# Archived Binding energy and deuterium nucleus

#### alicia113

1. 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?

2. 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

3. 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.

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#### James R

Homework Helper
Gold Member
As you pointed out, a value like 1.12 MeV/c^2 has units of mass, not energy. So, really this number is a mass defect. The equivalent binding energy is simply $E=(\Delta m)c^2$, which in this example would be 1.12 MeV.

Assuming this number is correct (the reasoning seems to be ok), the ratio of this binding energy to the binding energy of the electron is then simply the ratio $1.12 \times 10^6 / 13.6$.

"Binding energy and deuterium nucleus"

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