The no. of electrons satisfying n+l=7 for Lu (atomic number:71)

Answer

To solve the problem of finding the number of electrons satisfying the condition \( n + l = 7 \) for Lutetium (Lu, atomic number 71), we can follow these steps:

### Step 1: Determine the Electronic Configuration of Lutetium
The electronic configuration of Lutetium (Lu) is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 4d^{10} \, 5s^2 \, 5p^6 \, 4f^{14} \, 5d^1 \, 6s^2 \]

### Step 2: Identify the Values of \( n \) and \( l \)
In the electronic configuration, we need to calculate \( n + l \) for each subshell:
- For \( 1s \): \( n = 1, l = 0 \) → \( n + l = 1 + 0 = 1 \)
- For \( 2s \): \( n = 2, l = 0 \) → \( n + l = 2 + 0 = 2 \)
- For \( 2p \): \( n = 2, l = 1 \) → \( n + l = 2 + 1 = 3 \)
- For \( 3s \): \( n = 3, l = 0 \) → \( n + l = 3 + 0 = 3 \)
- For \( 3p \): \( n = 3, l = 1 \) → \( n + l = 3 + 1 = 4 \)
- For \( 3d \): \( n = 3, l = 2 \) → \( n + l = 3 + 2 = 5 \)
- For \( 4s \): \( n = 4, l = 0 \) → \( n + l = 4 + 0 = 4 \)
- For \( 4p \): \( n = 4, l = 1 \) → \( n + l = 4 + 1 = 5 \)
- For \( 4d \): \( n = 4, l = 2 \) → \( n + l = 4 + 2 = 6 \)
- For \( 5s \): \( n = 5, l = 0 \) → \( n + l = 5 + 0 = 5 \)
- For \( 5p \): \( n = 5, l = 1 \) → \( n + l = 5 + 1 = 6 \)
- For \( 4f \): \( n = 4, l = 3 \) → \( n + l = 4 + 3 = 7 \)
- For \( 5d \): \( n = 5, l = 2 \) → \( n + l = 5 + 2 = 7 \)
- For \( 6s \): \( n = 6, l = 0 \) → \( n + l = 6 + 0 = 6 \)

### Step 3: Identify the Subshells that Satisfy \( n + l = 7 \)
From the calculations, we find that the subshells that satisfy \( n + l = 7 \) are:
- \( 4f \) with \( n + l = 7 \) and contains \( 14 \) electrons.
- \( 5d \) with \( n + l = 7 \) and contains \( 1 \) electron.

### Step 4: Calculate the Total Number of Electrons
Now, we sum the electrons in the subshells that satisfy the condition:
- Electrons in \( 4f \): \( 14 \)
- Electrons in \( 5d \): \( 1 \)

Total number of electrons satisfying \( n + l = 7 \):
\[ 14 + 1 = 15 \]

### Final Answer
The number of electrons satisfying \( n + l = 7 \) for Lutetium (Lu) is **15**.

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