- #1

MagneticNeutron

- 1

- 2

- Homework Statement
- Calculate the binding energy of 11-Na-18

- Relevant Equations
- E = mc^2

Semi-Empirical Mass Formula = SEE CODE BELOW

Hi,

I am calculating the binding energy of 11- Na- 18

Stats:

Table of nuclides has BE/A = 6.202276 ± 0.006249 MeV

m used = 18.026878252 amu

E= mc^2 answer = 5.894327537620224 MeV

Semi Empirical Answer = 5.919667778950925 MeV

Including excess mass in E = mc^2 method

1 - Calculate energy of mass defect

(18.026878252 amu - (11*mP + 7*mN))*(convKilogram) * c^2 => 106.097 MeV

2 - Add in energy of excess mass

106.097 MeV + 25.036931 MeV of excess mass from table

3 - Divide by A

BE/A = 131.134826 MeV / 18 = 7.2852 MeV/A

So I am looking for hints as to why my SEM formula inputs of A = 18 and Z = 11 are coming out so close to an E =m*c^2 calculation that does not include the excess mass, why they are both off by about 0.3 MeV, and why the inclusion of the excess mass in the E = m*c^2 formula comes out way off.I also understand that the SEM formula is a just a model, and that it gets more accurate as the nuclei get larger and heavier. If what I am seeing is simply the error, cool

Code:

I am calculating the binding energy of 11- Na- 18

Stats:

Table of nuclides has BE/A = 6.202276 ± 0.006249 MeV

m used = 18.026878252 amu

E= mc^2 answer = 5.894327537620224 MeV

Semi Empirical Answer = 5.919667778950925 MeV

Including excess mass in E = mc^2 method

1 - Calculate energy of mass defect

(18.026878252 amu - (11*mP + 7*mN))*(convKilogram) * c^2 => 106.097 MeV

2 - Add in energy of excess mass

106.097 MeV + 25.036931 MeV of excess mass from table

3 - Divide by A

BE/A = 131.134826 MeV / 18 = 7.2852 MeV/A

So I am looking for hints as to why my SEM formula inputs of A = 18 and Z = 11 are coming out so close to an E =m*c^2 calculation that does not include the excess mass, why they are both off by about 0.3 MeV, and why the inclusion of the excess mass in the E = m*c^2 formula comes out way off.I also understand that the SEM formula is a just a model, and that it gets more accurate as the nuclei get larger and heavier. If what I am seeing is simply the error, cool

Code:

Python:

```
from semiEmpBind import semiEmpBind
massProton = 1.00728;
massNeutron = 1.00867;
lightSpeed = 2.99792458*10**8;
# 11 - Na - 18
massExcess = ((25.036931 * 10**6) / (lightSpeed**2))
massConst = 11*massProton + 7*massNeutron;
massDefect = 18.026878252 - massConst;
massDefect = massDefect * 1.66054*10**-27; #in kg
bindingEnergy = (-1) * massDefect * (lightSpeed)**2;
#in electron Volts
BE = bindingEnergy * 6.242*10**18 / (10**6); #add in mass excess in beginning
perNucleonBE = (25.036931 + BE)/ 18
print("E= mc^2 answer = " , perNucleonBE)semiEmpAns = semiEmpBind(18.0,11.0)/18.0
print("Semi Empirical Answer = " , semiEmpAns)
```

Python:

```
def semiEmpBind(A, Z):
N = A-Z; #this is the number of neutrons
aV = 15.76
aSurf = 17.81
aC = 0.711
aA = 23.702
aP = 11.18
remA = A %2
remZ = Z%2
remN = N%2
#determine paring term
if (remA == 0 and remN == 0 and remZ == 0):
delNot = 34/(A**0.75)
elif (remA == 0 and remN == 1 and remZ == 1):
delNot = -34/(A**0.75)
else:
delNot = 0
bindEnergy = aV*A - aSurf*(A**(2/3)) - aC*((Z*(Z-1))/(A**(1/3))) - aA*(((A -2*Z)**2)/A) + delNot
return bindEnergy
```