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Chapter 36: Problem 9

Why is there no spin-orbit splitting in hydrogen's ground state?

Short Answer

Expert verified
Hydrogen's ground state has an azimuthal quantum number (\(l\)) of zero, meaning it has no orbital angular momentum. Without orbital angular momentum, no spin-orbit interaction occurs, hence why there is no spin-orbit splitting in the ground state of hydrogen.

Step by step solution

01

Understanding Spin-Orbit Interaction

Spin-Orbit Interaction is a term used in quantum physics to describe the interaction of the spin of a particle with its orbital motion. In the case of an atom, the electrons follow an orbital path around the nucleus and possess spin. When these two are coupled, they lead to a slight energy variation in atomic levels, known as spin-orbit splitting.
02

Identify the Ground State of Hydrogen

The ground state of hydrogen is defined as the lowest energy state of an atom, which is described as 1s. In this state, the electron is located in the closest orbital to the nucleus. The quantum number that characterizes the orbital angular momentum, also known as the angular momentum quantum number or azimuthal quantum number (\(l\)), in this state is zero.
03

Explaining the Absence of Spin-Orbit Splitting

Considering that for the ground state shell of hydrogen, the azimuthal quantum number (\(l\)) is zero, the absence of spin-orbit splitting can now be explained. Simply, because for spin-orbit interaction to occur, there must exist a non-zero orbital angular momentum. However, in hydrogen's ground state where \(l = 0\), there is no orbital angular momentum. Therefore, there is no spin-orbit interaction nor splitting.

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Most popular questions from this chapter

Elements \(A\) and \(B\) have atomic numbers \(Z_{A}\) and \(Z_{B}=2 Z_{A} .\) How do you expect element \(B^{\prime}\) s \(K \alpha\) X-ray energy to compare with that of element \(A\) ? a. \(B^{\prime}\) s \(K \alpha\) energy should be about one-fourth that of \(A\) b. \(B^{\prime}\) s \(K \alpha\) energy should be about half that of \(A\). c. \(B^{\prime}\) s \(K \alpha\) energy should be about twice that of \(A\) d. \(B^{\prime}\) s \(K \alpha\) energy should be about four times that of \(A\)

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Give a symbolic description for the state of the electron in a hydrogen atom with total energy -1.51 eV and orbital angular morentum \(\sqrt{6} \hbar.\)
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