Question

For 3s orbital of hydrogen atom, the normalised wave function is
$\mathbf{\psi }{{~}_{3s}}=\dfrac{1}{81\sqrt{3\pi }}{{(\dfrac{1}{{{a}_{0}}})}^{\dfrac{3}{2}}}\,\,[27-\dfrac{18r}{{{a}_{0}}}+\dfrac{2{{r}^{2}}}{{{a}_{0}}^{2}}]\,\,\,{{e}^{\dfrac{-r}{3{{a}_{0}}}}}$
If the distance between radial nodes is d, calculate the value of $\dfrac{d}{1.73{{a}_{0}}}$.

Answer 1

In the 3s orbital, the value of $\mathbf{\psi }{{~}_{3s}}=0$. So equate it with 0 and obtain a suitable value of r by solving the discriminant. Then manipulate the values and equations to obtain the desired answer.

Complete answer:
The wave function Ψ could be a mathematical expression. It is responsible for carrying crucial information about the electron it's associated with: from the wave function we obtain the electron's energy, momentum, and orbital orientation within the shape of the quantum numbers n, l, and ml.
The wave function can have a positive or negative sign. The sign is very important in calculations. It's also important when the wave functions of two or more atoms combine to create a molecule.
Wave functions with like signs (waves in phase) will interfere constructively(called constructive interference), resulting in the chance of bonding. Wave functions with unlike signs (waves out of phase) will interfere destructively(called destructive interference).
Now at nodal point the value of wave function for the 3s orbital electrons will be 0. Equating, we get
$\mathbf{\psi }{{~}_{3s}}=\dfrac{1}{81\sqrt{3\pi }}{{(\dfrac{1}{{{a}_{0}}})}^{\dfrac{3}{2}}}\,\,[27-\dfrac{18r}{{{a}_{0}}}+\dfrac{2{{r}^{2}}}{{{a}_{0}}^{2}}]\,\,\,{{e}^{\dfrac{-r}{3{{a}_{0}}}}}=0$
Which means that
$\,\,[27-\dfrac{18r}{{{a}_{0}}}+\dfrac{2{{r}^{2}}}{{{a}_{0}}^{2}}]\,\,=0$
$2{{r}^{2}}-18{{a}_{0}}r+27{{a}_{0}}^{2}=0$
On calculating the discriminant, we have
$r=\dfrac{-(-18{{a}_{0}})\pm \sqrt{{{(18{{a}_{0}})}^{2}}-4\times 2\times 27{{a}_{o}}^{2}}}{2\times 2}$
$r=\dfrac{18{{a}_{0}}\pm 10.39{{a}_{0}}}{4}$( we obtain 2 values of r)
So, the value of $\dfrac{d}{1.73{{a}_{0}}}$ can be said as
$\dfrac{d}{1.73{{a}_{0}}}=\dfrac{2\times 10.39{{a}_{0}}}{4\times 1.73{{a}_{0}}}$
as we obtain two values of r and on subtracting both we get the distance(d)=$\dfrac{2\times 10.39{{a}_{0}}}{4}$
So, on calculating ${{a}_{0}}$ cancels out and we get the value of $\dfrac{d}{1.73{{a}_{0}}}$ as 3.

Note:
As in order to proceed with the question, we needed the value of r, in order to find the distance d. That is why we had equated $\,\,[27-\dfrac{18r}{{{a}_{0}}}+\dfrac{2{{r}^{2}}}{{{a}_{0}}^{2}}]\,\,=0$ and not $\dfrac{1}{81\sqrt{3\pi }}{{(\dfrac{1}{{{a}_{0}}})}^{\dfrac{3}{2}}}\,=0$. It is assumed that the distances are in the SI units i.e metre(m).