Carbon Potential
If the electric potential inside a carbon atom can be expressed as
V ( r ) = a π ϵ 0 Z e ( r 1 − c R b + d R 3 r 2 )
where R is the radius of the carbon atom, calculate Z + a + b + c + d .
Notes
1) e is the charge of a proton
2) ϵ 0 is the permittivity of free space
3) V ( r ) should not be confused with the potential energy, which would be of the form U ( r ) = − a π ϵ 0 Z e 2 ( r 1 − c R b + d R 3 r 2 )
The answer is 17.
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2 solutions
Finally, the explanation I was looking for! The Rutherford expression can be derived using Gauss' law and some boundary conditions. Only integration is required.
Inside an atom the positive charge is concentrated at the center and the electrons occupy the whole volume. We can use this logic to solve the problem. Z=6 because there are 6 electrons ( and protons) in a carbon atom and a=4. The first term in the bracket is the potential due the the protons. The rest two terms can be compared by R 3 K Q ( 0 . 5 r 2 − 1 . 5 R 2 ) which is the potential at a distance r(less than R) for a sphere of radius R having a negative charge Q in its bulk( the formula can be easily derived using the basic relation dV=-E.dr). Sum Z+a+b+c+d=17.
Who says b = 3 and c= 2, why aren't they 30 and 20 e.g.?
Moreover, this reasoning doesn't give the first ionisation potential well (which I tried using in tackling this problem), b/c equals 1,41 (roughly). Admitted, ionisation potential take quantum mechanical effects into account.
I arrived at Z = 6, a = 4 and d= 2 though , the last one by assuming the electric field tends tot zero at r close to R.
Hmmm... Actually, the expression of the electric potential inside the carbon atom shown in the question looks similar to Rutherford's calculation of the electric field at a distance r from the center of the atom to the spherical Gaussian surface, which led him to the mathematical expression for the electric potential. The formula is as follows...