Question 25.2: Binding Energy of Helium Use data in Appendix D to find the ...
Binding Energy of Helium
Use data in Appendix D to find the binding energy of the { }^{4} He atom. Compare the result with the binding energy of { }^{12} C computed in the text.
ORGANIZE AND PLAN Equation 25.3 gives the binding energy: E_{ b }=(\Delta m) c^{2}=\left(Z m_{p}+N m_{n}+Z m_{e}-M\left({ }^{A} X\right)\right) c^{2} \text {. For the }{ }^{4} He nucleus, Z=2 and N = 2 . Also needed is the atomic mass for the { }^{4} He atom, from Appendix D: 4.0026 u.
E_{ b }=(\Delta m) c^{2}=\left(Z m_{p}+N m_{n}+Z m_{e}-M\left({ }^{A} X\right)\right) c^{2} (Binding energy;SI unit: J) (25.3).
\text { Known: } M\left({ }^{4} He \right)=4.0026 u ; m_{n}=1.00866 u ; m_{p}=1.00728 u ;m_{e}=0.00054858 u ; Z=2 ; N=2 ; 1 u \cdot c^{2}=931.5 MeV.
Using the known values in Equation 23.2,
\lambda_{\text {med }} T=4.107 \times 10^{-3} m \cdot K (Median wavelength; SI unit: m.K) (23.2).
E_{ b }=((2)(1.00728 u )+(2)(1.00866 u )+(2)(0.00054858 u )-4.0026 u ) c^{2}=0.0304 u \cdot c^{2}.
\text { Using } u \cdot c^{2}=931.5 MeV \text {, this is } 28.3 MeV.
REFLECT Our result is little less than one-third the binding energy of { }^{12} C -not surprising, since { }^{12} C contains three times as many protons, neutrons, and electrons as { }^{4} He Again, the total binding energy is vastly greater than the electron-volt-level energies of atomic electrons.