gamma rays. Each type of radiation has a characteristic wavelength or frequency which determines its position on the spectrum.

• Radio Waves: Longest wavelength and lowest energy.
• Infrared: Beyond visible red light, heats objects.
• Visible Light: Detected by the human eye, ranging from violet (shortest wavelength) to red (longest wavelength).
• Ultraviolet: Beyond violet, causes sunburns.
• X-rays: Penetrate various materials, used in medical imaging.
• Gamma Rays: Highest energy, emitted from nuclear reactions.

Photon emissions discussed, such as the one with a wavelength of 320 nm, fall into the ultraviolet region of this spectrum. Recognizing where specific wavelengths lie within the spectrum helps scientists identify different physical processes and properties.

Hydrogen-like Atom

A hydrogen-like atom is a simple atomic model used to describe ions or atoms with only one electron orbiting around a nucleus. Despite differences in nuclear charge, which is represented by \( Z \), these atoms exhibit behavior similar to hydrogen.

For a hydrogen-like atom, the energy associated with each electron level is given by \[E_{n} = -\frac{Z^{2}}{n^{2}} imes (2.179 \times 10^{-18} \text{ J})\]where \(n\) is the principal quantum number and \(Z\) is the nuclear charge. This formula reflects that the energy levels depend significantly on both the nuclear charge and the electron's position in different orbitals. For instance, in the He⁺ ion, \(Z\) is 2, doubling the charge compared to hydrogen. This higher nuclear charge results in stronger electronic attractions and alters energy levels compared to hydrogen, explaining why He⁺ still behaves similarly, albeit with a multiplied energy level.