# Homework Help: What is the work required to push three protons to form an nucleus?

1. Sep 16, 2014

### Jonsson

Hello there,

Imagine that a nucleus consists of three atoms arranged in a equilateral triangle with the length of each side, $R=2 \rm fm$.

Our protons starts infinitely far away. What is the work required to push these protons together in order to overcome the electric force between them?

I imagine that at any point in time, the force between each pair of protons is given by:

$$F_E = k\frac{q_0^2}{r^2}$$

Since each proton sees force from two protons, each at an angle $\theta = 60^\circ$ relative to the other since the triangle has angles $60^\circ$. The magnitude of the sum of these two force vectors must be given by:

$$\|\mathbf{F}\| = 2\cdot\cos(\theta/2)F_E = 3^{1/2}k\frac{q_0^2}{r^2}$$

We can then compute the work by letting $F_{net} = 3 \, \|\mathbf{F}\|$:
$$W_{\infty\to R} = \int_\infty^R\,F_{net}\,{\rm d}l = 3^{3/2}kq_0\int_\infty^R \frac{1}{r^2}\,{\rm d}l$$

which is incorrect.

What of the previous is incorrect?

Kind regards,
Marius

Last edited: Sep 16, 2014
2. Sep 16, 2014

### Orodruin

Staff Emeritus
You have used r variably as the distance between the protons and as the direction of motion. Why do you not just compute the individual potentials between each pair of protons and add them up?

Note: A bit of nomenclature, a nucleus does not consist of atoms, it consists of nucleons (protons and neutrons). An atom has a nucleus which is in a bound state with a number of electrons. A molecule consists of atoms.

3. Sep 16, 2014

### Jonsson

You are right, silly me. I fixed that now.

I could do as you say, but I am trying to learn what is wrong with my understanding of physics since I end up getting the wrong answer.

Are you able to help me work it out? :)

Kind regards,
Marius

4. Sep 16, 2014

### nrqed

You have to do in three steps. First you bring in the first proton from infinity. That requires no work. Then you bring in the second one. What is the work needed to bring that second one near the first one (note that there is only repulsion from ONE proton in that step). Finally you bring in the third one. You then add up the works for the three steps (the first work being trivially zero)

5. Sep 16, 2014

### Jonsson

Thanks!

6. Sep 16, 2014

### BiGyElLoWhAt

I'm not getting the same force equation that you have. Can you draw a picture and show how you oriented your system with respect to your axes?