- #1

Haths

- 33

- 0

[tex]

$ \psi (x,y,z) = Ae^{- \alpha ( x^{2} + y^{2} + z^{2} ) }$

[/tex]

Find the probability that a particle is in a dr shell of space.

For what value of r is the probability of finding this particle greatest, and is this the same r value as;

[tex]

$ | \psi (x,y,z) |^{2} $

[/tex]

Right well. Having looked at the question baulked, then scratched my head to wonder how I could go about this question I have made the assumption that if wavefunction

**psi**is converted into spherical polar co-ordinate co-ords and intergrated as a volume integral with limits: r+dr and r therefore r+dr - r gives the dr shell of probability.

Taking that line, the integral of the function becomes;

[tex]

$ \int_{r}^{r+dr} \int_{0}^{2 \pi} \int_{0}^{\pi} Ae^{- \alpha r^{2} } \cdot r^{2} sin( \theta) d \theta d \phi dr$

[/tex]

Intergrating...

[tex]

$ \int_{r}^{r+dr} \int_{0}^{2 \pi} 2r^{2} Ae^{- \alpha r^{2} } d \phi dr$

[/tex]

[tex]

$ \int_{r}^{r+dr} 4r^{2} \pi Ae^{- \alpha r^{2} } dr$

[/tex]

Using intergration by parts at this stage...

[itex]

\frac{ \pi Ae^{ - \alpha r^{2} }} { - \alpha} - \frac{2 \pi Ae^{ - \alpha r^{2} }}{ \alpha^{2} r} |^{r+dr}_{r}

[/itex]

However try as I might I can't see a way to simplify this equation down to evaluate it for dr, hence perhaps I am barking up the wrong tree.

Secondly the second part of the question kind of confused me, because I thought that;

[tex]

\psi(x,y,z)

[/tex]

Having been called 'normalised' would mean that this is the square of the wavefunction already. Hence allowing for the intergration because it was in the form of;

[tex]

$ \int_{b}^{a}| \psi(x,y,z) |^{2} dxdydz = p$

[/tex]

where p here would be the probability for the particle to be in the shell bound by the two limits a and b.

This would lead me to conclude that there is no difference in the r value, mindyou the question to me appears to be wanting some difference because it asks to 'explain any differences', thus presupposing that there are some.

I would guess intuitively that when r is zero the probability is highest to find the particle, but my solution above doesn't show that explicitly. Hence I can't say that with confidence, even though it feels right as the wavefuntion is an exponentially decreasing spherical ripple in space. Perhaps my reading of the question is wrong. Eitherway I don't feel like I've completed this question sufficiently.

Haths