The wave function of `3s` electron is given by
`Psi_(3s)=1/(81sqrt(3)prod)(1/a_(0))^(3//2)[27-18(r/a_(0))+2(r/a_(0))^(2)]e^(-r//3a_(0))`
It has a node at `r=r_(0)`, Find out the relation between `r_(0)` and `a_(0)`

Explanation

To find the relation between \( r_0 \) and \( a_0 \) for the given wave function of the \( 3s \) electron, we start by analyzing the wave function:

\[ \Psi_{3s} = \frac{1}{81\sqrt{3}} \left( \frac{1}{a_0} \right)^{3/2} \left[ 27 - 18\left(\frac{r}{a_0}\right) + 2\left(\frac{r}{a_0}\right)^2 \right] e^{-\frac{r}{3a_0}} \]

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

\[ 27 - 18\left(\frac{r}{a_0}\right) + 2\left(\frac{r}{a_0}\right)^2 = 0 \]

### Step 2: Substitute \( x = \frac{r}{a_0} \)
Let \( x = \frac{r}{a_0} \). Then, the equation becomes:

\[ 27 - 18x + 2x^2 = 0 \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:

\[ 2x^2 - 18x + 27 = 0 \]

### Step 4: Apply the quadratic formula
We can solve this quadratic equation using the quadratic formula:

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

Where \( a = 2 \), \( b = -18 \), and \( c = 27 \). Plugging in these values:

\[ x = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 2 \cdot 27}}{2 \cdot 2} \]

### Step 5: Calculate the discriminant
Calculating the discriminant:

\[ (-18)^2 - 4 \cdot 2 \cdot 27 = 324 - 216 = 108 \]

### Step 6: Solve for \( x \)
Now substituting back into the formula:

\[ x = \frac{18 \pm \sqrt{108}}{4} \]

Calculating \( \sqrt{108} = 10.39 \) (approximately):

\[ x = \frac{18 \pm 10.39}{4} \]

This gives us two possible values for \( x \):

1. \( x_1 = \frac{28.39}{4} = 7.0975 \)
2. \( x_2 = \frac{7.61}{4} = 1.9025 \)

### Step 7: Relate back to \( r_0 \)
Recalling that \( x = \frac{r_0}{a_0} \), we can express \( r_0 \) in terms of \( a_0 \):

1. For \( x_1 \):
\[ r_0 = 7.0975 a_0 \]

2. For \( x_2 \):
\[ r_0 = 1.9025 a_0 \]

### Conclusion
Thus, the relation between \( r_0 \) and \( a_0 \) can be expressed as:

\[ r_0 = \left( \frac{18 \pm \sqrt{108}}{4} \right) a_0 \]