Select the correct answer and explain your reasoning. An electron having \(n=3\) and \(m_{\ell}=0\) (a) must have \(m_{s}=+\frac{1}{2} ;(\mathbf{b})\) must have \(\ell=1 ;(\mathbf{c})\) may have \(\ell=0,1\) or \(2 ;\) (d) must have \(\ell=2\).

Short Answer

Expert verified
The correct option is (c), because when \(n=3\), the electron can have \(\ell=0,1\) or \(2\).

Step by step solution

01

Analyze Option A

Option (a) refers to the spin quantum number \(m_{s}\). It does not depend on either \(n\) or \(m_{\ell}\), so we can't say it must have \(m_{s}=+\frac{1}{2}\). It can have either \(+\frac{1}{2}\) or \(-\frac{1}{2}\). Therefore, option (a) is incorrect.
02

Analyze Option B

Option (b) states the electron must have \(\ell=1\). However, since \(n=3\), the value of \(\ell\) can range from 0 to 2. Therefore, it doesn't have to be that \(\ell=1\). So, option (b) is also incorrect.
03

Analyze Option C

Option (c) suggests the electron may have \(\ell=0,1\) or \(2\). This is logically correct because for \(n=3\), the possible values of \(\ell\) can indeed range from 0 to 2. Therefore, option (c) is correct. Nevertheless, we should analyze option (d) to complete our examination.
04

Analyze Option D

Option (d) states that the electron must have \(\ell=2\). However, as in option (b), we have already established that the value of \(\ell\) can range from 0 to 2 for \(n=3\). So, option (d) is incorrect.

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