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.