For a 3s - orbital, value of `Phi` is given by following realation:
`Psi(3s)=(1)/(9sqrt(3))((1)/(a_(0)))^(3//2)(6-6sigma+sigma^(2))e^(-sigma//2)," where " sigma=(2r.Z)/(3a_(0))`
What is the maximum radial distance of node from nucleus?

Options

A. `((3+sqrt(3))a_(0))/(Z)`
B. `(a_(0))/(Z)`
C. `(3)/(2)((3+sqrt(3))a_(0))/(Z)`
D. `(2a_(0))/(Z)`

Answer

To find the maximum radial distance of the node from the nucleus for a 3s orbital, we start with the given wave function:

\[ \Psi(3s) = \frac{1}{9\sqrt{3}} \left(\frac{1}{a_0}\right)^{3/2} (6 - 6\sigma + \sigma^2) e^{-\sigma/2} \]

where

\[ \sigma = \frac{2Zr}{3a_0} \]

### Step 1: Identify the condition for a node
A node occurs where the wave function \(\Psi(3s)\) is equal to zero. Therefore, we set the polynomial part of the wave function to zero:

\[ 6 - 6\sigma + \sigma^2 = 0 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:

\[ \sigma^2 - 6\sigma + 6 = 0 \]

### Step 3: Apply the quadratic formula
To solve for \(\sigma\), we can use the quadratic formula:

\[ \sigma = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 1\), \(b = -6\), and \(c = 6\):

\[ \sigma = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ = \frac{6 \pm \sqrt{36 - 24}}{2} \]
\[ = \frac{6 \pm \sqrt{12}}{2} \]
\[ = \frac{6 \pm 2\sqrt{3}}{2} \]
\[ = 3 \pm \sqrt{3} \]

### Step 4: Determine the maximum value of \(\sigma\)
The two values of \(\sigma\) are:

\[ \sigma_1 = 3 + \sqrt{3}, \quad \sigma_2 = 3 - \sqrt{3} \]

To find the maximum radial distance, we take the larger value:

\[ \sigma_{\text{max}} = 3 + \sqrt{3} \]

### Step 5: Relate \(\sigma\) to radial distance \(r\)
Recall the relationship:

\[ \sigma = \frac{2Zr}{3a_0} \]

Setting \(\sigma = 3 + \sqrt{3}\):

\[ 3 + \sqrt{3} = \frac{2Zr}{3a_0} \]

### Step 6: Solve for \(r\)
Now we solve for \(r\):

\[ r = \frac{3a_0}{2Z} (3 + \sqrt{3}) \]

### Final Answer
Thus, the maximum radial distance of the node from the nucleus is:

\[ r = \frac{3 + \sqrt{3}}{2} \cdot \frac{3a_0}{Z} \]