Question

For a certain orbital $\Psi(r) = \frac{1}{16\sqrt{4}a_0^{3/2}}(1-\sigma)(\sigma^2 - 8\sigma + 12)e^{-\sigma/2}, \sigma = \frac{2r}{na_0}\{a_0 = constant\}$

The distance of nearest radial node from nucleus is $xa_0$ while farthest radial node from nucleus is $ya_0$.

Find $(\frac{y}{x})$.

Answer 1

6

Answer 2

The radial nodes of an atomic orbital occur at distances from the nucleus where the radial part of the wave function is zero, excluding $r=0$ and $r \to \infty$.

The given wave function is $\Psi(r) = \frac{1}{16\sqrt{4}a_0^{3/2}}(1-\sigma)(\sigma^2 - 8\sigma + 12)e^{-\sigma/2}$, where $\sigma = \frac{2r}{na_0}$.

The radial part of the wave function is proportional to $(1-\sigma)(\sigma^2 - 8\sigma + 12)e^{-\sigma/2}$.

The exponential term $e^{-\sigma/2} = e^{-r/na_0}$ is never zero for finite $r$, so the radial nodes are determined by the polynomial part: $(1-\sigma)(\sigma^2 - 8\sigma + 12) = 0$.

This equation is satisfied if either $(1-\sigma) = 0$ or $(\sigma^2 - 8\sigma + 12) = 0$.

Case 1: $1-\sigma = 0 \implies \sigma = 1$.

Case 2: $\sigma^2 - 8\sigma + 12 = 0$. This is a quadratic equation. We can factor it or use the quadratic formula. We look for two numbers that multiply to 12 and add up to -8. These numbers are -2 and -6. So, the equation can be factored as $(\sigma - 2)(\sigma - 6) = 0$. This gives $\sigma = 2$ or $\sigma = 6$.

The values of $\sigma$ for which the radial wave function is zero are $\sigma = 1, 2, 6$. These correspond to the radial nodes.

We are given the relation $\sigma = \frac{2r}{na_0}$. We can use this to find the distances $r$ corresponding to these $\sigma$ values: $r = \frac{na_0 \sigma}{2}$.

For $\sigma = 1$, the distance is $r_1 = \frac{na_0 \cdot 1}{2} = \frac{na_0}{2}$. For $\sigma = 2$, the distance is $r_2 = \frac{na_0 \cdot 2}{2} = na_0$. For $\sigma = 6$, the distance is $r_3 = \frac{na_0 \cdot 6}{2} = 3na_0$.

The distances of the radial nodes from the nucleus are $\frac{na_0}{2}$, $na_0$, and $3na_0$. The nearest radial node is at the minimum of these distances, which is $\frac{na_0}{2}$. The farthest radial node is at the maximum of these distances, which is $3na_0$.

We are given that the distance of the nearest radial node from the nucleus is $xa_0$. So, $xa_0 = \frac{na_0}{2}$. This implies $x = \frac{n}{2}$.

We are given that the distance of the farthest radial node from the nucleus is $ya_0$. So, $ya_0 = 3na_0$. This implies $y = 3n$.

We need to find the ratio $\frac{y}{x}$. $\frac{y}{x} = \frac{3n}{n/2} = \frac{3n \cdot 2}{n} = 6$.

The value of $n$ cancels out in the ratio.

The final answer is $\boxed{6}$.