The Schrodinger wave equation for hydrogen atom is `Psi_(2s) = (1)/(4sqrt(2pi)) ((1)/(a_(0)))^(3//2) (2 - (r)/(a_(0))) e^(-r//a_(0))` , where `a_(0)` is Bohr's radius . If the radial node in 2s be at `r_(0)` , then `r_(0)` would be equal to :

Options

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

Answer

To find the radial node \( r_0 \) for the 2s orbital of the hydrogen atom, we start with the given wave function:

\[ \Psi_{2s} = \frac{1}{4\sqrt{2\pi}} \left(\frac{1}{a_0}\right)^{3/2} \left(2 - \frac{r}{a_0}\right) e^{-r/a_0} \]

### Step 1: Identify the condition for a radial node
A radial node occurs at a point where the probability density of finding an electron is zero. This is determined by setting the radial part of the wave function equal to zero. The radial part of the wave function for the 2s orbital is given by:

\[ R(r) = 2 - \frac{r}{a_0} \]

### Step 2: Set the radial part equal to zero
To find the radial node, we set \( R(r) = 0 \):

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

### Step 3: Solve for \( r \)
Rearranging the equation gives:

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

Multiplying both sides by \( a_0 \) gives:

\[ r = 2a_0 \]

### Step 4: Conclusion
Thus, the radial node \( r_0 \) for the 2s orbital is:

\[ r_0 = 2a_0 \]

### Final Answer
The radial node \( r_0 \) is equal to \( 2a_0 \).

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