How do I prove that the wave function R_{10}(r) is normalized?

[karkas]

Homework Statement

Its simple, I've encountered this problem in Beiser's book and it doesn't seem right:

Prove that the function $$R_{10}(r) = \frac{2}{a_{0}^{3/2}}\cdot e^{-\frac{r}{a_0}}$$ is normalized.

Homework Equations

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

The Attempt at a Solution

I figured it's simply just taking the integral
$$R_{10} (r)=\int^{\infty}_{0}( \frac{2}{a_{0}^{3/2}}\cdot e^{-\frac{r}{a_0}})^2dr = 1$$
but the result is not 1, it's $$2/a_0^2$$

 

[Simon Bridge]

Don't you need r²dr instead of just dr?
(volume integral, spherical-polar coords.)

 

[karkas]

Yes it appears that solves the problem, thanks!

 

[karkas]

I would enjoy a pointer as to where this comes from exactly, maybe a link?

edit: just standard triple integration in 3 coordinates?

 

[asteriatic]

To prove that the wave function R_{10}(r) is normalized, we must show that the integral of its square over all space is equal to 1. This can be expressed mathematically as:

∫∞0 |R_{10}(r)|^2 dr = 1

To solve this integral, we can use the substitution u = r/a_0, which transforms the integral into:

∫∞0 \frac{4}{a_0^3} e^{-2u} du = \frac{4}{a_0^3} ∫∞0 e^{-2u} du

Using the integral rule ∫ e^x dx = \frac{1}{n} e^x + C, we can solve for this integral and substitute back in for u:

\frac{4}{a_0^3} (-\frac{1}{2} e^{-2u}) \Big|_0^∞ = \frac{4}{a_0^3} (-\frac{1}{2} e^{-2r/a_0}) \Big|_0^∞ = \frac{4}{a_0^3} (-\frac{1}{2}) = \frac{2}{a_0^3}

This result shows that the square of the wave function is equal to 1, thus proving that R_{10}(r) is normalized.