Semi-Empirical Mass and E=mc^2 close, but both off? Binding Energy calc

[MagneticNeutron]

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:

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)

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

 

[shjacks45]

Might point out that "excess mass" (binding energy) for an unstable nucleus like 18Na is less per nucleon than stable nuclei like 18O. I don't see a problem with excess mass ~ 0. Try 22Na (well known) as a comparison. 18Na is also near the double proton decay line.