Consider the charge density of an electron in the ground...

Consider the charge density of an electron in the ground state of the hydrogen atom, ignoring relativistic and effective mass effects. Writing the radius in Bohr units for simplicity, it is: ρ(r) = - (e/π) exp(-2r). Verify using integration-by-parts that this density integrates over all space to -e, as expected. Using Gauss' Law, calculate the electric field E(r) due to ρ(r). Plot its behavior. Calculate the electrostatic potential Φ(r) corresponding to this E(r), and plot its behavior. i) Calculate the electrostatic energy cost of placing the charge density ρ(r) in the potential Φ(r). ii) Calculate the energy stored in the electric field E(r), which is due to ρ(r). Explain the reason for the difference, if any, between your results for calculations i) and ii). Calculate the energy density (i.e. pressure) of the electric field, as a function of radius.

Transcript

00:01 Okay, so we're doing chapter 22, problem 59 here.

00:04 And this one says neutral hydrogen can be modeled as a positive point charge here, which we'll call q, where q is plus 1 .6 times 10 to the negative 19 coulums.

00:19 This is surrounded by a distribution of negative charge with a volume density given here.

00:26 And a not is the bore radius which is just a constant and a is a constant such that the total negative charge is equal to negative 1 .6 times 10 to the negative 19 columns or equivalent to the charge of one electron okay so part a asks us to figure out what is the strength of the electric or sorry what is the net charged inside of a sphere of a radius a knot.

01:05 So we want to draw a gaussian surface here of radius a not and figure out how much charge is enclosed.

01:14 So we should know that the charge enclosed will be given as q negative at r equaling a not plus 1 .6 times 10 of the negative 19 columns.

01:35 Now first we got to figure out what this constant a is.

01:40 And to do that, we take the normalization criterion that all the negative charge sums up to being as much as an electron.

01:50 So that means q negative is the same as integrating our charge density over all space.

02:13 If we now convert this into just an r and we want to integrate r from zero to infinity, dr, and we're left with 4 pi r squared from converting dv to d r and now our row is negative a e to the minus 2r over a not so that's our integral if we solve this we should see that this comes out to being well first let's pull out what the constants here so we have negative 4 pi a and it goes from zero to infinity d r r squared e to the minus 2 r over a not.

02:57 Okay, and this equals negative 4 pi a times the result of this integral is 2 factorial over 2 over a not.

03:14 If we simplify this out, this is just negative pi a, a, a not cubed.

03:25 And this equals negative e if we take e being a constant of 1 .6 times.

03:37 Times 10 to the negative 19 coulums.

03:41 So if we rearrange that equation, we see that a is e over pi a0.

03:54 I'm going to add a new page here.

03:56 Now that we know that a is e over pi a not cubed, now we can figure out how much charge we have.

04:09 So we know that the charge enclosed is given as e, the constant e, so this is a constant e.

04:16 Charge, so this is the positive charge enclosed, plus our negative enclosed.

04:22 And to get this, we have to integrate from 0 to a not, row of e, row e of r, dv...

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