Schaum's Outline of Quantum Mechanics 8.11

[Jimmy Snyder]

Homework Statement

If a hydrogen atom is in the state n = 2, l = 0, what is the probability that an electron has a value of r that is smaller than the Bohr radius.

Homework Equations

eqn (8.8)
\psi_{200}(r, \theta, \phi) = R_{20}Y_0^0(\theta, \phi)

Page 305
Y_0^0(\theta, \phi) = \frac{1}{\sqrt{4\pi}}

eqn (8.33)
R_{20} = 2(2a_0)^{-3/2}(1 - \frac{r}{2a_0})e^{-r/2a_0}

The Attempt at a Solution

P = \int_0^{2\pi}d\phi \int_0^{\pi}d\theta\int_0^{a_0}dr r^2|\psi(r, \theta, \phi)|^2sin(\theta)
= \int_0^{2\pi}d\phi \int_0^{\pi}d\theta sin(\theta) (Y_0^0(\theta, \phi))^2 \int_0^{a_0}dr r^2 4(2a_0)^{-3}(1 - \frac{r}{2a_0})^2e^{-r/a_0}
Now let z = r/a_0; dr = a_0dz then
P = \frac{4\pi}{4\pi}\frac{4}{(2a_0)^3}\int_0^1 a_0dz a_0^2z^2(1 - \frac{z}{2})^2e^{-z}
= \frac{1}{2}\int_0^1 dz (z^2 - z^3 + \frac{z^4}{4})e^{-z}
= (-1 -z -\frac{z^2}{2} - \frac{z^4}{8})e^{-z}|_0^1
= 1 - \frac{21}{8e} = 0.0343

The book gives the answer .176

I calculated that the probability of r being less than 4 times the Bohr radius is .176

Placing n = 2, l = 0 in equation 8.42 gives that the mean value for r is 6 times the Bohr radius. I find it hard to believe that the electron could spend nearly 18% of the time at less than 1 Bohr radius and yet average out to 6 Bohr radii.

 

[dextercioby]

It's + z^4/4 inside the bracket under the integral sign. And are you sure about the integrations ?

 

[Jimmy Snyder]

It's + z^4/4 inside the bracket under the integral sign. And are you sure about the integrations ?

Thank you. I fixed that. I am as sure about them as I can be, but if I were 100% sure, I wouldn't have posted. By the way, I also solved the integral numerically using a calculator and got the same 0.0343 result. I believe that if I am wrong, it is because I am using the wrong approach, not because of an error in calculation.