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Same wavelength is observed with respect to \(\mathrm{Li}^{2+}\) from a transition of \(\mathrm{n}_{2}=6\) to \(\mathrm{n}_{1}=3\) a. \(4 \rightarrow 2\) of \(\mathrm{Be}^{+3}\) b. \(3 \rightarrow 1\) of \(\mathrm{H}\)-atom c. \(4 \rightarrow 2\) of \(\mathrm{He}^{+}\) d. \(4 \rightarrow 2\) of He-atom

Short Answer

Expert verified
None of the options match the wavelength of Li虏鈦 transition.

Step by step solution

01

Formula for Wavelength

The wavelength observed for a transition in a hydrogen-like atom is given by the Rydberg formula: \[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \( \lambda \) is the wavelength, \( R \) is the Rydberg constant, \( Z \) is the atomic number, and \( n_1, n_2 \) are the initial and final energy levels respectively.
02

Calculate for Li虏鈦 Transition

For \( \mathrm{Li}^{2+} \), the atomic number \( Z = 3 \). Substitute \( Z = 3 \), \( n_1 = 3 \), and \( n_2 = 6 \):\[ \frac{1}{\lambda} = R (3)^2 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = R \cdot 9 \left( \frac{1}{9} - \frac{1}{36} \right) \]
03

Simplify the Li虏鈦 Transition Formula

Calculate the values within the parentheses:\[ \frac{1}{9} - \frac{1}{36} = \frac{4}{36} = \frac{1}{9} \]Thus,\[ \frac{1}{\lambda} = R \cdot 1 \Rightarrow \lambda = \frac{1}{R} \]
04

Compare with Other Transitions

We need to find transitions that give \( \lambda = \frac{1}{R} \).
05

Evaluate Option a: Be鈦郝 Transition

For \( \mathrm{Be}^{+3} \), \( Z = 4 \), \( n_2 = 4 \), and \( n_1 = 2 \):\[ \frac{1}{\lambda} = R (4)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 16R \left( \frac{1}{4} - \frac{1}{16} \right) \]Simplify:\[ \frac{1}{4} - \frac{1}{16} = \frac{3}{16} \]\[ \frac{1}{\lambda} = 16R \times \frac{3}{16} = 3R \]\( \lambda = \frac{1}{3R} \).
06

Evaluate Option b: H Transition

For \( \mathrm{H} \), \( Z = 1 \), \( n_2 = 3 \), and \( n_1 = 1 \):\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) \]Simplify:\[ 1 - \frac{1}{9} = \frac{8}{9} \]\[ \frac{1}{\lambda} = \frac{8R}{9} \]\( \lambda = \frac{9}{8R} \).
07

Evaluate Option c: He鈦 Transition

For \( \mathrm{He}^{+} \), \( Z = 2 \), \( n_2 = 4 \), and \( n_1 = 2 \):\[ \frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) \]Simplify:\[ \frac{1}{4} - \frac{1}{16} = \frac{3}{16} \]\[ \frac{1}{\lambda} = 4R \times \frac{3}{16} = \frac{3R}{4} \]\( \lambda = \frac{4}{3R} \).
08

Evaluate Option d: He Transition

For helium (He), \( Z = 2 \), but it is not a hydrogen-like ion unless it's ionized. Helium atom in a \( 4 \rightarrow 2 \) transition doesn't follow the hydrogen-like Rydberg formula directly as it has two electrons.
09

Conclusion: Comparing Wavelengths

The transition that matches the wavelength for \( \mathrm{Li}^{2+} \) (where \( \lambda = \frac{1}{R} \)) is none of the provided options, based on their calculated wavelengths.

