Question

An early model of the atom, proposed by Rutherford after his discovery of the atomicnucleus, had a positive point charge +Ze(the nucleus) at the center of a sphere ofradiusRwith uniformly distributed negative charge−Ze.Zis the atomic number, thenumber of protons in the nucleus and the number of electrons in the negative sphere.
a. Show that the electric field inside this atom is
b. What is the electric field at the surface of the atom? Is this the expected value? Explain.
c. A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field at r = R/2?

Answer 1

Answer:

a)  E = k Ze (1- r³ / R³)  1/r², b) E=0, c)   E = -6.62 10¹⁰  N / C

Explanation:

a) For this we can use the law of Gauus

         Ф = E- dA = $q_{int}$ / ε₀

where we take a sphere as a Gaussian surface, so that the electric field lines and the radii of the sphere are parallel, consequently the dot product is reduced to the algebraic product

       E A =q_{int} / ε₀

the area of ​​a sphere  

      A = 4π r²

      E 4π r² = q_{int} / ε₀

      E = 1 / 4πε₀   q_{int} / r²

       k = 1 /4π ε₀

       E = k q_{int} / r²       (1)

       

let's analyze the charge inside the gaussian sphere,

let's use the concept of density for electrons, since they indicate that the charge is evenly distributed

     ρ = Q / V

where the volume of the sphere is

    V = 4/3 πr³

     Qe = ρ V

     Qe = ρ 4 / 3π r³

the density of the electrons is

     ρ = Ze 3 / (4π R³)

where R is the atomic radius

we substitute

       Qe = Ze r³/ R³

for protons they are in a very small space, the atomic nucleus, so we can superno that they are a point charge.

The net charge inside our Gaussian surface, the charge of the protoens plus the charge of the electroens (Qe)

     q_{int} = q_proton + Q_electron

     q_{int} = + Ze - Qe

     q_{int} = + Ze - Ze r³ / R³

     q_{int} = Ze (1- r³ / R³)

   

  we substitute in equation 1

     E = k Ze (1- r³ / R³)  1/r²

b) on the surface of the atom r = R

therefore the electric field is zero

      E = 0

c) Calculate the electric field for the Uranium for

       r = R / 2 = 0.10 10⁻⁹ / 2 = 0.05 10⁻⁹ = 5 10⁻¹¹ m

     

       E = 8.99 10⁹ 92 1.6 10⁻¹⁹ (1-1/2)³   1/ (5 10⁻¹¹)²

       E = -6.62 10¹⁰  N / C