Find the value of x+2y+z
x=no. of radial nodes in `3p_x` , y=no. of angular nodes in 6s
z=the maximum no. of electrons in `._24Cr` with n=3 and s=`+1/2` and orbital angular momentum `sqrt6ħ`

Answer

To solve the problem, we need to find the values of \( x \), \( y \), and \( z \) based on the definitions provided in the question.

### Step 1: Calculate \( x \) (Number of Radial Nodes in \( 3p_x \))
- The formula for calculating the number of radial nodes is given by:
\[ x = n - l - 1 \]
where \( n \) is the principal quantum number and \( l \) is the azimuthal quantum number.
- For \( 3p_x \):
- \( n = 3 \) (since it's in the 3rd shell)
- \( l = 1 \) (for p orbitals)
- Plugging in the values:
\[ x = 3 - 1 - 1 = 1 \]

### Step 2: Calculate \( y \) (Number of Angular Nodes in \( 6s \))
- The formula for calculating the number of angular nodes is given by:
\[ y = l \]
- For \( 6s \):
- \( l = 0 \) (for s orbitals)
- Therefore:
\[ y = 0 \]

### Step 3: Calculate \( z \) (Maximum Number of Electrons in \( _{24}Cr \) with \( n=3 \) and \( s=\frac{1}{2} \))
- The electronic configuration of chromium (atomic number 24) is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5 \, 4s^1 \]
- We need to consider only the orbitals with \( n=3 \) and spin \( +\frac{1}{2} \):
- \( 3s \): 2 electrons (1 with \( +\frac{1}{2} \))
- \( 3p \): 6 electrons (3 with \( +\frac{1}{2} \))
- \( 3d \): 5 electrons (5 with \( +\frac{1}{2} \) since they can be filled up to \( +\frac{1}{2} \) first)
- Thus, the total number of electrons with \( n=3 \) and spin \( +\frac{1}{2} \):
\[ z = 1 + 3 + 5 = 9 \]

### Step 4: Calculate \( x + 2y + z \)
- Now we can substitute the values we found into the expression:
\[ x + 2y + z = 1 + 2(0) + 9 = 1 + 0 + 9 = 10 \]

### Final Answer
\[ \text{The value of } x + 2y + z = 10 \]

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