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Key Concepts

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

Hydrogen-Like Atoms
Hydrogen-like atoms are a special class of ions that have only one electron, similar to the hydrogen atom. Despite often being associated solely with atomic hydrogen, the term "hydrogen-like" applies to any ion with a single electron orbiting a nucleus with more than one proton. These ions include species like - \( \mathrm{Li}^{2+} \)- \( \mathrm{He}^+ \), and even highly ionized heavier elements. They are called "hydrogen-like" because their energy levels and the spectral lines they produce can be described using the same formulae developed for hydrogen. The key feature of hydrogen-like atoms is that their electron energy levels depend specifically on the atomic number \( Z \), giving rise to unique spectral lines that can be described using the Rydberg formula. This formula demonstrates how the interaction between the single electron and the heightened positive charge of the nucleus (compared to hydrogen's single-proton nucleus) affects the atom's energy levels.
Transition Wavelengths
Transition wavelengths are the specific distances between the peaks of the visible light waves emitted or absorbed when an electron changes its energy state within an atom. These transitions occur when an electron moves from one electron shell, with a specific quantum number \( n \), to another. - In hydrogen-like atoms, the wavelengths of these transitions can be predicted using the Rydberg formula.- The difference in energy between initial (\( n_2 \)) and final (\( n_1 \)) states determines the photon's wavelength emitted.Using the Rydberg formula:\[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]The Rydberg constant \( R \) quantifies the scale of the wavelengths of electromagnetic radiation emitted by electrons transitioning between levels. Because \( Z \), the atomic number, appears in the formula, it shows how the nucleus's charge strength significantly influences these wavelengths. For a higher \( Z \), the nucleus holds the electron more tightly, leading to shorter wavelengths.
Atomic Number
The atomic number \( Z \) is a fundamental property of an element that represents the number of protons in an atom's nucleus. It defines the element's identity and position on the periodic table. - In the context of hydrogen-like atoms, the atomic number plays an essential role in the Rydberg formula for calculating transition wavelengths.- The value \( Z^2 \) in the Rydberg formula influences the emitted photon's wavelength; a larger \( Z \) results in a shorter wavelength due to increased electrostatic attraction between the electron and the nucleus.This is why different chemical elements or ions with higher atomic numbers exhibit distinct spectral lines, even if they are "hydrogen-like," vastly different from those seen in atomic hydrogen, which has \( Z = 1 \). Understanding the atomic number's influence is key to exploring atomic behavior and characterizing substances based on their spectral signatures.

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Most popular questions from this chapter

Which of the following statement(s) is/are correct? a. Planck's study of black-body radiation led to the hypothesis that radiation is emitted in quantas of energy b. Photons of lower frequency radiation have lower energies. c. Bohr's model of the hydrogen atom showed emission of energy by atoms occurs when an electron moves to a higher orbit. d. The lowest energy state of an electron in an atom is known as the ground state

Transition of an electron from \(\mathrm{n}=3\) level to \(\mathrm{n}=1\) level results in: a. Band spectrum b. IR spectrum c. Emission spectrum d. X-ray spectrum

Which of the following match is/are correct? a. \(4 \mathrm{f}\) orbital \(\quad\) (maximum 2 electrons) b. 4 s orbital \(\quad(2\) radial nodes) c. 3 dxy orbital \(\quad(2\) nodal planes) d. \(5 \mathrm{px}\) orbital ( 2 radial nodes)

The Balmer-Rydberg equation can be extended to ions with only one electron, such as \(\mathrm{He}^{+}\). In that case it has the form: \(1 / \lambda=\mathrm{Z}^{2} \mathrm{R}\left(1 / \mathrm{m}^{2}-1 / \mathrm{n}^{2}\right)\), here \(\mathrm{Z}\) is the atomic number. What is the energy of the photon required to promote an electron in He' from a 1 s-orbital to a \(2 \mathrm{p}\)-orbital? a. 12 hcR b. 6 hcR c. 3 hcR d. \((3 / 4)\) hcR

List all the elements that have a ground state configuration with five unpaired electrons in the \(3 \mathrm{~d}\) subshell. a. Mn b. \(\mathrm{Cr}\) c. \(\mathrm{Cr}\) and \(\mathrm{Mn}\) d. \(\mathrm{Mn}, \mathrm{Fe}, \mathrm{Co}\),

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