#### 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 selectrons 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.

#### Solution

(a)

Let us consider the three surfaces as three concentric spheres A, B and C.

Let us take q = 1.6× 10^{-19} C .

Sphere A is the nucleus; so, the charge on sphere A, q_{1} = 4_{q}

Sphere B is the sphere enclosing the nucleus and the 2 1s electrons; so charge on this sphere, q_{2} = 4_{q} -2_{q} = 2_{q}

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 × 10^{-11} m

Radius of sphere C, r_{3} = 5.2 × 10^{-11} m

As the point 'P' is just inside the spherical cloud 1s, its distance from the centre

x = 1.3 × 10-^{11} m

Electric field,

`"E" = q/(4 pi ∈_0 "x"^2)`

Here, the charge enclosed is due to the charge of the 4 protons inside the nucleus . So,

`"E" = (4 xx (1.6 xx 10^-19))/(4 xx 3.14 xx (8.85 xx 10^-12)xx ( 1.3 xx 10^-11)^2 ) `

E = 3.4× 10^{13} N/C

(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,

qen = 2q = 2× (1.6 ×10^{-19}) C

Hence, electric field,

`"E" = ("q"_"en")/ (4 pi ∈_0"x"^2)`

x = 5.2× 10^{-11} m

`"E" = (2 xx (1.6 xx 10^-19))/(4 xx 3.14 xx ( 8.85 xx 10^-12) xx ( 5.2 xx 10^-11)^2)`

E = 1.065 × 10^{12} N/C

Thus , E =1.1× 10^{12} N/C