Question

The wave function $(\Psi)$ of $2 s$ is given by $\Psi_{2 n }=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{r}{a_0}\right) e^{-\frac{r}{2 a_0}}$ At $r=r_0$, radial node is formed Thus, $r_0$ in terms of $a_0$

• A. $r_0=\frac{a_0}{2}$
• B. $r_0=2 a_0$
• C. $r_0=4 a_0$
• D. $r_0=a_0$

Answer 1

The Correct Option is B

To find the value of \(r_0\) where a radial node is formed in the wave function of the \(2s\) orbital, we will follow these steps:

1. Understand Wave Function for \(2s\) Orbital:
2. Identify the Radial Node Condition:
3. Solve the Equation for \(r\):
4. Conclusion:
   • \(r_0 = 2a_0\)
5. Explanation of Other Options:
   • \(r_0 = \frac{a_0}{2}\): This would not make the term \(\left(2 - \frac{r}{a_0}\right)\) equal to zero.
   • \(r_0 = 4a_0\): Similarly, this value would not satisfy the node condition.
   • \(r_0 = a_0\): This also does not satisfy the condition for zeroing the radial part.

Thus, the correct answer is \(r_0 = 2a_0\).