The wave function () of 2s is given by 2s = 12 2 ( 1a0)^1/2 ( 2 - ra0) e^-r/2a0 At r = r0, a radial node is formed. Thus, r0 in terms of a0 is:

The wave function () of 2s is given by 2s = 12 2 ( 1a0)^1/2 ( 2 -… — Solution | JEE Chemistry

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The wave function ( $\Psi$ ) of $2s$ is given by

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

At $r = r_0$ , a radial node is formed. Thus, $r_0$ in terms of $a_0$ is:

Correct answer:D

Standard Method

Given: The wave function for the $2s$ orbital is

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

Find: The value of $r_0$ where a radial node is formed.

At the node, the wave function $\Psi_{2s} = 0$ . Thus,

2 - \frac{r_0}{a_0} = 0

\Rightarrow \frac{r_0}{a_0} = 2

\Rightarrow r_0 = 2a_0

Therefore, the correct option is D, and the required value is $r_0 = 2a_0$ .

Setting the exponential term equal to zero. This is wrong because $e^{-r/2a_0}$ is never zero for any finite value of $r$ . The node comes from the factor $\left(2 - \frac{r}{a_0}\right)$ becoming zero.
Confusing the option label from the solution with the actual listed value. The solution says option B, but the worked result clearly gives $r_0 = 2a_0$ , which matches option D in the provided options. Always trust the derived value.
Using the idea of radial probability distribution instead of the given wave function directly. Here the question asks for the node from $\Psi_{2s}$ , so the correct step is to set the wave function equal to zero.

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