User:Redragon104

$$\Psi^2 = 1/(\pi*a_o^3) * e^{-2*r/a_0}$$

thats at a point, so on a sphere we mutliply by $$4 \pi r^2$$

This gives us $$P_{shell}(r) = 4 * r^2 / a_0^3 * e^{-2*r/a_0}$$

Now integrate the right side, and the left side = .9 because we are looking for the a 90% probability. we get

$$.9 = 4 * a_0^3 * \int^R_0 r^2 e^{-2*r/a_0} dr$$

Simplify by letting a_0 = 1, then are answer x will be simpler and we just multiply by a_0.

$$.9 = 4 * \int^R_0 r^2 e^{-2*r} dr$$