# The probability of an electron being found in a certain region - probably easy?

1. Oct 22, 2009

### jeebs

the hydrogen ground state has the following wavefunction

$$\psi$$(r,$$\theta$$,$$\phi$$) = $$(\frac{1}{a^3^/^2\sqrt{\pi}})e^-^r^/^a$$

where a is the bohr radius a = 5.29x10-11m
I have to calculate the probability of finding the electron within the nucleus (radius R = 10-15m), and I am given an identity:

$$\int x^2e^-^xdx = b^3/3$$ when this integral is evaluated between 0 and b.

I know that the probability is given by P(r) = $$\int |\psi|^2 dV$$ = $$\int |\psi|^2 4\pi r^2dr$$ evaluating between 0 and R.

so when I put $$\psi$$ into the probability equation, I get

P(r) = $$(4/a^3)\int r^2 e^-^2^r^/^a dr$$ and I then try to use that identity, by making the substitution that x=kr (where k = 2/a)
so that dx = kdr and x2 = (kr)2

which gives me $$k^3\int r^2 e^-^k^r dr = R^3/3$$

which yields P(r) = $$(4/a^3)\int r^2 e^-^k^r dr = (4/a^3)R^3/3k^3$$ which has units of m3.

clearly I am going wrong because I am getting units of volume. this is frustrating me now, where am I going wrong?

thanks.

2. Oct 22, 2009

### latentcorpse

i think u used the wrong substitution....
your integral isn't of the same form as that given in the hint.

try $y=\frac{2r}{a}$. hopefully that will help.

3. Oct 22, 2009

### jeebs

even with your substitution this still gives me units of m3...