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?
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
The Attempt at a Solution
a) Deuterium-electron=mass of nucleus
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
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:
Then convert eV/c^2 to kg:
1eV/c^2= 1.78266173 × 10-36 kg
11155500eV/c^2= 1.98855916 e-30kg
E=(1.98855916 e-30kg)(2.998 e8)^2
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.