Yukawa Potential
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Let us consider a charge distribution with a spherical symmetry around a stationary point O.
At a certain distance \(r=OM\), the potential is expressed by (Yukawa potential) :
\(V(r) = \frac{q}{{4\pi \varepsilon _0 }}\frac{1}{r}\exp \left( { - \frac{r}{a}} \right)\)
Where \(q\) and \(a\) are positive constants.
Question
Calculate the field \(E\) at a given distance \(r\) from O.
Study the two particular cases \(r<<a\) and \(r>a\).
What is the significance of \(a\) ?
Indice
Use the relation between the field and the potential :
\(\vec E = - \overrightarrow {grad\;} V\)
Solution
The electric field is expressed by :
\(E(r) = - \frac{{dV(r)}}{{dr}} = \frac{q}{{4\pi {\varepsilon _0}}}\left( {1 + \frac{r}{a}} \right)\frac{1}{{{r^2}}}\exp \left( { - \frac{r}{a}} \right)\)
By noticing that, for \(r<<a\) :
\(E(r) \approx \frac{q}{{4\pi {\varepsilon _0}}}\frac{1}{{{r^2}}}\)
It is the electric field created by a punctual charge \(q\) placed in point O. This charge is the proton charge.
And, for \(r>>a\) :
\(E(r) \approx 0\)
From “far away”, the field is equal to zero and the charge distribution is globally neutral.
Here, \(a\) represents the order of magnitude of the dimension of the atom. It is Bohr's radius.
Question
Calculate \(\varphi (r)\), the surface integral of the field exiting a sphere of radius \(r\).
Deduce that the charge distribution can be equivalent to a punctual charge placed in O and a charge located in space and defined by its volumic charge distribution \(\rho (r)\).
Find the density of charge \(\rho (r)\) that is equivalent to the distribution.
Indice
Calculate the flux and use Gauss' law.
Solution
In spherical symmetry, the surface integral of the electric field (the flux) is given by :
\(\varphi (r) = 4\pi {r^2}E(r)\)
Hence :
\(\varphi (r) = \frac{q}{{{\varepsilon _0}}}\left( {1 + \frac{r}{a}} \right)\exp \left( { - \frac{r}{a}} \right)\)
Let's consider the volume between two spheres of radius \(r\) and \(r+dr\).
Gauss' law applied to this volume gives :
\(\varphi (r + dr) - \varphi (r) = \frac{1}{{{\varepsilon _0}}}4\pi {r^2}dr\rho (r)\)
We can notice :
\(\varphi (r + dr) - \varphi (r) = \frac{{d\varphi (r)}}{{dr}}dr \)
Thus :
\(\frac{{d\varphi (r)}}{{dr}} = \frac{1}{{{\varepsilon _0}}}4\pi {r^2}\rho (r)\)
Hence the density of charges, supposed to represent the electron in a semi-quantic manner, is expressed by :
\(\rho (r) = \frac{{{\varepsilon _0}}}{{4\pi {r^2}}}\frac{{d\varphi (r)}}{{dr}}\)
Thus :
\(\rho (r) = -\frac{q}{{4\pi }}\frac{1}{{{ra^2}}}\exp \left( { - \frac{r}{a}} \right)\)
Remarque :
The charge \(dq(r)\) located in the volume between two spheres of radius \(r\) and \(r+dr\) is :
\(dq(r) = 4\pi {r^2}\rho (r)dr = -q\frac{{{r}}}{{{a^2}}}\exp \left( { - \frac{r}{a}} \right)dr\)
We can define a linear density of charges, which is equivalent to the density of probability in quantum mechanics :
\(f(r) = \frac{{dq(r)}}{{dr}} = -q\frac{{{r}}}{{{a^2}}}\exp \left( { - \frac{r}{a}} \right)\)
Let's compute its derivative :
\(f'(r) = - q\frac{1}{{{a^2}}}\left( {1 - \frac{r}{a}} \right)\exp \left( { - \frac{r}{a}} \right)\)
We notice that \(f'(r)=0\) when \(r=a\). We can verify that it is a maximum.
We see that the electron is essentially distributed around Bohr's radius.
This radius corresponds to the classical circular trajectory in the planetary model of the atom.