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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?

Short Answer

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

Step by step solution

01

Understand the Ion

The ion mentioned is He鈦, 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.
02

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.
03

Calculate Energy at n=5

Insert the values for n = 5 and Z = 2 into the energy equation: \[E_{5} = -\frac{2^{2}}{5^{2}} \left(2.179 \times 10^{-18} \text{ J}\right) = -\frac{4}{25} \times 2.179 \times 10^{-18} \text{ J}\] \[E_{5} = -3.4864 \times 10^{-19} \text{ J}\]
04

Calculate Energy at n=3

Insert the values for n = 3 and Z = 2 into the energy equation: \[E_{3} = -\frac{2^{2}}{3^{2}} \left(2.179 \times 10^{-18} \text{ J}\right) = -\frac{4}{9} \times 2.179 \times 10^{-18} \text{ J}\] \[E_{3} = -9.6844 \times 10^{-19} \text{ J}\]
05

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.
06

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: \[3.20 \times 10^{-7} \text{ m} = 320 \text{ nm}\].
07

Determine Part of Electromagnetic Spectrum

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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聽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鈦 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鈦 still behaves similarly, albeit with a multiplied energy level.

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Most popular questions from this chapter

Give the \(n, \ell,\) and \(m_{\ell}\) values for (a) Each atomic orbital in the \(6 f\) sublevel. (b) Each atomic orbital in the \(n=5\) level.

Microwave ovens, commonly used to heat water in bevcrages and foods, cmit radiation with a wavelength of \(12.2 \mathrm{~cm}\) (a) Calculate the amount (moles) of photons of this microwave radiation required to raise the temperature of \(230.0 \mathrm{~g}\) water (such as in a cup of coffee, which is mainly water) from \(24.0^{\circ} \mathrm{C}\) to \(55.0^{\circ} \mathrm{C}\) (b) As noted in Chapter \(4,\) the watt. \(W\), is a unit of power: \(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\). If the microwave oven is rated at \(800 \mathrm{~W}\) calculate the time needed to heat the water in part (a). Assume that all the energy is delivered to the water.

(a) What is the electron configuration for an atom of tin? (b) What are the electron configurations for \(\mathrm{Sn}^{2+}\) and \(\mathrm{Sn}^{4+}\) ions?

Write the electron configurations of chromium: \(\mathrm{Cr}\), \(\mathrm{Cr}^{2+}\), and \(\mathrm{Cr}^{3+}\). Use atomic orbital box diagrams to determine the number of unpaired electrons for each species.

According to a relationship developed by Niels Bohr, for an atom or ion that has a single electron, the total energy of an electron in a stable orbit of quantum number \(n\) is \(E_{n}=-\left[Z^{2} / n^{2}\right]\left(2.179 \times 10^{-18} \mathrm{~J}\right)\) where \(Z\) is the atomic number. Calculate the ionization energy for the electron in a ground-state hydrogen atom.

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