Radial Wavefunction of a Hydrogen Atom | Quantum Science Philippines

## Radial Wavefunction of a Hydrogen Atom

Gibson T. Maglasang and John Paul Aseniero

In this article, we outlined the necessary steps in calculating the radial wavefunctions [eq]R_{nl}[/eq] for the Hydrogen atom. Thus, the radial wavefunctions particularly [eq]R_{30 [/eq], [eq]R_{31 [/eq] and [eq]R_{32 [/eq] are easily obtained  without bothering to normalize it.

We use the formula below to find the wavefunction,

[eq]R_{nl}=\frac{1}{r}\rho^{l+1}e^{-\rho}\nu(\rho),[/eq]                                ( 1)

where

[eq]\nu(\rho)=\sum_{j=0}^{\infty}c_j\rho^j,[/eq]                                         (2)

while [eq]c_j[/eq] is determined by the recursion formula given by,

[eq]c_{j+1}=\frac{2(j+l+1-n)}{(j+1)(j+2l+2)}c_j,[/eq]                          (3)

and

[eq]\rho=\frac{r}{na}.[/eq]                                                            (4)

(i) Now, finding [eq]R_{30}[/eq]

Using equation 1, we need to solve first the coefficient [eq]c_j[/eq] from (eqn. 3), with [eq]n=3[/eq] and [eq]l=0[/eq].

[eq]c_1=\frac{2(0+1-3)}{1(0+0+2)}c_0=-2c_0[/eq]

[eq]c_2=\frac{2(1+1-3)}{2(1+0+2)}c_1=-\frac{2}{3}c_0[/eq]

[eq]c_3=\frac{2(2+1-3)}{3(2+2)}c_2=0.[/eq]

Knowing the value of [eq]c_j[/eq], (eqn. 2) can now be easily determined,

[eq]\nu(\rho)=c_0\rho^0+c_1\rho^1+c_2\rho^2+c_3\rho^3.[/eq]                         (5)

Substituting the value of the calculated coefficients to (eqn. 5), we then have

[eq]\nu(\rho)=c_0-2c_0+\frac{2}{3}\rho^2c_0.[/eq]                                         (6)

Thus,

[eq]R_{30}(\nu)=\frac{1}{r}\Big(\frac{r}{3a}\Big)e^{-\rho}[c_0-2c_0+\frac{2}{3}c_0\rho^2].[/eq]              (7)

Plugging in (eqn. 4) to (eqn. 7), we finally have

[eq]R_30=\frac{c_0}{3a}\bigg[1-2\Big(\frac{r}{3a}\Big)+\frac{2}{3}\Big(\frac{r}{3a}\Big)\rho^2\bigg]e^{-(r/3a)}.[/eq]

Following the same process in (i), the rest of the wavefunctions are just straightforward.

(ii) For [eq]R_{31}[/eq]

Determining first the coefficients, with [eq]n=3[/eq] and [eq]l=1[/eq],

[eq]c_1=\frac{2(1+1-3)}{1(0+2+2)}c_0=-\frac{1}{2}c_0[/eq]

[eq]c_2=\frac{2(1+1+1-3)}{2(1+2+2)}c_0=0[/eq]

[eq]c_3=0[/eq]

Then,

[eq]\nu(\rho)=c_0-\frac{1}{2}c_0\rho[/eq].

Thus,

[eq]R_{31}=\Big(\frac{r}{9a^2}\Big)\bigg[1-\frac{r}{6a}\bigg]e^{-r/3a}[/eq]

(iii) We have the coefficients for [eq]R_{32}[/eq] with [eq]n=3[/eq] and [eq]l=2[/eq],

[eq]c_1=\frac{2(2+1-3)}{1(4+2)}c_0=0[/eq]

[eq]c_2=0[/eq]

[eq]c_3=0[/eq]

Using again (eqn. 1), we have

[eq]R_{32}=\frac{r^2}{27a^3}e^{-r/2a}c_0[/eq]

We finally generated the radial wavefunctions ([eq]R_{30}[/eq], [eq]R_{31}[/eq], [eq]R_{32}[/eq]) for the hydrogen atom which is the main aim of this paper.

### 2 Responses to “Radial Wavefunction of a Hydrogen Atom”

1. SEO Coach Says:

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2. Abraham Says:

what is the value of $c_0$ ?