How Do Binding Energies Compare Between Nucleons and Electrons in Deuterium?

[chemphys]

Homework Statement

a) Calculate the average binding energy per nucleon in the deuterium nucleus.
b) The energy that binds an orbiting electron to the hydrogen nucleus is 13.4eV. Calculate the ratio of the binding energy per nucleon to the binding energy per electron in deuterium. Which particle is held more tightly, the electron or the neutron?

Homework Equations

Given:

m of deuterium= 1876.12MeV/c^2
m of electron = 0.511MeV/c^2
m of neutron = 939.57 MeV/c^2
m of proton = 938.27 MeV/c^2
*all masses are also given in kg and u, but the example my text gives its answer in MeV/c^2
E=mc^2

The Attempt at a Solution

a) Deuterium-electron=mass of nucleus
(1876.12MeV/c^2)-(0.511MeV/c^2)=1875.609MeV/c^2

Proton+Neutron= (938.27 MeV/c^2)+(939.57 MeV/c^2) = 1877.84MeV/c^2

(proton+neutron)-(mass of nucleus)= (1877.84MeV/c^2)-(1875.609MeV/c^2)= 2.231MeV/c^2

(2.231MeV/c^2)/2nucleons=1.1155

Therefore the average binding energy per nucleon in the deuterium nucleus is 1.12MeV/c^2 (to 3 sigfigs)

I know this is a mass and not an energy but the example earlier in my book gave the average binding energy per nucleon in MeV/c^2, kg, or u so I'm inclined to leave my answer as a mass.

b) I think I should multiply 1.1155MeV/c^2 by 1000000 to eliminate the M:

(1.1155MeV/c^2)x(1000000)= 1115500eV/c^2

Then convert eV/c^2 to kg:

1eV/c^2= 1.78266173 × 10-36 kg
11155500eV/c^2= 1.98855916 e-30kg

E=mc^2
E=(1.98855916 e-30kg)(2.998 e8)^2
E=1.78731777 e-13J

Convert Joules to eV:

1J = 6.24150974 e18 eV
1.78731777 e-13J = 1115556.126eV

Ratio of binding energy per nucleon to electron= 1115556.126eV/13.4eV= 83250:1

Therefore the neutron is held more tightly than the electron as expressed by the ratio 8.32e4:1.
I'm not sure if this is done correctly. Did I convert from MeV/c^2 to eV properly? I think the answer is wrong because my binding energy in both eV/c^2 and eV are both 1.12 e6.

 

[anathema]

Is this because the binding energy is the same for both nucleons and electrons?

Your calculations for part a) seem correct. The binding energy per nucleon is indeed a mass, not an energy, so your answer of 1.12 MeV/c^2 is correct.

For part b), you are on the right track, but your conversion from MeV/c^2 to eV is incorrect. The correct conversion factor is 1 MeV/c^2 = 1.6021766208e-13 J. So your calculation should be:

1.12 MeV/c^2 x 1.6021766208e-13 J/MeV/c^2 = 1.7922495457e-13 J

Then, converting from Joules to eV using the conversion factor you provided, we get:

1.7922495457e-13 J x 6.24150974e18 eV/J = 1.118520497e6 eV

So the binding energy per nucleon in deuterium is 1.118520497 MeV, which is slightly different from your previous calculation. This is because the binding energy per nucleon is not exactly 1.12 MeV, but rather 1.118520497 MeV.

For the second part of b), your calculation is correct. The ratio of the binding energy per nucleon to the binding energy per electron is indeed 83250:1, indicating that the neutron is held more tightly than the electron in the deuterium nucleus. Good job!