Question

Question: Verify the correctness of the normalization constant of the radial wave function given in Table 7.4 as

$\frac{\mathbf{1}}{{\left(2a_0\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}\sqrt{\mathbf{3}}{\mathbf{a}}_{\mathbf{0}}}$

Short answer

Answer

It has been proved that the normalization for the case 2pstateis correct.

Step-by-step solution

Given data

The radial wave function corresponding to the state is,

$R_{n/l}\left(r\right)=\frac{r{e}^{-\frac{r}{2a_0}}}{{\left(2a_0\right)}^{3/2}\sqrt{3}{a}_0}$

Normalization condition

The radial part of the Hydrogen atom wave function should satisfy the condition as,

$\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\left\{R_{n/l}\left(r\right)\right\}}^{\mathbf{2}}{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r}\mathbf{=}\mathbf{1}$

Determining whether the given normalization constant is correct

Check equation (I) with the given wave function as:

$$\begin{gathered} =\frac{1}{{\left(2a_0\right)}^{3}3a_0}\underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-\frac{r}{a_0}}dr \\ =\frac{1}{24a_0^3}\underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-\frac{r}{a_0}}dr \end{gathered}$$

Let us assume,

$$\begin{gathered} r/a_0=z \\ r=a_{0}z \\ dr=a_{0}dz \end{gathered}$$

Then the integral becomes

$$\begin{gathered} =\frac{1}{24a_0^5}{a}_0^5\underset{0}{\overset{\infty }{\int }}{z}^4{e}^{-z}dz \\ =\frac{\Gamma \left(5\right)}{24} \\ =\frac{4!}{24} \\ =\frac{24}{24} \\ =1 \end{gathered}$$

Here $\Gamma \left(5\right)$ is the gamma function.

Thus the normalization is correct.