(5-16)
① The overall electron waveform is stationary, so there must be equal contributions from individual components with positive and negative momentum vectors. Therefore the mean value of the momentum, <p> = 0.
② Since <p> = 0, we can write that:
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③ Extending Pythagoras theorem into 3 dimensions, we can write:
<p2> = <p(x)2> + <p(y)2> + <p(z)2>
(If you're not used to Pythagoras in 3 dimensions, consider this:
Applying Pythagoras’ theorem first in the xy plane, we get
pxy2 = px2 + py2.
Then applying it again, this time in the plane defined by pxy and pz, we get
p2 = pxy2 + pz2
So, combining these, we get:
p2 = px2 + py2 + pz2
Symmetry dictates that <p(x)2> = <p(y)2> = <p(z)2>, so we can rewrite this result as:
<p2> = 3<p(x)2>
④ The kinetic energy is given by:
Ek = <p2>/2m = 3<(p(x))2>/2m
From ② we know that σp(x)2 = <(p(x))2>
and, therefore:
Ek = 3σp(x)2/2m
⑤ This is very similar to what we found with the momentum. Since the symmetry dictates that <x> = 0:
⑥ The root mean square value of a parameter, √(<x2>), is not the same as the mean magnitude of the distance, <lxl>.
You can see this with an example: if you have four values of x, of -3, -1, 1 and 3:
√(<x2>) = √[(9 + 1 + 1 + 9) / 4] = √5 = 2.24
<lxl> = [3 + 1 + 1 + 3] / 4 = 2.0
As this example illustrates, the squaring tends to amplify the bigger values in the distribution, relative to the smaller ones, so that √(<x2>) comes out a little bit bigger than <lxl>. The difference is not huge, however, so if we are looking for an order of magnitude answer rather than high precision, √(<x2>) will be a reasonable approximation to <lxl>, and hence to <r> .
⑦ We know from ④ that Ek = 3σp(x)2/2m.
We want to find an expression for Ek in terms of <r>, the mean electron distance from the nucleus. Our strategy will be to use the uncertainty principle to rewrite the Ek expression in terms of σx2 and then relate this to <r>.
We are looking to find the minimum radius to which an electron wave packet can be confined. So rather than write the uncertainty principle as the usual inequality (which allows for additional sources of uncertainty), we'll write it as a simple equality: σxσp(x) = h/4π
We can square and rearrange this equation, to get σp(x)2 = h2/16π2σx2
Now from ⑤ we know that σx2 = <x2>
And in ⑥, we decided that <x2> is a reasonable approximation to <r>2
So we can substitute into the σp(x)2 expression, to get: σp(x)2 = h2/16π2<r>2
Which, finally, allows us to substitute into the Ek expression, Ek = 3σp(x)2/2m, to get what we want:
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