# Distribution of charge in hydrogen atom

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1. Oct 20, 2015

### Peter Velkov

Suppose the hydrogen atom consists of a positive point charge (+e), located in the center of the atom, which is surrounded by a negative charge (-e), distributed in the space around it.

The space distribution of the negative charge changes according to the law p=Ce^(−2r/R), where C is a constant, r is the distance from the center of the atom, and R is Bohr's radius.

Find the value of the constant C by using the electrical neutrality of the atom.

I don't think I understand the charge distribution very well. I tried integrating the total negative charge of the sphere ( atom ), since I know it's equal to ( -e ).

Last edited: Oct 20, 2015
2. Oct 20, 2015

### Orodruin

Staff Emeritus
Please show us what you did when you integrated the distribution.

Edit: the model is really, really bad by the way, but for the sake of the problem, let us assume it is not.

3. Oct 20, 2015

### Peter Velkov

Sorry about the format. I don't know if the last line makes sense.
In the way I understand it the negative charge in a point should be p = Ce^(-2r/R). However I think i am wrong. Thank you in advance.

4. Oct 20, 2015

### Orodruin

Staff Emeritus
You cannot do it like an integral in one dimension, the distribution is three dimensional.

I see now that you really meant e^(-2r/R) with e being the base of the natural logarithm and not multiplication by the charge e. This is normally denoted by ^ or if you do not find that symbol by writing out "exp" for "exponential function".

5. Oct 20, 2015

### Peter Velkov

Yes, it's three dimensional, for a three dimensional point. But can't I integrate it for the whole radius, and then use the standard volume formula?
Furthermore, I know Bohr's radius is the mean of the orbit, but can it be used to derive the radius of the atom.

6. Oct 22, 2015

### blue_leaf77

The modulus square of a wavefunction, $p(r)$ in your notation, describes the probability density of the electron and it has a dimension of inverse volume. You can therefore build the charge density $\rho(r)$ by multiplying $p(r)$ with the electron charge $e$, so $\rho(r) = e p(r)$. The total charge is then just the integral of this quantity over all space, not just until certain radius like you did there.

7. Dec 20, 2017

### Lucas Silva

Did anyone get an answer for that? P L E A S E

8. Dec 20, 2017

### Staff: Mentor

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