We are going to make a very rough estimate of how much...

We are going to make a very rough estimate of how much pressure must be applied to a typical solid to compress it to the point where the potential energy reference level $u_0$ of the individual atoms becomes positive. Ordinarily, for a typical solid, $u_0$ is around -0.2 eV . (a) If the interatomic spacings are typically 0.2 nm , how many atoms are there per cubic meter? (b) Roughly, how much work (in joules) must be done on one cubic meter of this solid to raise $u_0$ to zero? (c) Work is equal to force times distance parallel to the force ( $F \mathrm{~d} x$ ) but, by multiplying and dividing by the perpendicular surface area, this can be changed into pressure times volume ( $-p \mathrm{~d} V$ ). Because solids are elastic, the change in volume is proportional to the change in applied pressure, $\mathrm{d} V=-C \mathrm{~d} p$, and the constant $C$ is typically $10^{-17} \mathrm{~m}^5 / \mathrm{N}$. With this background, calculate the work done on a solid as the external pressure is increased from 0 to some final value $p_f$. (d) With your answer to part (c) above, estimate the pressure that must be exerted on a typical solid to compress it to the point where $u_0$ becomes positive (Figure 4.4, top right). (e) What is a typical value for the variation of $u_0$ with pressure, $\partial u_0 / \partial p$, at constant temperature and atmospheric pressure in a solid?

We are going to make a very rough estimate of how much pressure must be applied to a typical solid to compress it to the point where the potential energy reference level $u_0$ of the individual atoms becomes positive. Ordinarily, for a typical solid, $u_0$ is around -0.2 eV .
(a) If the interatomic spacings are typically 0.2 nm , how many atoms are there per cubic meter?
(b) Roughly, how much work (in joules) must be done on one cubic meter of this solid to raise $u_0$ to zero?
(c) Work is equal to force times distance parallel to the force ( $F \mathrm{~d} x$ ) but, by multiplying and dividing by the perpendicular surface area, this can be changed into pressure times volume ( $-p \mathrm{~d} V$ ). Because solids are elastic, the change in volume is proportional to the change in applied pressure, $\mathrm{d} V=-C \mathrm{~d} p$, and the constant $C$ is typically $10^{-17} \mathrm{~m}^5 / \mathrm{N}$. With this background, calculate the work done on a solid as the external pressure is increased from 0 to some final value $p_f$.
(d) With your answer to part (c) above, estimate the pressure that must be exerted on a typical solid to compress it to the point where $u_0$ becomes positive (Figure 4.4, top right).
(e) What is a typical value for the variation of $u_0$ with pressure, $\partial u_0 / \partial p$, at constant temperature and atmospheric pressure in a solid?

Transcript

00:02 So in this question, we are trying to calculate the pressure at absolute zero.

00:07 So when a molecule is at absolute zero, its electron motion is going to exert a pressure at where it's a container, where the item is contained.

00:21 So this amount of pressure can be calculated by pressure equals negative, derivative of e of the energy total energy over volume.

00:36 And the purpose of this question is it wants us to calculate what this energy is.

00:45 So the first thing, let's first find out what e total is.

00:51 So e total, one thing to note is that this is not the same as fermi's energy.

00:57 This is equal to 3n over 5.

01:03 Fermi's energy, zero point energy.

01:08 So because 3 over 5, zero point energy is the average energy of the entire thing at 0 degree.

01:20 And then we want to plug in the equation for fermi energy.

01:25 Zero point fermi energy is pi 43, 2 bar square over 2m.

01:32 And then we have n 2 over 3, but then n is equal to the total number over the volume.

01:44 Yeah, because small n is like the number density.

01:48 So if we replace small n, the number density, by the total number over the volume, and then we can combine this total number with this and over here.

01:58 So if we organize it a little bit, it becomes 3 over 5, 3, 2, 3, 5, 5, 4 over 3, hbar square over 2m times big n 5 over 3 and v negative 2 over 3.

02:14 And then we can find the pressure by taking the derivative of this.

02:23 E total, i just write e total.

02:28 So we can see all of this will not change, it will remain the same.

02:45 But the last volume, which is what the derivative is taking it, it will become negative 2 over 3, v.

02:56 Negative 5 over 3.

02:59 So now it becomes 5 over 3 as well.

03:02 And n is 5 over 3.

03:04 So we can put them together again...