# Homework Help: Hydrogen Wave Function

1. Sep 24, 2010

### wavingerwin

1. The problem statement, all variables and given/known data
Calculate the probability of finding the electron in a hydrogen within the angle $$\pm30\circ$$ from the x-y plane.The hydrogen is in the (2,1,1) state.

2. Relevant equations
$$probability = \int\int\int\left|R_{2,1,1}\right|^{2} \left|Y^{1}_{1}\right|^{2} r^{2} sin(\theta) dr d\phi d\theta$$

$$Y^{1}_{1} = -\frac{1}{2}\sqrt{\frac{3}{2\pi}}sin(\theta)e^{i\phi}$$

3. The attempt at a solution
$$\int\left|R_{2,1,1}\right|^{2} r^{2} dr = 1$$
because limits are 0 to infinity.

limit for $$\theta$$ is $$\frac{\pi}{3}$$ to $$\frac{2\pi}{3}$$
limit for $$\phi$$ is $$0$$ to $$2\pi$$

so...

$$probability = \int\int \left|Y^{1}_{1}\right|^{2} d\phi d\theta$$

$$probability = \int\int \frac{1}{4}\frac{3}{2\pi}sin^{3}\theta e^{\phi} d\phi d\theta$$

$$= \frac{3}{8\pi}\int\int sin^{3}\theta e^{\phi} d\phi d\theta$$
$$= \frac{3}{8\pi}\int sin^{3}\theta \left[e^{\phi}\right]^{2\pi}_{0} d\theta$$

$$= \frac{3}{8\pi} \left(e^{2\pi}-1\right) \int sin^{3}d\theta$$

and after some algebra..

$$\int^{\frac{2\pi}{3}}_{\frac{\pi}{3}} sin^{3}d\theta = 1-\frac{1}{12}$$

and so

$$probability = \frac{3}{8\pi} \left(e^{2\pi}-1\right) \left(1-\frac{1}{12}\right)$$

Now, $$e^{2\pi}$$ is more than $$500$$
which makes the probability equal to $$58.5$$

?????

2. Sep 25, 2010

### wavingerwin

... Nevermind.

Found what's wrong...

3. Sep 25, 2010

### zak8000

Can you plz tell me whats wrong with it . Cos I have a similar problem . Thanks

4. Sep 27, 2010

### wavingerwin

$$probability = \int\int \frac{1}{4}\frac{3}{2\pi}sin^{3}\theta e^{\phi} d\phi d\theta$$

Should be

$$probability = \int\int \frac{1}{4}\frac{3}{2\pi}sin^{3}\theta d\phi d\theta$$