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. `sqrt2 a_(0)`
D. `(a_(0))/(sqrt2)`

Answer

To find the radial node in the 2s orbital of the hydrogen atom, we start with the given wave function:

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

### Step 1: Understand the concept of radial nodes
A radial node is a point where the probability density of finding an electron is zero. This occurs when the wave function \(\Psi\) is equal to zero.

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

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

This leads us to the equation:

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

### Step 3: Solve for \(r\)
From the equation \(2 - \frac{r}{a_0} = 0\), we can solve for \(r\):

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

Multiplying both sides by \(a_0\):

\[ r = 2a_0 \]

### Step 4: Conclusion
Thus, the radial node \(r_0\) is:

\[ r_0 = 2a_0 \]

### Final Answer
The radial node in the 2s orbital is at \(r_0 = 2a_0\).

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