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In what ways should the emission spectra of \(\mathrm{H}\) and \(\mathrm{He}^{+}\) be alike, and in what ways should they be different?

Short Answer

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Question: Compare the emission spectra of hydrogen (H) and helium with a +1 charge (He鈦). Explain any similarities or differences in their patterns and wavelength of the emitted light. Answer: The emission spectra of H and He鈦 are similar in their patterns due to both being one-electron systems. However, they differ in the exact wavelengths of the emitted light because of the difference in their Rydberg constants, with He鈦 having shorter wavelengths (higher energy) compared to hydrogen.

Step by step solution

01

Understand Atomic Structure

Hydrogen has one electron orbiting around the nucleus, while helium has two electrons. He鈦 has one electron removed, which results in one electron orbiting the nucleus - similar to hydrogen. Both H and He鈦 are one-electron systems.
02

Emission Spectra Formation

When an electron in an atom absorbs energy, it jumps from a lower energy level (n1) to a higher energy level (n2). When the electron falls back to its original energy level, it releases energy in the form of light, creating emission spectra. The energy of the emitted light corresponds to the energy difference between the two energy levels.
03

Calculate Wavelength (Energy) of Emitted Light

To calculate the wavelength of the emitted light, we use the Rydberg formula for one-electron systems: $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$ Where \(\lambda\) is the wavelength of emitted light, \(R_H\) is the Rydberg constant for hydrogen \((1.097\times10^7\, \mathrm{m}^{-1})\), and \(n_1, n_2\) are integers corresponding to the initial and final energy levels (with \(n_2 > n_1\)).
04

The Analysis of H and He鈦 Emission Spectra

1. Similarities: Both H and He鈦 are one-electron systems, so their energy levels and the energy differences between these levels follow the same pattern. 2. Differences: The Rydberg constant for He鈦 is not the same as for H. The constant for He鈦 is given by the relation: $$R_{He^+} = R_H\cdot Z^2$$ where \(Z\) is the atomic number of the element (1 for hydrogen and 2 for helium). Since the value differs, while their patterns will be similar, the exact wavelengths of the emission spectra will be different. For He鈦, the lines will appear at shorter wavelengths (higher energy) compared to hydrogen. In summary, the emission spectra of H and He鈦 are alike in their patterns due to their similar one-electron systems. However, they differ in the exact wavelengths of the emitted light because of the difference in their Rydberg constants, with He鈦 having shorter wavelengths (higher energy) compared to hydrogen.

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Key Concepts

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

One-Electron Systems
Both hydrogen (\(\mathrm{H}\)) and helium ion (\(\mathrm{He}^{+}\)) can be understood as one-electron systems. In simple terms, a one-electron system is an atom or ion that has just a single electron revolving around its nucleus, akin to the structure of hydrogen. For hydrogen, there is naturally only one electron. For \(\mathrm{He}^{+}\), it originally has two electrons, but when one electron is removed (making it a helium ion), it behaves like hydrogen with just one electron.
This similarity influences the basic structure of their atomic emission spectra, as both only consider interactions between a single electron and the nucleus in the formulae we use to describe energy transitions. Hence, the way their energy levels and electron behaviors are calculated becomes remarkably similar, even though \(\mathrm{He}^{+}\) has a different nuclear charge due to more protons in its nucleus.
Rydberg Formula
The Rydberg formula is crucial for understanding and analyzing the atomic emission spectra of one-electron systems. This formula helps us to calculate the wavelengths or energy of light emitted when an electron transitions between energy levels in an atom. For hydrogen, the formula is:
  • \[ \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]
Here, \(\lambda\) is the wavelength of emitted light, \(R_H\) is the Rydberg constant specific to hydrogen, and \(n_1\) and \(n_2\) are integers representing the initial and final energy levels (with \(n_2 > n_1\)).
The calculated wavelength provides insights into the energy released during these electron transitions. However, for the helium ion (\(\mathrm{He}^{+}\)), the Rydberg formula slightly differs since its Rydberg constant (\(R_{He^+}\)) accounts for its higher nuclear charge using the relation:
  • \[ R_{He^+} = R_H \cdot Z^2 \]
where \(Z\) is the atomic number (2 for helium). This equation modification results in the helium ion having emission lines at shorter wavelengths compared to hydrogen since the energy levels are more widely spaced.
Energy Levels
Energy levels play a significant role in determining an atom's emission spectrum. An energy level refers to the specific energies an electron can have when it occupies a specific orbit within an atom. These levels are quantized, meaning electrons can't have values of energy between these levels. In one-electron systems, this quantization helps in simplifying the formulation of energy differences during electron transitions.
For both hydrogen and \(\mathrm{He}^{+}\), due to their single electron composition, energy levels are calculated using similar principles. The differences, however, arise from the nuclear charge. Helium鈥檚 nuclear charge being greater means its one-electron ion has stronger interactions with its nucleus, leading to energy levels more tightly bound compared to those in hydrogen.
When electrons transition from a higher to a lower energy level, they emit light, and the energy difference between levels dictates the light's wavelength. Thus, though hydrogen and helium ions follow the same basic transitional pattern, the actual wavelengths differ due to variations in their specific energy level calculations.

