When you apply the quantum mechanical Schrödinger equation for a hydrogen atom, you need to put together the solutions for small r and large r. The Schrödinger equation gives you a solution to the radial Schrödinger equation for a hydrogen atom as follows:
![image0.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634454_396774.image0.png)
where f(r) is some as-yet-undetermined function of r. Your next task is to determine f(r), which you do by substituting this equation into the radial Schrödinger equation, giving you the following:
![image1.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634448_396775.image1.png)
Performing the substitution gives you the following differential equation:
![image2.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea63444e_396776.image2.png)
Quite a differential equation, eh? But just sit back and relax — you solve it with a power series, which is a common way of solving differential equations. Here’s the power-series form of f(r) to use:
![image3.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5d38584e222594c200_396800.image1.png)
Substituting the preceding equation into the one before it gives you
![image4.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634484_396778.image4.png)
Changing the index of the second term from k to k – 1 gives you
![image5.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634481_396779.image5.png)
Because each term in this series has to be zero, you have
![image6.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634461_396780.image6.png)
Dividing by rk–2 gives you
![image7.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea63445e_396781.image7.png)
This equation gives the recurrence relation of the infinite series,
![image8.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634484_396778.image4.png)
That is, if you have one coefficient, you can get the next one using this equation. What does that buy you? Well, take a look at the ratio of ak/ak–1:
![image9.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea63447c_396783.image9.png)
Here’s what this ratio approaches as k goes to infinity:
![image10.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634464_396784.image10.png)
This resembles the expansion for ex, which is
![image11.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634478_396785.image11.png)
As for e2x, the ratio of successive terms is
![image12.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea6344a4_396786.image12.png)
And in the limit
![image13.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea63449e_396787.image13.png)
the ratio of successive expansion coefficients of e2xapproaches 2/k:
![image14.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea63449b_396788.image14.png)
That’s the case for e2x. For f(r), you have
![image15.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634464_396784.image10.png)
Comparing these two equations, it’s apparent that
![image16.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634472_396790.image16.png)
The radial wave function, Rnl(r), looks like this:
![image17.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea63448f_396791.image17.png)
where
![image18.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d46b02d09890803049581_397905.image1.png)
Plugging the form you have for f(r),
![image19.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea63446e_396793.image19.png)
Okay, should you be overjoyed? Well, no. Here’s what the wave function
![image20.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5d38584e222594c230_396799.image0.png)
looks like:
![image21.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634498_396795.image21.png)
And substituting in your form of Rnl(r) from this equation gives you
![image22.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a5f3597b078ea634487_396796.image22.png)