The wave function of 3s and `3p_(z)` orbitals are given by :
`Psi_(3s) = 1/(9sqrt3) ((1)/(4pi))^(1//2) ((Z)/(sigma_(0)))^(3//2)(6-6sigma+sigma)e^(-sigma//2)`
`Psi_(3p_(z))=1/(9sqrt6)((3)/(4pi))^(1//2)((Z)/(sigma_(0)))^(3//2)(4-sigma)sigmae^(-sigma//2)cos0,`
`sigma=(2Zr)/(nalpha_(0))`
where`alpha_(0)=1st` Bohr radius , Z= charge number of nucleus, r= distance from nucleus.
From this we can conclude:

Options

A. Number of nodal surface for `3p_(z)` & `3s` orbitals is equal.
B. The angualr nodal surface of `3p_(z)` orbital has the equation `theta=(pi)/(2)`
C. The radial nodal surfaces of 3 orbital and `3p_(z)` orbital are at equal distance from the nucleus.
D. 3s electron have greater penetrating power into the nucleus in comparison to 3p electrons.

Explanation

To solve the problem regarding the wave functions of the 3s and 3p_z orbitals, we will analyze the given wave functions to determine the number of radial and angular nodes for each orbital.

### Step-by-Step Solution:

1. **Understanding the Wave Functions**:
The wave functions for the 3s and 3p_z orbitals are given as:
\[ \Psi_{3s} = \frac{1}{9\sqrt{3}} \left(\frac{1}{4\pi}\right)^{1/2} \left(\frac{Z}{\sigma_0}\right)^{3/2} (6 - 6\sigma + \sigma)e^{-\sigma/2} \]
\[ \Psi_{3p_z} = \frac{1}{9\sqrt{6}} \left(\frac{3}{4\pi}\right)^{1/2} \left(\frac{Z}{\sigma_0}\right)^{3/2} (4 - \sigma)\sigma e^{-\sigma/2} \cos \theta \]
where \(\sigma = \frac{2Zr}{n\alpha_0}\).

2. **Finding Radial Nodes for 3s Orbital**:
- Set the wave function \(\Psi_{3s} = 0\):
\[ 6 - 6\sigma + \sigma = 0 \]
- This simplifies to:
\[ \sigma^2 - 6\sigma + 6 = 0 \]
- Using the quadratic formula:
\[ \sigma = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{6 \pm \sqrt{12}}{2} = 3 \pm \sqrt{3} \]
- Thus, the radial nodes for the 3s orbital are at:
\[ \sigma = 3 + \sqrt{3} \quad \text{and} \quad \sigma = 3 - \sqrt{3} \]

3. **Finding Radial Nodes for 3p_z Orbital**:
- Set the wave function \(\Psi_{3p_z} = 0\):
\[ (4 - \sigma)\sigma = 0 \]
- This gives:
\[ \sigma = 0 \quad \text{or} \quad \sigma = 4 \]
- Since \(\sigma = 0\) corresponds to \(r = 0\), we discard it. Thus, the radial node for the 3p_z orbital is at:
\[ \sigma = 4 \]

4. **Finding Angular Nodes for 3p_z Orbital**:
- The angular part of the wave function involves \(\cos \theta\):
\[ \cos \theta = 0 \implies \theta = \frac{\pi}{2} \]
- This indicates one angular node.

5. **Summary of Nodes**:
- For the **3s orbital**:
- Radial nodes: 2 (from the two solutions for \(\sigma\)).
- Angular nodes: 0.
- For the **3p_z orbital**:
- Radial nodes: 1 (from \(\sigma = 4\)).
- Angular nodes: 1 (from \(\cos \theta = 0\)).

### Conclusion:
- The total number of nodal surfaces for the 3s orbital is 2 (both radial), and for the 3p_z orbital, it is 2 (1 radial and 1 angular).