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

Explanation

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

### Step 1: Understand the condition for radial nodes
Radial nodes occur where the probability of finding an electron is zero. This means that the wave function, \(\Phi(3s)\), must equal zero.

### Step 2: Set the wave function equal to zero
The given wave function for the 3s orbital is:
\[ \Phi(3s) = \frac{1}{a\sqrt{3}} \left(\frac{1}{a_0}\right)^{3/2} (6 - 6\sigma + \sigma^2) e^{-\sigma/2} \]
To find the radial nodes, we set the part that depends on \(\sigma\) equal to zero:
\[ 6 - 6\sigma + \sigma^2 = 0 \]

### Step 3: Solve the quadratic equation
This is a quadratic equation in \(\sigma\). We can rearrange it as:
\[ \sigma^2 - 6\sigma + 6 = 0 \]
Using the quadratic formula, \(\sigma = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = 6\):
\[ \sigma = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ \sigma = \frac{6 \pm \sqrt{36 - 24}}{2} \]
\[ \sigma = \frac{6 \pm \sqrt{12}}{2} \]
\[ \sigma = \frac{6 \pm 2\sqrt{3}}{2} \]
\[ \sigma = 3 \pm \sqrt{3} \]

### Step 4: Choose the positive root
For the maximum radial distance, we take the positive root:
\[ \sigma = 3 + \sqrt{3} \]

### Step 5: Substitute \(\sigma\) into the equation for \(r\)
We know that:
\[ \sigma = \frac{2rZ}{3a_\sigma} \]
Rearranging this gives:
\[ r = \frac{3a_\sigma \sigma}{2Z} \]
Substituting \(\sigma = 3 + \sqrt{3}\):
\[ r = \frac{3a_\sigma (3 + \sqrt{3})}{2Z} \]

### Step 6: Final expression for maximum radial distance
Thus, the maximum radial distance of the node from the nucleus is:
\[ r = \frac{3a_\sigma (3 + \sqrt{3})}{2Z} \]

### Summary
The maximum radial distance of the node from the nucleus for the 3s orbital is given by:
\[ r = \frac{3a_\sigma (3 + \sqrt{3})}{2Z} \]

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