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 ×10⁻¹¹ m from the nucleus, whereas the two 2 s electrons make another spherical cloud at an average distance of 5⋅2 × 10⁻¹¹ m from the nucleus. Find three electric fields at (a) a point just inside the 1 s cloud and (b) a point just inside the 2 s cloud.

Answer 1

(a)

Let us consider the three surfaces as three concentric spheres A, B and C.
Let us take q = $1.6\times {10}^{-19}\mathrm{C}$ .

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, $r_1={10}^{-15}\mathrm{m}$
Radius of sphere B, $r_2=1.3\times {10}^{-11}\mathrm{m}$
Radius of sphere C, $r_3=5.2\times {10}^{-11}\mathrm{m}$
As the point 'P' is just inside the spherical cloud 1s, its distance from the centre
$x=1.3\times {10}^{-11}\mathrm{m}$
Electric field,
$E=\frac{q}{4\mathrm{\pi }{\mathit{\in }}_0{x}^2}$
Here, the charge enclosed is due to the charge of the 4 protons inside the nucleus. So,

$$\begin{gathered} E=\frac{4\times \left(1.6\times {10}^{-19}\right)}{4\times 3.14\times \left(8.85\times {10}^{-12}\right)\times {\left(1.3\times {10}^{-11}\right)}^2} \\ E=3.4\times {10}^{13}\mathrm{N}/\mathrm{C} \end{gathered}$$


(b)

For a point just inside the 2s cloud, the total charge enclosed will be due to the 4 protons and 2 electrons. Charge enclosed,

$q_{\mathrm{en}}=2q=2\times \left(1.6\times {10}^{-19}\right)\mathrm{C}$

Hence, electric field,

$E=\frac{q_{\mathrm{en}}}{4\mathrm{\pi }{\mathit{\in }}_0{x}^2}$
x = 5⋅2 × 10⁻¹¹ m
$$\begin{gathered} E=\frac{2\times \left(1.6\times {10}^{-19}\right)}{4\times 3.14\times \left(8.85\times {10}^{-12}\right)\times {\left(5.2\times {10}^{-11}\right)}^2} \\ E=1.065\times {10}^{12}\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, $E=1.1\times {10}^{12}\mathrm{N}/\mathrm{C}$