The wave function $(\Psi)$ of 2 s is given by $\Psi_{2 \mathrm{~s}}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{\mathrm{a}_{0}}\right)^{1 / 2}\left(2-\frac{\mathrm{r}}{\mathrm{a}_{0}}\right) \mathrm{e}^{-\mathrm{r} / 2 \mathrm{a}_{0}}$ At $r=r_{0}$, radial node is formed. Thus, $r_{0}$ in terms of $a_{0}$

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Atomic Structure

The wave function

(Ψ)

of 2 s is given by

Ψ_{2 s} = \frac{1}{2 2 π} (\frac{1}{a _{0}})^{1/2} (2 - \frac{r}{a _{0}}) e^{- r /2 a_{0}}

r = r_{0}

, radial node is formed. Thus,

r_{0}

in terms of

a_{0}

(A) $r_{0} = a_{0}$
(B) $r_{0} = 4 a_{0}$
(C) $r_{0} = \frac{a _{0}}{2}$
(D) $r_{0} = 2 a_{0}$ ✓ Correct

Answer: (D)

Solution

A radial node is a distance at which the wave function, and hence the probability of finding the electron, drops to zero. The constant prefactor and the exponential

e^{- r /2 a_{0}}

can never vanish, so the function is zero only where the polynomial factor is zero:

2 - \frac{r _{0}}{a _{0}} = 0.

Solving places the node at

r_{0} = 2 a_{0}

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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