For a 3s-orbital
`Phi(3s)=(1)/(asqrt(3))((1)/(a_(0)))^(3//2)(6-6sigma+sigma^(2))in^(-sigma//2)`
where `sigma=(2rZ)/(3a_(sigma))`
What is the maximum radial distance of node from nucleus?

Options

A. `((3+sqrt(3))a_(sigma))/(Z)`
B. `(a_(sigma))/(Z)`
C. `(3)/(2)((3+sqrt(3))a_(sigma))/(Z)`
D. `(3a_(sigma))/(Z)`

Explanation

To find the maximum radial distance of the node from the nucleus for a 3s orbital, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Wave Function**: The wave function for a 3s orbital is given by:
\[ \Phi(3s) = \frac{1}{a\sqrt{3}} \left(\frac{1}{a_0}\right)^{3/2} (6 - 6\sigma + \sigma^2) e^{-\sigma/2} \]
where \(\sigma = \frac{2rZ}{3a_\sigma}\).

2. **Set the Wave Function to Zero**: To find the nodes, we need to set the wave function equal to zero:
\[ 6 - 6\sigma + \sigma^2 = 0 \]

3. **Rearranging the Equation**: Rearranging the quadratic equation gives us:
\[ \sigma^2 - 6\sigma + 6 = 0 \]

4. **Using the Quadratic Formula**: We can solve for \(\sigma\) using the quadratic formula:
\[ \sigma = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -6\), and \(c = 6\):
\[ \sigma = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ \sigma = \frac{6 \pm \sqrt{36 - 24}}{2} \]
\[ \sigma = \frac{6 \pm \sqrt{12}}{2} \]
\[ \sigma = \frac{6 \pm 2\sqrt{3}}{2} \]
\[ \sigma = 3 \pm \sqrt{3} \]

5. **Choosing the Positive Root**: We take the positive root for \(\sigma\):
\[ \sigma = 3 + \sqrt{3} \]

6. **Relate \(\sigma\) to Radial Distance \(r\)**: Recall that:
\[ \sigma = \frac{2rZ}{3a_\sigma} \]
Rearranging gives:
\[ r = \frac{3a_\sigma \sigma}{2Z} \]

7. **Substituting \(\sigma\)**: Substitute \(\sigma = 3 + \sqrt{3}\) into the equation for \(r\):
\[ r = \frac{3a_\sigma (3 + \sqrt{3})}{2Z} \]

### Final Answer:
The maximum radial distance of the node from the nucleus is:
\[ r = \frac{3a_\sigma (3 + \sqrt{3})}{2Z} \]