Question

For a certain orbital

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

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 orbital occur at distances $r$ from the nucleus where the radial part of the wave function, $\Psi(r)$, 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 $\sigma = \frac{2r}{na_0}$.

For $\Psi(r)$ to be zero, the terms that can become zero must be zero. The constant pre-factor is non-zero. The exponential term $e^{-\sigma/2}$ is non-zero for finite values of $\sigma$ (and thus finite values of $r$). The term $\sigma = \frac{2r}{na_0}$ is zero only at $r=0$, which is not a radial node.

Thus, the radial nodes occur when the polynomial part of the wave function is zero: $(1-\sigma)(\sigma^2 - 8\sigma + 12) = 0$.

We can factor the quadratic term $\sigma^2 - 8\sigma + 12$. We look for two numbers that multiply to 12 and add up to -8. These numbers are -2 and -6. So, $\sigma^2 - 8\sigma + 12 = (\sigma - 2)(\sigma - 6)$.

The equation for the nodes becomes: $(1-\sigma)(\sigma - 2)(\sigma - 6) = 0$.

This equation is satisfied when:

1. $1 - \sigma = 0 \implies \sigma = 1$
2. $\sigma - 2 = 0 \implies \sigma = 2$
3. $\sigma - 6 = 0 \implies \sigma = 6$

These are the values of $\sigma$ at which radial nodes occur. Now we convert these $\sigma$ values back to radial distances $r$ using the definition $\sigma = \frac{2r}{na_0}$.

For $\sigma = 1$: $1 = \frac{2r}{na_0} \implies r = \frac{na_0}{2}$.

For $\sigma = 2$: $2 = \frac{2r}{na_0} \implies r = \frac{2na_0}{2} = na_0$.

For $\sigma = 6$: $6 = \frac{2r}{na_0} \implies r = \frac{6na_0}{2} = 3na_0$.

The radial nodes are located at distances $\frac{n}{2}a_0$, $na_0$, and $3na_0$ from the nucleus.

To determine the value of $n$, we examine the structure of the wave function. The polynomial part is of degree 3 in $\sigma$. The number of radial nodes for an orbital with principal quantum number $n$ and azimuthal quantum number $l$ is $n-l-1$. The number of roots of the polynomial is 3, so $n-l-1 = 3$, which means $n-l = 4$. The general form of the radial wave function for a hydrogenic atom is proportional to $(r/a_0)^l \times (\text{polynomial in } \sigma) \times e^{-\sigma/2}$. The given wave function does not have the $(r/a_0)^l$ term, which implies $l=0$. If $l=0$, then $n-0 = 4$, so $n=4$. This means the orbital is a 4s orbital ($n=4, l=0$). The definition of $\sigma = \frac{2r}{na_0}$ with $n=4$ is $\sigma = \frac{2r}{4a_0} = \frac{r}{2a_0}$. Let's check the node distances using $n=4$: $r_1 = \frac{4a_0}{2} = 2a_0$. $r_2 = 4a_0$. $r_3 = 3(4)a_0 = 12a_0$.

The distances of the radial nodes from the nucleus are $2a_0$, $4a_0$, and $12a_0$. The nearest radial node is at distance $xa_0$. From the calculated distances, the nearest is $2a_0$. So, $xa_0 = 2a_0 \implies x = 2$.

The farthest radial node is at distance $ya_0$. From the calculated distances, the farthest is $12a_0$. So, $ya_0 = 12a_0 \implies y = 12$.

We are asked to find the ratio $(\frac{y}{x})$. $\frac{y}{x} = \frac{12}{2} = 6$.