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Most popular questions from this chapter

Between 1999 and 2007 the Far Ultraviolet Spectroscopic Explorer satellite analyzed the spectra of emission sources within the Milky Way. Among the satellite's findings were interplanetary clouds containing oxygen atoms that have lost five electrons. a. Write an electron configuration for these highly ionized oxygen atoms. b. Which electrons have been removed from the neutral atoms? c. The ionization energies corresponding to removal of the third, fourth, and fifth electrons are \(4581 \mathrm{kJ} / \mathrm{mol}$$7465 \mathrm{kJ} / \mathrm{mol},\) and \(9391 \mathrm{kJ} / \mathrm{mol},\) respectively. Explain why removal of each additional electron requires more energy than removal of the previous one. d. What is the maximum wavelength of radiation that will remove the fifth electron from an O atom?

In which sub-shell are the highest-energy electrons in a ground-state atom of the isotope \(^{131}\) I? Are the electron configurations of \(^{131} \mathrm{I}\) and \(^{127} \mathrm{I}\) the same?

Two helium ions (He') in the \(n=3\) excited state emit photons of radiation as they return to the ground state. One ion does so in a single transition from \(n=3\) to \(n=1 .\) The other does so in two steps: \(n=3\) to \(n=2\) and then \(n=2\) to \(n=1 .\) Which of the following statements about these two pathways is true? a. The sum of the energies lost in the two-step process is the same as the energy lost in the single transition from \(n=3\) to \(n=1\) b. The sum of the wavelengths of the two photons emitted in the two-step process is equal to the wavelength of the single photon emitted in the transition from \(n=3\) to \(n=1\) c. The sum of the frequencies of the two photons emitted in the two-step process is equal to the frequency of the single photon emitted in the transition from \(n=3\) to \(n=1\) d. The wavelength of the photon emitted by the He \(^{+}\) ion in the \(n=3\) to \(n=1\) transition is shorter than the wavelength of a photon emitted by an \(\mathrm{H}\) atom in an \(n=3\) to \(n=1\) transition.

How rapidly would each of the following particles be moving if they all had the same wavelength as a photon of red light \((\lambda=750 \mathrm{nm}) ?\) a. an electron of mass \(9.10938 \times 10^{-28} \mathrm{g}\) b. a proton of mass \(1.67262 \times 10^{-24} \mathrm{g}\) c. a neutron of mass \(1.67493 \times 10^{-24} \mathrm{g}\) d. an \(\alpha\) particle of mass \(6.64 \times 10^{-24} \mathrm{g}\)

The hydrogen atomic emission spectrum includes a UV line with a wavelength of \(92.3 \mathrm{nm} .\) a. Is this line associated with a transition between different excited states or between an excited state and the ground state? b. What is the value of \(n_{1}\) of this transition? c. What is the energy of the longest wavelength photon that a ground-state hydrogen atom can absorb?

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