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Chapter 8: Problem 32

What is the total number of states with orbital angular momentum \(l=0\) ? (Ignore electron spin.)

Short Answer

Expert verified
There are an infinite number of states with orbital angular momentum \(l=0\), disregarding the electron spin.

Step by step solution

01

Identify the relevant quantum numbers.

Here, n and l are the quantum numbers we will work with. n represents the energy level of the electron, and l represents the orbital angular momentum. n can take any positive integer value (1, 2, 3, ...), while l can take integer values from 0 to n-1.
02

Find the relation between quantum numbers for l=0.

We know that l = 0, so we need to find the restrictions on the principal quantum number n. Since l can be as high as n-1, as long as n ≥ 1, l=0 is a valid solution.
03

Calculate the total number of states with l=0.

Since there are no restrictions on the values of n as long as n is a positive integer, there is no upper limit, and infinite energy levels can be occupied by electrons with l=0. Hence, the total number of states with orbital angular momentum l=0 is infinite. In conclusion, there are an infinite number of states with orbital angular momentum l=0, disregarding the electron spin.

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Most popular questions from this chapter

(a) If one subshell of an atom has nine electrons in it, what is the minimum value of \(l\) ? (b) What is the spectroscopic notation for this atom, if this subshell is part of the \(n=3\) shell?

For \(n=1,\) write all the possible sets of quantum numbers \(\left(n, l, m, m_{s}\right)\).

List all the possible values of \(s\) and \(m_{s}\) for an electron. Are there particles for which these values are different?

What are the possible polar orientations of the spin momentum vector for an electron?

Compare the electron configurations of the elements in the same column of the periodic table.
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