- #1
jeebs
- 314
- 5
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.
\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.