# Is the wave function normalized?

## Homework Statement

The ground state wave function for the electron in a hydrogen atom is:

$$\psi(r) = \frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o}$$

where r is the radial coordinate of the electron and a_o is the Bohr radius.

Show that the wave function as given is normalized.

## Homework Equations

Any wave function satisfying the following equation is said to be normalized:

$$\int_{-\infty}^{+\infty} |\psi|^2\dx = 1$$

## The Attempt at a Solution

Because the sum of all probabilities over all values of r must be 1,

$$\int_{-\infty}^{+\infty} (\frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o})^2 dr = \frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = 1$$

Since the integral can be expressed as the sum of two integrals, we have,

$$\frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = \frac{2}{\pi a_o^3} \int_0^{+\infty} e^\frac{-2r}{a_o^2} dr = 1$$

After integrating, I obtain,

$$\frac{1}{\pi a_o^3} = 1$$

which is definitely incorrect. However, I do not see any other way to proceed. could someone give some assitance.

jg370

$\int_0^{\infty} e^{-\frac{2r}{a_0^2}} dr = [-\frac{a_0^2}{2} e^{-\frac{2r}{a_0^2}]_0^{\infty}=\frac{a_0^2}{2}$
which gives $\frac{1}{2 \pi a_0}=1$