The hydrogen-like radial wave function of the 3s orbital is given as R3,0=1/(9√(3)) (Z/(a₀))3/2 (6-2ρ+(ρ²)/9)… · ABC26PH0331

Subject: Physical Chemistry · Chapter: Atomic Structure · Topic: Atomic Structure – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

The hydrogen-like radial wave function of the 3s orbital is given as
\[ \mathrm{R}_{3,0}=\frac{1}{9\sqrt{3}}\left(\frac{\mathrm{Z}}{\mathrm{a}_{0}}\right)^{3/2}\left(6-2\rho+\frac{\rho^{2}}{9}\right) \mathrm{e}^{-\rho/6} \]
where $\rho=2\mathrm{Zr}/\mathrm{a}_{0}$; Z = atomic number; r = distance from the nucleus and $\mathrm{a}_{0}$ = Bohr radius. Positions of the radial nodes (in units of $\mathrm{a}_{0}$) of the 3s orbital are at
(a)$\frac{3+3 \sqrt{3}}{2 Z}, \frac{3-3 \sqrt{3}}{2 Z}$
(b)$\frac{9+3 \sqrt{3}}{2 Z}, \frac{9-3 \sqrt{3}}{2 Z}$
(c)$\dfrac{3+\sqrt{3}}{2\mathrm{Z}}, \dfrac{3-\sqrt{3}}{2\mathrm{Z}}$
(d)$\dfrac{6+3\sqrt{3}}{2\mathrm{Z}}, \dfrac{6-3\sqrt{3}}{2\mathrm{Z}}$
Answer
Answer: B ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
The nodes come from 6 − 2ρ + ρ²/9 = 0, i.e. ρ² − 18ρ + 54 = 0, so ρ = 9 ± 3√3. With ρ = 2Zr/a₀, r = (9 ± 3√3)a₀/2Z = 1.90 and 7.10 a₀ for H, the known 3s nodes (b). The printed key (c) is wrong.

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