Simple expression for Coloumb energy

[unscientific]

How did they arrive at this expression? I understand the $Q^2$ term in the nominator and the $r = r_0 (A)^{\frac{1}{3}}$ term in the denominator. Where did the $\frac{\alpha}{2}$ term come from?

Putting in Z=40, A=100 gives $3.3 \times 10^{14} J$, not $65 MeV$.

The energy term should be of the form $\frac{1}{4\pi \epsilon_0} \frac{Q^2}{R}$.

Also, shouldn't the energy barrier $E_{barrier}$ decrease with increasing nucleon number?

However, the text seems to imply the opposite.

 

[mfb]

This could be related to the chosen unit system. Alternatively, it is the difference between a single sphere and the two spheres. Where does the text come from?

How did you get 10¹⁴ J? That does not work in any unit system.

The coulomb energy shown there increases with increasing A and Z as you have more and more protons but the distance does not grow that much.
The full energy barrier decreases, to see that you have to include the original coulomb energy (as a single nucleus) - you'll see it is larger than the energy for two separated halves.

 

[unscientific]

This could be related to the chosen unit system. Alternatively, it is the difference between a single sphere and the two spheres. Where does the text come from?

How did you get 10¹⁴ J? That does not work in any unit system.

The coulomb energy shown there increases with increasing A and Z as you have more and more protons but the distance does not grow that much.
The full energy barrier decreases, to see that you have to include the original coulomb energy (as a single nucleus) - you'll see it is larger than the energy for two separated halves.

Using $\alpha = \frac{1}{137}$, $r_0 = 1.2 \times 10^{-15} m$, $Z=40$ and $A=100$ gives $10^{14}$.

I'm confused. You say: "The coulomb energy shown there increases with increasing A and Z..." and you then say "The full energy barrier decreases..". Does the energy barrier increase or decrease?

 

[mfb]

Using $\alpha = \frac{1}{137}$, $r_0 = 1.2 \times 10^{-15} m$, $Z=40$ and $A=100$ gives $10^{14}$.

No, it gives 10¹⁴/m. Which shows you got the units wrong, or there is some prefactor missing. Either way, you cannot simply put some arbitrary unit of energy (like Joule) to it.

I'm confused. You say: "The coulomb energy shown there increases with increasing A and Z..." and you then say "The full energy barrier decreases..". Does the energy barrier increase or decrease?

The expression there increases, but the energy barrier for splitting decreases.

 

[unscientific]

The expression there increases, but the energy barrier for splitting decreases.

Since $E_{barrier} \propto Z^2 A^{-\frac{1}{3}}$, shouldn't the energy barrier increase with addition of protons?

 

[mfb]

This is not the energy needed to split the nucleus.

 

[unscientific]

This is not the energy needed to split the nucleus.

Ok, if the energy barrier is not the energy needed to split the nucleus, then:

1. What does the energy barrier mean?
2. What then, is the energy required to split the nucleus?

 

[mfb]

1. What does the energy barrier mean?

That's what I asked in post 2: where is the text from?

2. What then, is the energy required to split the nucleus?

The difference in total energy between the two discussed states, including the surface term that has been mentioned in post 1 and the electrostatic potential difference between initial and final state. That is not exact, but it will should give a reasonable approximation.

 

[unscientific]

That's what I asked in post 2: where is the text from?
The difference in total energy between the two discussed states, including the surface term that has been mentioned in post 1 and the electrostatic potential difference between initial and final state. That is not exact, but it will should give a reasonable approximation.

This came from our lecture notes.

That splitting energy makes sense. The notes are confusing, as they say:

"For larger values of A, the energy barrier (7.1) continues to increase relative to final state of two well-separated daughters. "

Shouldn't (7.1) decrease with A?!

 

[mfb]

That looks more like the energy barrier for fusion (because it is compared two two well-separated nuclei). I don't know.