1 Let\'s build a helium atom. Start with a nucles of 2 protons and 2 neutrons (a

Question

1 Let's build a helium atom. Start with a nucles 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 × 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 2'm fron it), e = 1.60 × 10-19.

Explanation / Answer

Given  

helium atom with 2 protons and 2 neutrons

when we bring an electron (first) from infinity to a point r _He = 3.1*10^-11 m , the work done stored is in the form of energy

the potential energy of the system of two charges is  

U = kq1*q2/r

U1 = k*2q*q/r_He

U1 = (9*10^9*2*1.6*10^-19*1.6*10^-19)/(3.1*10^-11) J

U1 = 1.486452*10^-17 J

and when bringing second charge , we have to do more work against the two fields now

due to two protons and one electrons so  

U2 = k*2q1*q2/r + kq^2/r

here q1=q2 = q

U2 = (kq^2/r_He)(2+1)

U2 = ((9*10^9*1.6*10^-19*1.6*10^-19)/(3.1*10^-11))(3)

U2 = 2.229677*10^-17 J

now the net work done W = U1+U2 = 1.486452*10^-17+2.229677*10^-17 J

W = 3.716129*10^-17 J

so the energy gained by the system is 3.716129*10^-17 J

because what ever work done on the electrons to bring form infinity to that point inside the electric field will be stored in the form of electrostatic potential energy

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