What is the Probability of Finding an Electron in a Specific Angle in Hydrogen?

[wavingerwin]

Homework Statement

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.

Homework 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}

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

?

Please help and thanks in advance

 

[wavingerwin]

... Nevermind.

Found what's wrong...

 

[zak8000]

Can you please tell me what's wrong with it . Cos I have a similar problem . Thanks

 

[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