The wave function orbital of H-like atoms is given as onder
`psi_(2s) = (1)/(4sqrt(2pi)) Z^(3//2) (2 - Zr)^(Zr//2)`
Given that the radius is in `Å` then which of the following is the radius for nodal surface for `He^(Theta)` ion ?

Answer

To find the radius for the nodal surface for the He\(^+\) ion using the given wave function orbital, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Wave Function**: The wave function for the 2s orbital of H-like atoms is given by:
\[ \psi_{2s} = \frac{1}{4\sqrt{2\pi}} Z^{3/2} (2 - Zr)^{Zr/2} \]
where \( Z \) is the atomic number and \( r \) is the radius in Ångströms.

2. **Identify the Condition for Nodal Surfaces**: For nodal surfaces, the wave function must equal zero. Therefore, we set:
\[ \psi_{2s} = 0 \]

3. **Set the Wave Function to Zero**: From the wave function, we have:
\[ 2 - Zr = 0 \]
This implies:
\[ 2 = Zr \]

4. **Substitute the Atomic Number for Helium**: The atomic number \( Z \) for helium (He) is 2. Substituting this value into the equation gives:
\[ 2 = 2r \]

5. **Solve for \( r \)**: Rearranging the equation to find \( r \):
\[ r = \frac{2}{2} = 1 \text{ Å} \]

6. **Conclusion**: The radius for the nodal surface for the He\(^+\) ion is:
\[ r = 1 \text{ Å} \]

### Final Answer:
The radius for the nodal surface for the He\(^+\) ion is **1 Å**.