Question

1. The neutral Helium atom is made of a nucleus (containing two protons and two neutrons) and two electrons. The electrons have an average separation from the nucleus of r =31pm. Note: 1pm = 10^-12m (pico-meter)
   
   
   
   Consider the average configuration of the Helium atom shown above, treating the nucleus as a single charge of the appropriate size. What is the magnitude of the total electric force acting on one of the electrons?

Answer 1

Magnitude of electrostatic force is given by

|F| = k|Q1*Q2|/d^2

Force will be attractive if both charge have different signs, and force will be repulsive if both charge have same signs.

Force F1, on electron due to nucleus will be attractive and towards the nucleus

Force F2, on electron due to other electron will be repulsive and in opposite direction of F1

So net force on q3 will be

Fnet = F1 - F2

Fnet = k*q1*q2/d12^2 - k*q1*q3/d13^2

q1 = q3 = charge on electron = -e = -1.6*10^-19 C

q2 = charge on nucleus = 2e = 2*1.6*10^-19 C

d13 = distance between both electrons in Helium = 2*31 pm = 62 pm = 62*10^-12 m

d12 = distance between electrons and nucleus = 31 pm = 31*10^-12 m

k = 1/(4*pi*e0) = 9*10^9

So Using these values:

Fnet = k*q1*[q2/d12^2 - q3/d13^2]

Fnet = 9*10^9*1.6*10^-19*[3.2*10^-19/(31*10^-12)^2 - 1.6*10^-19/(62*10^-12)^2]

Fnet = 4.1956*10^-7 N

in 2 significant figures, So

Fnet = magnitude of Force on electron = 4.2*10^-7 N towards the nucleus

Let me know if you've any query.