# Please explain the following example given in the snapshot below?

Jul 13, 2017

The answers are (B) $r = 2 {a}_{0}$ and (D) 0.423 Å.

#### Explanation:

Ex. 1

Psi_text(2s) = 1/(4sqrt(2π)a_0^"3/2")[2-r/a_0]e^("-"r/(2a_0)) = 0

The zeroes of this function occur at

$2 - \frac{r}{a} _ 0 = 0$ and ${e}^{\text{-} \frac{r}{2 {a}_{0}}} = 0$

e^(-r/(2a_0)) → 0 only as r → ∞.

This should be no surprise, because we know that a wave function becomes vanishingly small as distance from the nucleus increases.

If $2 - \frac{r}{a} _ 0 = 0$

then $\frac{r}{a} _ 0 = 2$

and $r = 2 {a}_{0}$.

Ex. 2

R(r) = 1/(9sqrt6)(Z/a_0)^(3/4)(4 - σ)σe^("-"σ/2)

The zeroes of this function occur at

4 - σ = 0, σ = 0, and e^("-"σ/2) = 0.

σ = 0 corresponds to zero probability at the nucleus (no surprise here!).

e^("-"σ/2) = 0 corresponds to zero probability at an infinite distance from the nucleus (again, no surprise!).

If 4 - σ = 0

then σ = 4

and $\frac{Z r}{a} _ 0 = 4$

r = (4a_0)/Z = (color(red)(color(black)(4)) × "0.529 Å")/color(red)(color(black)(5)) = "0.423 Å"