The schrodinger wave equation for hydrogen atom is
$Ψ (radial)= \frac{1}{16 \sqrt{4}} {(\frac{z}{a_{0}})}^{3 / 2} [(σ - 1) (σ^{2} - 8 σ + 12)] e^{- σ / 2}$
where a₀ and z are the constant in which answer can be expressed and $σ = \frac{2 zr}{a_{0}}$ minimum and maximum position of radial nodes from nucleus are

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$\frac{a_{0}}{2 Z}, \frac{3 a_{0}}{Z}$

$\frac{a_{0}}{2 Z}, \frac{4 a_{0}}{Z}$

$\frac{a_{0}}{2 Z}, \frac{a_{0}}{Z}$

$\frac{a_{0}}{Z}, \frac{3 a_{0}}{Z}$

answer is C.

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Detailed Solution

(c) Probability of finding e^– is zero implies that
$ψ^{2} = 0 or ψ = 0$
$\Rightarrow (σ - 1) = 0 \Rightarrow σ = 1$
$or r_{1} = \frac{a_{0}}{2 z} or (σ^{2} - 8 σ + 12) = 0$
and $(σ - 6) (σ - 2) = 0$
$σ =6, r = \frac{3 a_{0}}{z}; σ = 2, r = \frac{a_{0}}{z}; r_{2} = \frac{3 a_{0}}{z}$