I am trying to wrap my head around the Schrodinger's quantum mechanical model of an atom.

According to the NCERT$^1$, the Schrodinger's equation is given by:

$$\hat{H}\Psi=E\Psi$$ where $\hat{H}$ is an operator called hamiltonian.

Schrödinger gave a recipe of constructing this operator from the expression for the total energy of the system. The total energy of the system takes into account the kinetic energies of all the sub-atomic particles (electrons, nuclei), attractive potential between the electrons and nuclei and repulsive potential among the electrons and nuclei individually. Solution of this equation gives $E$ and $\Psi$.

To get more details about what $\Psi$ was I looked up in Wikipedia$^2$:

In addition to l and m, a third integer n > 0, emerges from the boundary conditions placed on R. The functions R and Y that solve the equations above depend on the values of these integers, called quantum numbers. It is customary to subscript the wave functions with the values of the quantum numbers they depend on. The final expression for the normalized wave function is:

$$\Psi_{nlm}=R_{nl}(r)Y_{lm}(\theta,\phi)$$ $$R_{n \ell} (r) = \sqrt {{\left ( \frac{2 Z}{n a_{\mu}} \right ) }^3\frac{(n-\ell-1)!}{2n{(n+\ell)!}} } e^{- Z r / {n a_{\mu}}} \left ( \frac{2 Z r}{n a_{\mu}} \right )^{\ell} L_{n-\ell-1}^{(2\ell+1)} \left ( \frac{2 Z r}{n a_{\mu}} \right )$$

Not quoting further as my question comes now -

I am given two particular proportionalities for $\Psi$ as below:

$$\Psi_{nlm} \propto \bigg(\frac{Z}{a_o}\bigg)^{\frac{3}{2}}e^{-\big(\frac{Zr}{a_o}\big)}\tag{1}$$

$$\Psi_{nlm} \propto \bigg(\frac{Z}{a_o}\bigg)^{\frac{5}{2}}re^{-\big(\frac{Zr}{2a_o}\big)}\cos\theta\tag{2}$$

I am trying to find which orbital these two wave functions belong to.

For the first one, $n=1$, it should be $1s$ orbital - as it is one to exist for sure. This is for hydrogen atom.

But for $\ce{He+}$ ion, the second one is correct. I am not able to understand why, given that:

  1. according to exponential term, n=2
  2. but according to the power of $\frac{Z}{a_o}$, n=1

Can anyone please explain?


This question is actually based on a JEE Advanced problem, from the year 2017 (set by IIT Madras which does give great Chemistry questions), which goes as follows:

The wave function $\Psi_{nlm}$ is a mathematical function whose value depends upon spherical polar coordinates $(\phi\text{, θ, r)}$ of the electron and characterized by the quantum numbers $n$, $l$ and $m_l$. Here $r$ is distance from nucleus, $θ$ is colatitude and $\phi$ is azimuth. In the mathematical functions given in the Table, Z is atomic number and $a_o$ is Bohr radius. Match the following columns according to the information given -

JEE Adv 2017 - problem

enter image description here

This is the question I was trying to solve.

  • 1
    $\begingroup$ But according to the cos function n=2, l=1. $\endgroup$
    – Poutnik
    Commented Apr 10 at 6:08
  • 1
    $\begingroup$ You do need to put $n$ and $\ell$ into the full equation because for the generalised Laguerre polynomial (L) and this depends on $n$ in two places, as $(n-\ell -1)$ and as $2Z/na$. Also you know from the spherical harmonic $\cos(\theta)$ that $\ell=1$ and this means that $n\ne 1$ $\endgroup$
    – porphyrin
    Commented Apr 10 at 7:13
  • $\begingroup$ So what does that mean @porphyrin? That is, I should also consider cos function as 'l' and decide for n accordingly? $\endgroup$ Commented Apr 10 at 7:44
  • 1
    $\begingroup$ I guess that the OP meant 'one of them to exist' is hydrogen... anyway $\endgroup$ Commented Apr 10 at 7:46
  • 2
    $\begingroup$ You know that there are limits on $n,ell$ and $m$ so list the possible values and substitute into the equations and simplify all terms. The Mcquarrie & Simon 'Physical Chemistry" text book gives all the details and so do many other texts. $\endgroup$
    – porphyrin
    Commented Apr 10 at 8:25

1 Answer 1


If we look at a table of equations for the radial component of a hydrogen atom:

enter image description here

You will notice that all the equations have a factor of $\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}$, irrespective of whether the electron belongs to $n=1$, $n=2$ or $n=3$

The power of $\frac{Z}{a_0}$ does not tell us anything about the subshell that the electron is in, it simply helps us identify which atom the wave equation belongs to.

Coming to your question, the wave equation given in column B is

$$\psi_{n, l, m_l} \propto\left(\frac{Z}{a_o}\right)^{\frac{5}{2}} r e^{-\left(\frac{Z r}{2 a_o}\right)} \cos \theta$$

The power of $\frac{Z}{a_0}$ is $\frac{5}{2}$, implying that it belongs to the helium atom. The denominator in the exponential term tells us $n=2$. Also, the equation does not equal zero for any nonzero value of r, which tells us that the orbital has zero radial nodes and $$n-l-1 = 0$$ Therefore, $n=2$, $l=1$, so the radial equation describes an electron in the $2p$ orbital, not $1s$. So option (C) is the incorrect combination

References: The Schrödinger Wave Equation for the Hydrogen Atom - Chemistry LibreTexts


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