To solve the problem of finding the maximum number of electrons in aluminum (Al) for which \((l \cdot m)/n = 0\), we will follow these steps:
### Step 1: Determine the Atomic Number and Electronic Configuration
The atomic number of aluminum (Al) is 13. This means that aluminum has 13 electrons. The electronic configuration of aluminum can be determined using the Aufbau principle, which states that electrons fill atomic orbitals in order of increasing energy levels.
**Electronic Configuration of Al:**
- 1s²
- 2s²
- 2p⁶
- 3s²
- 3p¹
So, the complete electronic configuration is: **1s² 2s² 2p⁶ 3s² 3p¹**.
### Step 2: Identify Quantum Numbers
For each electron, we need to consider the quantum numbers:
- \(n\) (principal quantum number)
- \(l\) (azimuthal quantum number)
- \(m\) (magnetic quantum number)
### Step 3: Analyze Each Subshell
1. **1s²:**
- \(n = 1\), \(l = 0\) (s orbital)
- For both electrons, \((l \cdot m)/n = (0 \cdot m)/1 = 0\) (since \(l = 0\)).
- Contribution: 2 electrons.
2. **2s²:**
- \(n = 2\), \(l = 0\) (s orbital)
- For both electrons, \((l \cdot m)/n = (0 \cdot m)/2 = 0\).
- Contribution: 2 electrons.
3. **2p⁶:**
- \(n = 2\), \(l = 1\) (p orbital)
- Possible values of \(m\) are -1, 0, +1.
- For 6 electrons, \((l \cdot m)/n\) will not be zero for all electrons. However, when \(m = 0\), \((1 \cdot 0)/2 = 0\) for 2 electrons (one for each pair of degenerate orbitals).
- Contribution: 2 electrons (when \(m = 0\)).
4. **3s²:**
- \(n = 3\), \(l = 0\) (s orbital)
- For both electrons, \((l \cdot m)/n = (0 \cdot m)/3 = 0\).
- Contribution: 2 electrons.
5. **3p¹:**
- \(n = 3\), \(l = 1\) (p orbital)
- Possible values of \(m\) are -1, 0, +1.
- For 1 electron in 3p, it can occupy \(m = 0\), thus \((1 \cdot 0)/3 = 0\).
- Contribution: 1 electron.
### Step 4: Calculate Total Electrons
Now, we sum the contributions from each subshell:
- From 1s: 2 electrons
- From 2s: 2 electrons
- From 2p: 2 electrons (only when \(m = 0\))
- From 3s: 2 electrons
- From 3p: 1 electron (only when \(m = 0\))
Total = \(2 + 2 + 2 + 2 + 1 = 9\) electrons.
### Final Answer
The maximum number of electrons in aluminum for which \((l \cdot m)/n = 0\) is **9 electrons**.
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