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

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

 

[paranormal]

.



Hi there, great job on your attempt at solving this problem. Your solution looks correct to me. Just a few things to note:

1. The limits for \theta should be -\frac{\pi}{6} to \frac{\pi}{6} since we are looking for the electron within the angle \pm30\circ from the x-y plane. This means that the integral for sin^3\theta should be evaluated from -\frac{\pi}{6} to \frac{\pi}{6}.

2. In the expression for Y^{1}_{1}, there is a typo - it should be e^{i\phi} instead of e^{\phi}.

Other than those minor corrections, your solution looks good. Keep up the good work!