In light of your answer to ①, what does this simplify to in our case?
③ This analysis has focussed on the momentum in one dimension. An atom, however, is a three dimensional object ...
We could consider the one-dimensional momentum we have been considering so far, to be the resolved component of the overall momentum along one of the Cartesian axes: let's say the x-axis. So we could call it p(x). The symmetry of the atom means that there will be components, which on average will be of equal magnitude, along the y- and z-axes: p(y) and p(z).
With Pythagoras' cheerful assistance, write down an expression for the overall value of <p2>, in terms of p(x).
④ Now you can bring these results together to write down an expression for the kinetic energy, in terms of σp .
⑤ Next, we need to think about position within the atom. Symmetry tells us that it is equally likely that the electron could be found on either side of the nucleus and since the nucleus is at the origin (x=0), this means that <x> = 0.
What, therefore is the expression for σx , the standard deviation of x, ?
⑥ Because of the symmetry, what we deduce to be true along the x-axis will be equally true along any axis that passes through the nucleus. Therefore <lxl>, where lxl is the magnitude of x (ie the value, disregarding the sign) is equal to the mean distance from the nucleus, <r>. Unfortunately we have no easy access to <lxl>. Instead, the closest we can come is the root mean square distance, √(<x2>).
Is it reasonable to use √(<x2>) as an estimate of the mean magnitude of the distance, <lxl>? The easiest way to answer this is to think of simple examples, with just a few numbers in the distribution.
⑦ Now see if you can combine all these findings with the uncertainty principle to relate the kinetic energy, Ek, to the mean distance between the electron and the nucleus, <r>.
