# Probability of finding a particle in the solid angle $d\Omega$ at $\theta$ and $\phi$ [closed]

For a spinless particle with the wavefunction
$$\begin{equation} \psi(x,y,z)= K(x+y+2z)\exp(-\alpha r) \end{equation}$$ with $$r=\sqrt{x^2+y^2+z^2}$$ and K and $$\alpha$$ are real constants.

I have to calculate the probability of finding the particle in a solid angle $$d\Omega$$ at $$\theta$$ and $$\phi$$. I have already interpreted the wave function using the spherical harmonics and calculated the root of the expectation value of the squared orbital angular momentum, its z-component, and its probability, now all I have to do is find the probability of the function in the solid angle $$d\Omega$$. Can anyone give me a clue?

Edit: I have found this equation for the needed probability while searching for a solution: $$\begin{equation} W_{nl}(\theta,\phi)d\Omega= Y_{l}^{m*}(\theta,\phi) Y_{l}^{m}(\theta,\phi)d\Omega \int{R_{nl}(r)r^2 dr} \end{equation}$$ I still don't know how to go from here.

## closed as off-topic by Kyle Kanos, user191954, stafusa, Jon Custer, ZeroTheHeroOct 3 '18 at 23:59

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• Do you mean probability of finding a particle at position $x$, $y$, $z$? – K_inverse Oct 3 '18 at 10:23
• No, the problem says finding the particle in the solid angle $d\Omega$ around $\theta$ and $\phi$. – Mohamed Mossad Oct 3 '18 at 10:38

Probability of finding the particle in a certain solid angle will be a function of $$\theta$$ and $$\phi$$. The radial coordinate $$r$$ does not matter, since we need to find the probability in a certain direction in space, not in a patch of given finite distance. So you should integrate over $$r$$ (from $$0$$ to $$\infty$$).