How many electrons in `._(19)K` have `n = 3, l = 0` ?

Options

A. 1
B. 2
C. 4
D. 3

Explanation

To determine how many electrons in potassium-19 (represented as \( _{19}K \)) have quantum numbers \( n = 3 \) and \( l = 0 \), we will follow these steps:

### Step 1: Identify the Atomic Number
Potassium (K) has an atomic number of 19, which means it has 19 electrons.

### Step 2: Write the Electronic Configuration
We need to write the electronic configuration for potassium. The electrons fill the orbitals in the following order based on the Aufbau principle:

1. \( 1s^2 \) - 2 electrons
2. \( 2s^2 \) - 2 electrons
3. \( 2p^6 \) - 6 electrons
4. \( 3s^2 \) - 2 electrons
5. \( 3p^6 \) - 6 electrons
6. \( 4s^1 \) - 1 electron

So, the complete electronic configuration of potassium is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \]

### Step 3: Identify the Relevant Quantum Numbers
The principal quantum number \( n = 3 \) indicates that we are looking at the third energy level. The azimuthal quantum number \( l = 0 \) corresponds to the s-orbital (since \( l = 0 \) represents the s subshell).

### Step 4: Count the Electrons in the 3s Orbital
From the electronic configuration, we can see that the 3s subshell has:
\[ 3s^2 \]
This means there are 2 electrons in the 3s orbital.

### Final Answer
Thus, the number of electrons in \( _{19}K \) that have \( n = 3 \) and \( l = 0 \) is **2**.