Consider `psi` (wave function) of `2s` atomic orbital of H-atom is-
`psi_(2s)=(1)/(4sqrt(2pia_(0)^(3//2)))[2-(r )/(a_(0))]e^.(r )/(2a_(0)`
Find distance of radial node from nucleous in terms of `a_(0)`

Options

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

Answer

To find the distance of the radial node from the nucleus in terms of \( a_0 \) for the given wave function of the \( 2s \) atomic orbital of the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Wave Function**: The wave function for the \( 2s \) atomic orbital is given as:
\[ \psi_{2s} = \frac{1}{4\sqrt{2\pi a_0^{3/2}}} \left(2 - \frac{r}{a_0}\right) e^{-\frac{r}{2a_0}} \]
Here, \( a_0 \) is the Bohr radius.

2. **Identify the Condition for Radial Nodes**: A radial node occurs where the probability density is zero. The probability density is proportional to the square of the wave function, \( |\psi|^2 \). However, since the normalization constant and the exponential term cannot be zero, we focus on the term:
\[ 2 - \frac{r}{a_0} = 0 \]

3. **Set the Equation for the Node**: To find the radial node, we set the term equal to zero:
\[ 2 - \frac{r}{a_0} = 0 \]

4. **Solve for \( r \)**: Rearranging the equation gives:
\[ \frac{r}{a_0} = 2 \]
Multiplying both sides by \( a_0 \) yields:
\[ r = 2a_0 \]

5. **Conclusion**: The distance of the radial node from the nucleus in terms of \( a_0 \) is:
\[ r = 2a_0 \]

### Final Answer:
The distance of the radial node from the nucleus is \( 2a_0 \).