The Wave function (Psi) of 2s is given by:
`Psi_(2s)=(1)/(2sqrt2pi)(1/a_(0))^(1//2){2-(r)/a^(0)}e^(-r//2a_(0))`
At `r =r_(0)`, radial node is formed . Thus for 2s `,r_(0),` in terms of `a_(0)` is-

Options

A. `r_(0)=a_(0)`
B. `r_(0)=2a_(0)`
C. `r_(0)=a_(0)//2`
D. `r_(0)=4a_(0)`

Explanation

To find the value of \( r_0 \) in terms of \( a_0 \) for the 2s wave function, we start with the given wave function:

\[ \Psi_{2s} = \frac{1}{2\sqrt{2\pi}} \left( \frac{1}{a_0} \right)^{1/2} \left( 2 - \frac{r}{a_0} \right) e^{-\frac{r}{2a_0}} \]

### Step 1: Set the wave function equal to zero

To find the radial node, we need to set the wave function \( \Psi_{2s} \) equal to zero:

\[ \Psi_{2s} = 0 \]

This leads us to the equation:

\[ \frac{1}{2\sqrt{2\pi}} \left( \frac{1}{a_0} \right)^{1/2} \left( 2 - \frac{r}{a_0} \right) e^{-\frac{r}{2a_0}} = 0 \]

### Step 2: Analyze the equation

The exponential term \( e^{-\frac{r}{2a_0}} \) is never zero for any finite value of \( r \). Therefore, we can focus on the term \( \left( 2 - \frac{r}{a_0} \right) \):

\[ 2 - \frac{r}{a_0} = 0 \]

### Step 3: Solve for \( r \)

Rearranging the equation gives:

\[ \frac{r}{a_0} = 2 \]

Multiplying both sides by \( a_0 \) results in:

\[ r = 2a_0 \]

### Conclusion

Thus, the radial node \( r_0 \) in terms of \( a_0 \) is:

\[ r_0 = 2a_0 \]