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Chapter 36: Problem 44

Repeat Exercise 25 for the case where you know only that the principal quantum number is \(3 ;\) that is, \(l\) might have any of its possible values.

Short Answer

Expert verified
The possible sets of quantum numbers for \(n=3\) can be found by combining all possible values for the azimuthal quantum number \(l\), the magnetic quantum number \(m_l\), and the spin quantum number \(s\).

Step by step solution

01

Determine possible values for the azimuthal quantum number \(l\)

The azimuthal quantum number \(l\) can have any positive integer value from \(0\) to \(n-1\). So for \(n=3,\) the possible values for \(l\) could be \(0, 1\) and \(2\).
02

Determine possible values for the magnetic quantum number \(m_l\)

For each value of \(l\), the magnetic quantum number \(m_l\) can take on values ranging from \(-l\) to \(+l\) including \(0\). - For \(l=0, m_l\) can only be \(0\), - For \(l=1, m_l\) can be \(-1, 0, 1\), - For \(l=2, m_l\) can be \(-2, -1, 0, 1, 2\).
03

Determine possible values for the spin quantum number \(s\)

The spin quantum number (\(s\)) can be \(-\frac{1}{2}\) or \(\frac{1}{2}\). So for each combination of \(n\), \(l\), and \(m_l\), there are two possible values for \(s\).
04

List all the possible sets of quantum numbers

Using all possible combinations of \(n\), \(l\), \(m_l\) and \(s\) creates a list of possible sets of quantum numbers for \(n=3\).

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Most popular questions from this chapter

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

A hydrogen atom is in the \(2 s\) state. Find the probability that its electron will be found (a) beyond one Bohr radius and (b) beyond 10 Bohr radii.

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What's the orbital quantum number for an electron whose orbital angular momentum has magnitude \(L=\sqrt{30} \hbar ?\)

The electron in a hydrogen atom is somewhat like a particle confined to a three-dimensional box. In the atom, what plays the role of the confining box?
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