Let\'s build a helium atom. Start with a nucleus of 2 protons and 2 neutrons (an
ID: 2077042 • Letter: L
Question
Let's build a helium atom. Start with a nucleus of 2 protons and 2 neutrons (and a charge of +2e). Then bring in an electron from infinity to the mean radius of an electron in the helium atom. r_He = 3.1 times 10^-11 m. Then bring in a second electron to the same distance on the other side. Calculate the energy gained by the system constructing this atom. Remember that the first particle you bring in (the nucleus) has no energy change, since there are no other particles for it to interact with. The first electron interacts with the nucleus, while the second electron is both attracted to the nucleus, and repelled by the other electron (which is 2r_he from it). e = 1.60 times 10^-19.Explanation / Answer
Given Data
r = 3.1*10^-11 m
e = 1.6*10*^-19
Solution :-
Energy to being first electron is KQ1Q2 / r
= (9*10^9)*(-e)*(2e) / (3.1*10-11)
= (9*10^9)*(-1.6*10^-19)*(2*1.6*10^-19 ) / (3.1*10^-11)
= - 1.49*10^-17 J
Energy to being Second electron is KQ1Q2 / r + KQ^2 / 2*r
= - 1.49*10^-17 J + (9*10^9)*(1.6*10^-19)^2 / 2*(3.1*10^-11)
= - 1.49*10^-17 J + 3.72*10^-18 J
= - 1.12*10^-17 J
So Total Energy = - 1.49*10^-17 J + - 1.12*10^-17 J
= - 2.61*10^-17 J
(OR)
Energy = -2 K*Q1*Q2 / r^2
Here
Q1 = charge on electron
Q2 = charge on proton
Energy = -2 K*Q1*Q2 / r^2
= -2 * (9*10^9)*(1.6*10^-19)*(1.6*10^-19) / (3.1*10^-11)^2
= 4.8*10^-7 J
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