Some Nuclear Properties — Question 5

Question 5

Atomic masses give binding energy when the electron counts cancel

Use neutral-atom masses $m_H=1.007825\,\mathrm u$, $m_n=1.008665\,\mathrm u$, and $M({}^{4}\mathrm{He})=4.002603\,\mathrm u$. Take $1\,\mathrm u\,c^2=931.5\,\mathrm{MeV}$ and neglect electronic binding corrections.

1. Find the mass defect and total binding energy of the helium nucleus.

2. Find binding energy per nucleon and explain the choice of hydrogen-atom mass rather than bare proton mass.

Question 5 - Solution

Atomic masses give binding energy when the electron counts cancel

(a) Method and calculation

$$\boxed{\Delta m=2m_H+2m_n-M= .030377\,\mathrm u}.$$$$\boxed{B=\Delta mc^2=.030377(931.5)=28.296\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$\boxed{B/A=28.296/4=7.074\,\mathrm{MeV}}.$$$$2m_H\simeq2m_p+2m_e,\quad M\simeq m_{\mathrm{nucleus}}+2m_e\Rightarrow\boxed{2m_e\text{ cancels}}.$$

Physical check

Using atomic masses consistently avoids an electron-rest-mass error. The small difference in electronic binding energies is outside the precision of this exercise.