A hydrogen atom orbital has \(n=4\) and \(m_{l}=-2\). (a) What are the possible values of \(l\) for this orbital? (b) What are the possible values of \(m_{s}\) for the orbital?

Short Answer

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(a) The possible values of \(l\) for this orbital are 2 and 3. (b) The possible values of \(m_s\) for the orbital are always +1/2 and -1/2.

Step by step solution

01

(a) Possible values of l

To find the possible values of the angular momentum quantum number l, we need to recall the relationship between n, l and m_l: \(l\) can have integer values from 0 to n-1, and \(m_{l}\) can have integer values ranging from -l to +l. In our case, n = 4, and \(m_{l} = -2\). Since m_l = -2, we know that l must be at least 2, since m_l can only take values from -l to +l. Therefore, we will try each possible value of l = 2, 3, and check if \(m_l = -2\) is a valid option. For l = 2, \(m_l\) can take values -2, -1, 0, 1, 2, which includes \(m_l = -2\). For l = 3, \(m_l\) can take values -3, -2, -1, 0, 1, 2, 3, which also includes \(m_l = -2\). So, the possible values of l for this orbital are 2 and 3.
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(b) Possible values of m_s

To find the possible values of the spin quantum number \(m_s\), we recall that for an electron in an orbital, there are only two possible spin orientations available: 1. Spin-up (represented as +1/2) 2. Spin-down (represented as -1/2) So, for any electron in an orbital, the possible values of \(m_s\) are always +1/2 and -1/2.

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