Question

Problem $2.20$. Suppose you were to shrink Figure$2.7$until the entire horizontal scale fits on the page. How wide would the peak be?

Short answer

Fraction of the total width of the graph is around the hydrogen atom

$1\times {10}^{-12}\mathrm{m}$

Step-by-step solution

Distance between the two points

Schroeder shows that$\Omega$becomes, near the peak:

$\Omega ={\Omega }_{\max }·e^{-N{\left(\frac{2x}{q}\right)}^2}$

which is a Gaussian curve. The width can be defined as the distance in $x$between the points where the curve is $\frac{1}{e}$of its maximum value, so:

$\frac{{\Omega }_{\max }}{e}={\Omega }_{\max }·e^{-N{\left(\frac{2x}{q}\right)}^2}$

$e^1=e^{N{\left(\frac{2x}{q}\right)}^2}$

$1=N{\left(\frac{2x}{q}\right)}^2$

$x=\frac{q}{2\sqrt{N}}$

so the width of the curve is twice this, or:

$x=\frac{q}{\sqrt{N}}$

Peak is a fraction

Since we're assuming$q>>N$, this width is still a large number, but since the total width of the graph is $q+1$, the width of the peak as a fraction of the total width of the graph is around:

$\frac{q}{(q+1)\sqrt{N}}\approx \frac{1}{\sqrt{N}}$

which for any macroscopic value of $\mathrm{N}$, is vanishingly small. For example, if $N$is on the order of ${10}^{20}$, and we drew the graph so that the total width of the graph fits on a page $10\mathrm{cm}$wide, , the width of the central peak is around:

$\frac{10\times {10}^{-3}\mathrm{m}}{\sqrt{{10}^{20}}}=1\times {10}^{-12}\mathrm{m}$

which is about $frac1120$the size of a hydrogen atom.