# The wave function of an orbital of H-Atom is given by: psi=((sqrt q)/(81 sqrt pi))(1/a_0)^(3/2)(6-r/a_0)(r/a_0)e^(-r/(3a_0) )* sin theta sin phi Then find the orbital ?

## 1) $2 s$ 2) $3 {p}_{y}$ 3) $3 {d}_{{z}^{2}}$ 4) $2 {p}_{x}$ Is it the $2 s$???

Jul 13, 2017

There is no $2 s$ in here at all. If there were, then there would be an ${e}^{- r / 2 {a}_{0}}$ term, as well as a $\left(2 - \frac{r}{a} _ 0\right)$ term. Neither of those are in here.

CHECKING THE ANGULAR NODES

Furthermore, here's a relatively easy way to verify that you have a $p$ orbital:

This wave function given goes to zero when the angular component, which contains $\sin \theta \sin \phi$, uses the angles $\left(\theta , \phi\right) = \left({0}^{\circ} , {0}^{\circ}\right)$ or $\left(\theta , \phi\right) = \left({180}^{\circ} , {180}^{\circ}\right)$.

When $\theta = {0}^{\circ} , {180}^{\circ}$, we are along the $z$ axis. When $\phi = {0}^{\circ} , {180}^{\circ}$, we are along the $x$ axis. Together, the two angle combinations form a nodal plane in the $x z$ plane, as there would be for a $3 {p}_{y}$ atomic orbital.

VERIFYING WHAT ORBITAL THIS BELONGS TO

What you'll have to do is separate the wave function into its radial and angular components. This takes practice and you should reference your text to check if possible...

You gave:

$\psi = \left(\frac{\sqrt{\textcolor{red}{q}}}{81 \sqrt{\pi}}\right) {\left(\frac{1}{a} _ 0\right)}^{3 / 2} \left(6 - \frac{r}{a} _ 0\right) \left(\frac{r}{a} _ 0\right) {e}^{- r / 3 {a}_{0}} \cdot \sin \theta \sin \phi$

This has to be the $3 {p}_{y}$ atomic orbital wave function.

• It's the only wave function that contains $\sin \theta \sin \phi$ in its angular component.
• It contains a $\left(6 - \frac{r}{a} _ 0\right)$, unique to the $3 p$ radial wave function.
• It contains an $81$ in its normalization constant, unique to atomic orbitals of $n = 3$.

But let's check if it matches...

The actual 3py wave function for a hydrogen-like atom is (Inorganic Chemistry, Miessler et al., 5th ed.):

${\psi}_{3 , 1 , - 1} \left(r , \theta , \phi\right)$

$= {R}_{31} \left(r\right) {Y}_{1}^{- 1} \left(\theta , \phi\right)$

$= {\overbrace{\frac{1}{81 \sqrt{3}} {\left(\frac{2 Z}{{a}_{0}}\right)}^{3 / 2} \left(6 - \frac{Z r}{{a}_{0}}\right) \left(\frac{Z r}{{a}_{0}}\right) {e}^{- Z r / 3 {a}_{0}}}}^{{R}_{n l} \left(r\right) \left(\text{Radial")) cdot overbrace(1/2 sqrt(3/pi) sinthetasinphi)^(Y_(l)^(m_l)(theta,phi)("Angular}\right)}$

For hydrogen atom then, $Z = 1$, so this simplifies to:

$= {\overbrace{\frac{\cancel{2} \cdot \sqrt{2}}{81 \cancel{\sqrt{3}}} {\left(\frac{1}{{a}_{0}}\right)}^{3 / 2} \left(6 - \frac{r}{{a}_{0}}\right) \left(\frac{r}{{a}_{0}}\right) {e}^{- r / 3 {a}_{0}}}}^{{R}_{n l} \left(r\right) \left(\text{Radial")) cdot overbrace(1/cancel(2) sqrt(cancel(3)/pi) sinthetasinphi)^(Y_(l)^(m_l)(theta,phi)("Angular}\right)}$

= color(blue)(overbrace(sqrt2/(81 sqrt(pi)) (1/(a_0))^(3//2) (6 - (r)/(a_0))((r)/(a_0)) e^(-r//3a_0))^(R_(nl)(r) ("Radial")) cdot overbrace(sinthetasinphi)^(Y_(l)^(m_l)(theta,phi)("Angular"))

I didn't change anything; all I did was take two already-separated components from my textbook, combine them together, and cancel out terms that have ratios of $1$.

It's the same wave function that I first referenced, and evidently, this matches the $3 {p}_{y}$ wave function exactly. I also think you had a typo so that $\sqrt{q}$ was supposed to be $\sqrt{2}$.