# Homework Help: Binding energy, Coulomb energy

1. Apr 1, 2016

### Incand

1. The problem statement, all variables and given/known data
a) Calculate the difference in binding energy between the mirror nuclide $^{15}O$ and $^{15}N$.
b) Calculate the radius of both nuclide’s assuming the difference in binding energy exclusively depends on difference in Coulomb energy.

2. Relevant equations
$R = R_0A^{1/3}$.
Binding energy
$B = \left[ Zm(^1H)+Nm_n-m(^AX)\right]c^2$.
Binding energy due to Coulomb repulsion between protons
$E_c = -\frac{3}{5}\frac{e^2}{4\pi \epsilon_0 R_0} Z(Z-1)A^{-1/3}$.
Constants:
$m(^1H) = 1.007825u$
$m_n = 1.00866501u$
$m(^{15}O) = 15.003065u$
$m(^{15}N)=15.000109u$
$1 u = 931.502MeV/c^2$
$\frac{e^2}{4\pi \epsilon_0} = 1.439976MeV \cdot fm$.
3. The attempt at a solution
I think I solved part a) but I'm having trouble with part b).
The difference in binding energy should be
$\Delta B = \left[ m_n-m(^1H)+m(^{15}O)-m(^{15}N\right] = (0.00379601u)/c^2 = 3.536MeV$. (The answer says $3.532$ but I guess I worked with more decimals)

The difference in Coulomb energy should be (since $8\cdot 7 - 7\cdot 6 = 14$)
$\Delta B= 3/5 \frac{e^2}{4\pi \epsilon_0R_0} 14A^{-1/3}$
$R = R_0A^{1/3} = \frac{3\cdot 14 \cdot 1.439976}{5\cdot 3.536}fm = 3.420752fm$
The answer however says $R = 3.67fm$ so I did something wrong (using 3.532 instead doesn't change anything)

Last edited: Apr 1, 2016
2. Apr 1, 2016

### drvrm

check the calculation .
what value of R0 was taken? As i find people taking different values!

3. Apr 1, 2016

### Incand

As I understand the question I should compute $R_0$ from knowing the binding energy (actually $R_0A^{(1/3)}$).

The typical value for $R_0$ I've seen is $R_0 \approx 1.2fm$ but from my calculation I have a value of $R_0 =1.38$ (and from the answer I would need an even larger $R_0$. So perhaps I'm doing something wrong as I understand it $R_0$ should be about the same for a nuclide. I'm not entirely sure I'm using the right formulas either, they were simply the ones that seemed to apply to the question.

4. Apr 1, 2016

5. Apr 1, 2016

### Incand

Essentially I'm using the coulomb term out of the that formula. However the version I have is different from the one on wikipedia. My book says
$B = a_vA -a_sA^{(2/3)}-a_cZ(Z-1)A^{(-1/3)}-a_{\text{sym}}\frac{(A-2Z)^2}{A}+\delta.$
Note the difference in the Coulomb term.
If I use $Z^2$ I indeed get the correct result just correcting my earlier value with $3.42\cdot \frac{15}{14} = 3.67$ (since $8^2-7^2=15$).
Now I'm confused why there's $Z(Z-1)$ in the version In my book. The book argue that the term is proportional to $Z(Z-1)$ since each proton repels each of the other ones.

6. Apr 1, 2016

### drvrm

Nucleus Ro (in fermis)

B11 1.28

C13 1.34

N15 1.31

O17 1.26

F19 1.26

Ne21 1.25

Na23 1.22

Mg23 1.23

Al27 1.20

ref.
www.bhojvirtualuniversity.com/ss/sim/physics/msc_f_phy_p3u1.doc
The energy discrepancy is because of the use of classical principles instead of quantum mechanical principles in calculating the coulomb energy Ec..

7. Apr 1, 2016

### drvrm

The difference from wiki is that Wiki Formula is for B/A -binding energy per nucleon - but meanwhile i think its R0 value which changes with nucleids may be responsible for the difference.

8. Apr 1, 2016

### drvrm

actually its Z(Z-1) but approximated to Z^2 so when you use the former you are very correct. the culprit seems to be R0 value.

9. Apr 1, 2016

### Incand

Thanks for all the help! I believe the answer was calculated using the $Z^2$ approximation so that should explain everything. As for the $R_0$ value I'm not entirely sure. I don't use the actual value of $R_0$ in my calculations although I could compute it.
I'm guessing the difference in $R_0$(if I compute it) may be because the question only cares about the Coulomb term which isn't the entire truth.