Calculating Binding Energy of Deuterium Nucleus

[AClass]

Homework Statement

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

Homework Equations

E=mC^2

The Attempt at a Solution

a)

m(deuterium)=2.014102u
Has 2 nucleon
m(electron)=0.000549u
m(proton)=1.007276u
m(neuron)=1.008665u

m(Nucleus)=(2.014102u)-(0.000549u)=2.013553u
m(Nucleons)=(1.007276u)+(1.008665u)=2.015941u
[Delta]m=(2.015941u)-(2.013553u)=2.388x10^-3u
-Converting to Kg-
2.388x10^-3u = 3.965x10^-30kg

E(binding)=(3.965x10^-30kg)(3.0x10^8)^2=3.5685x10&-18J=2227282.752eV=2.227MeV
Eav(binding)=[2.227MeV]/[2 nucleons] =1.114MeV/Nucleon

The average binding energy per nucleon in the deuterium nucleus is 1.114 MeV/Nucleon

b)

E(Electron binding)=13.4eV/1 electron E(Nucleon binding)=1.114Mev/1 nucleon

[E(Electron binding)]/[E(Nucleon binding)] = [1.114MeV]/[13.4x10^-6MeV] =83134.33

The ratio of the binding energy per nucleon to the binding energy per electron in deuterium is 1:83134. More energy is required to remove the neutron than the electron.

I'm thinking my answer is correct, could someone kindly verify?

 

[anathema]

Thank you!

I can confirm that your calculations and reasoning are correct. The average binding energy per nucleon in the deuterium nucleus is indeed 1.114 MeV/Nucleon and the ratio of the binding energy per nucleon to the binding energy per electron in deuterium is approximately 1:83134. This means that the neutron is held more tightly than the electron in the deuterium nucleus. Great job on your calculations!