The hydrogen atom is composed of one proton in the nucleus and one electron, which moves about the nucleus. In the quantum theory of atomic structure, it is assumed that the electron does not move in a well-defined orbit. Instead, it occupies a state known as an $ orbital $, which may be thought of as a "cloud" of negative charge surrounding the nucleus. At the state of lowest energy, called the $ ground state $, or $ 1s-orbital $, the shape of this cloud is assumed to be a sphere centered at the nucleus. This sphere is described in terms of the probability density function

$ p(r) = \frac{4}{a^3_0} r^2 e^{\frac{-2r}{a_0}} $ $ r \ge 0 $

where $ a_0 $ is the Bohr radius $ (a_0 \approx 5.59 \times 10^{-11} m) $. The integral

$$ P(r) = \int_0^r \frac{4}{a^3_0} s^2 e^{\frac{-2r}{a_0}}\ ds $$

gives the probability that the electron will be found within the sphere of radius $ r $ meters centered at the nucleus.

(a) Verify that $ p(r) $ is a probability density function.

(b) Find $ lim_{r\to\infty} p(r) $. For what value of $ r $ does $ p(r) $ have its maximum value?

(c) Graph the density function.

(d) Find the probability that the electron will be within the sphere of radius $ 4_{a_0} $ centered at the nucleus.

(e) Calculate the mean distance of the electron from the nucleus in the ground state of the hydrogen atom.