Question

For the base state of the Hydrogen atom show that the probability P that the electron can be found in a sphere of radius R is
P = 1 - exp{(-2R/a0)(1 + 2R/a0 + 2R^2/a0^2)}

Answer 1

a_o is the Bohr radius?

Answer 2

Yes.

Answer 3

an impass :/

Are you allowed to start off with the normalized ground state wavefunction? I'm hoping you can just use it.... Then

\[ \\ \psi_{100}(r, \theta, \phi) = \frac{1}{\sqrt{\pi a_0^3}}e^{\frac{-r}{a_o}}
\\ \
\\ \
\int_{-\infty}^{+\infty} \psi^* \psi \ d^3 \textbf r = 1
\\ \
\\ \
\\ \ \textrm{use spherical}
\\ \
\\ \ \hspace{25px} d^3 \textbf r= r^2 \sin \theta \ dr \ d\theta \ d \phi\ \]

\[ P = \iiint\limits_V \psi^* \psi \ r^2 \sin \theta \ dr \ d\theta \ d \phi
\\ \
\\ \ = 4\pi \int_0^R \psi^* \psi \ r^2 \ dr
\\ \
\\ \
\\ \ = \frac{4}{ \ a_0^3} \int_0^R e^{\frac{-2r}{a_o}} r^2 \ dr \]

\[ \hspace{25px} \textrm{Integration by parts}
\\ \
\\ \hspace{35px} u = r^2 \hspace{15px} dv = e^{\frac{-2r}{a_o}}
\\ \
\\ \\ \hspace{35px} du = 2r dr \hspace{15px} v = \frac{-a_0}{2} e^{\frac{-2r}{a_o}} \]

\[ =\frac{4}{a_0^3} \left[ -\frac{r^2a_o}{2}e^{\frac{-2r}{a_o}} - \left( -\int a_o e^{\frac{-2r}{a_o}}r \ dr \right) \right]_0^R \]
\[ \textrm{Do that two more times and you get}\]
\[ P = \frac{4}{a_0^3} \left[ e^{\frac{-2r}{a_o}} \left( - \frac{r^2a_0}{2} - \frac{r^2a_0^2}{2} - \frac{a_0^4}{2}\right) \right]_0^R\]

Which is equivalent , then when you evalate at R you end up with the expression they're looking for. I think it's complete, but I dunno if all of the infinities are accounted for...
But!
So when R = 0 then P = 0, and when R = very very large the probability is almost 100& ^_^

Answer 4

last term in the brackets should be

\[ a_0^3 /4\]
mistype :P

Answer 5

and the second is
\[ ra_0^2/2 \]
:P

Answer 6

\[ P_{(R)} = \frac{4}{a^3_0} \left[ e^{\frac{-2r}{a_o}} \left( - \frac{r^2 a_0}{2} -\frac{r a_0^2}{2} -\frac{ a^3_0}{4} \right) \right]_0^R \]

P_{(R)} = \frac{4}{a^3_0} \left[ e^{\frac{-2r}{a_o}} \left( - \frac{r^2 a_0}{2} -\frac{r a_0^2}{2} -\frac{ a^3_0}{4} \right) \right]_0^R

Answer 7

I'm not sure that's a valid answer to the problem per se, as any normalized wavefunction will, by definition give you a probability of 0 when R=0 and 1 when R tends to infinity.

Answer 8

Right! If you're not looking for something there's no chance you'll find it find it, and if you look everywhere you have to find it. And that argument is true about the equation in the question, too. I could be wrong, but I thought this was what it asked for...

\[ P_{(R)} = \frac{4}{a^3_0} \left[ e^{\frac{-2r}{a_o}} \left( - \frac{r^2 a_0}{2} -\frac{r a_0^2}{2} -\frac{ a^3_0}{4} \right) \right]_0^R
\\ \
\\ \
\\ \
= \left[ e^{\frac{-2r}{a_o}} \left( - \frac{r^2}{2 a_0^2} -\frac{r}{2a_0 } -1 \right) \right]_0^R\
\\ \
\\ \
\\ \
= 1 -\left[ e^{\frac{-2R}{a_o}} \left( 1 + \frac{R}{2 a_0 } + \frac{R^2}{2 a_0^2} \right) \right] \]

I dunno ^_^

btw, ground state wavefunction was from Griffiths p. 151

Answer 9

Oh. xD That explains a lot. I actually used that same wave function for hydrogen yesterday to prove that the mean radius was 3a0/2, and never thought about expanding the probability that way to get the proper answer. Thanks!

Answer 10

^_^