Question
Consider the following very rough model of a beryllium atom. The nucleus has four protons and four neutrons confined to a small volume of radius $10^{-15} m$. The two 1 s electrons make a spherical charge cloud at an average distance of $1.3 \times 10^{-11} m$ from the nucleus, whereas the two 2 s electrons make another spherical cloud at an average distance of $5.2 \times 10^{-11} m$ from the nucleus. Find the electric field at:
  1. A point just inside the 1s cloud.
  2. A point just inside the 2s cloud.

Answer

  1. Let us consider the three surfaces as three concentric spheres A, B and C.
Let us take $q = 1.6 \times 10^{-19}C.$
Sphere A is the nucleus; so, the charge on sphere A, $q_1 = 4q$
Sphere B is the sphere enclosing the nucleus and the 2 1s electrons; so charge on this sphere, $q_2 = 4q - 2q = 2q.$
Sphere C is the sphere enclosing the nucleus and the 4 electrons of Be; so, the charge enclosed by this sphere, $q_3 = 4q - 4q = 0.$
Radius of sphere $A, r_1 = 10^{-15}m$
Radius of sphere $B, r_2 = 1.3 \times 10^{-11}m$
Radius of sphere $C, r_3 = 5.2 \times 10^{-11}m$
As the point 'P' is just inside the spherical cloud 1s, its distance from the centre
$x = 1.3 \times 10^{-11}m$
Electric field,
$\text{E}=\frac{\text{q}}{4\pi\in_0\text{x}^2}$
Here, the charge enclosed is due to the charge of the 4 protons inside the nucleus. So,
$\text{E}=\frac{4\times\big(1.6\times10^{-19}\big)}{4\times3.14\times\big(8.85\times10^{-12}\big)\times\big(1.3\times10^{-11}\big)^2}$
$\text{E}=3.4\times10^{13}\text{N/C}$
  1. For a point just inside the 2s cloud, the total charge enclosed will be due to the 4 protons and 2 electrons. Charge enclosed,
$q_{en}= 2q = 2 \times (1.6 \times 10^{-19})C$
Hence, electric field,
$\text{E}=\frac{\text{q}_{\text{en}}}{4\pi\in_0\text{x}^2}$
$\text{x}=5.2\times10^{-11}\text{m}$
$\text{E}=\frac{2\times\big(1.6\times10^{-19}\big)}{4\times3.14\times\big(8.5\times10^{-12}\big)\times\big(5.2\times10^{-11}\big)^2}$
$\text{E}=1.065\times\text{10}^{12}\text{N/C}$
Thus, $\text{E}=1.1\times\text{10}^{12}\text{N/C}$

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