Question 15.2: Using the atomic masses of the 1 1H and 4 2H e isotopes give...
Using the atomic masses of the ^{1}_{ 1}H and ^{4}_{ 2}H e isotopes given in Table 15.1 and the mass of the neutron given in Appendix A, calculate the binding energy of the ^{4}_{ 2}H e nucleus.
| TABLE 15.1 Some of the properties of a number of light isotopes. | |||||
| Name | Z | A | Atomic mass (u) | I | Natural abundance |
| H | 1 | 1 | 1.007825 | 1/2 | 99.989% |
| 2 | 2.014102 | 1 | 0.011% | ||
| He | 2 | 3 | 3.016029 | 1/2 | 1.37 × 10^{−4}% |
| 4 | 4.002603 | 0 | 99.99986% | ||
| Li | 3 | 6 | 6.015122 | 1 | 7.59% |
| 7 | 7.016004 | 3/2 | 92.41% | ||
| Be | 4 | 9 | 9.012182 | 3/2 | 100% |
| B | 5 | 10 | 10.012937 | 3 | 19.9% |
| 11 | 11.009306 | 3/2 | 80.1% | ||
| C | 6 | 12 | 12.000000 | 0 | 98.93% |
| 13 | 13.003355 | 1/2 | 1.07% | ||
| N | 7 | 14 | 14.003074 | 1 | 99.632% |
| 15 | 15.000109 | 1/2 | 0.368% | ||
| O | 8 | 16 | 15.994915 | 0 | 99.757% |
| 17 | 16.999132 | 5/2 | 0.038% | ||
| 18 | 17.999160 | 0 | 0.205% | ||
The reaction in which two hydrogen atoms combine with two neutrons to form the ^{4}_{ 2}H e atom is
2 ^{1}_{1}H+2n\rightarrow ^{4}_{2}He.The terms on each side of this equation contain two protons, two neutrons, and two electrons. To be concrete, we write the mass under each term in the above reaction to obtain
\begin{matrix}\begin{matrix}2 ^{1}_{1}H & +\\ 2(1.007825) u \end{matrix} &\begin{matrix} 2n &\rightarrow \\2(1.008665) u \end{matrix} & \begin{matrix} ^{4}_{2}He\\ 4.002602u \end{matrix} \end{matrix}The mass lost when two ^{1}_{ 1}H atoms combine with two neutrons to form the ^{4}_{ 2}H e isotope is
\Delta m = 2(1.007825) u + 2(1.008665) u − 4.002602 u = 0.030378 u.The binding energy of the helium nucleus is obtained by multiplying this mass loss by 931.5 MeV. We obtain
B(2, 2) = 28.297 MeV.
| Appendix A | |||
| Constants and conversion factors | |||
| Constants | |||
| Speed of light | c | 2.99792458 × 10^{8} m/s | |
| Charge of electron | e | 1.6021773 × 10^{−19} C | |
| Plank’s constant | h | 6.626076 × 10^{−34} J s | |
| 4.135670 × 10^{−15} eV s | |||
| \hbar=h/2π | 1.054573 × 10^{−34} J s | ||
| 6.582122 × 10^{−16} eV s | |||
| hc | 1239.8424 eV nm | ||
| 1239.8424 MeV fm | |||
| Hydrogen ionization energy | 13.605698 eV | ||
| Rydberg constant | 1.0972 × 10^{5} cm^{−1} | ||
| Bohr radius | a_{0} = (4π \epsilon _{0})/(me²) | 5.2917725 × 10^{−11} m | |
| Bohr magneton | μ_{B} | 9.2740154 × 10^{−24} J/T | |
| 5.7883826 × 10^{−5} eV/T | |||
| Nuclear magneton | μ_{N} | 5.0507865 × 10^{−27} J/T | |
| 3.1524517 × 10^{−8} eV/T | |||
| Fine structure constant | α = e^{2}/(4π\epsilon _{0} c \hbar) | 1/137.035989 | |
| e^{2}/4π\epsilon _{0} | 1.439965 eV nm | ||
| Boltzmann constant | k | 1.38066 × 10^{−23} J/K | |
| 8.6174 × 10^{−5} eV/K | |||
| Avogadro’s constant | N_{A} | 6.022137 × 10^{23} mole | |
| Stefan-Boltzmann constant | σ | 5.6705 × 10^{−8} W/m² K^{4} | |
| Particle masses | |||
| kg | u | MeV/c² | |
| Electron | 9.1093897 × 10^{−31} | 5.485798 × 10^{−4} | 0.5109991 |
| Proton | 1.6726231 × 10^{−27} | 1.00727647 | 938.2723 |
| Neutron | 1.674955 × 10^{−27} | 1.008664924 | 939.5656 |
| Deuteron | 3.343586 × 10^{−27} | 2.013553 | 1875.6134 |
| Conversion factors | |||
| 1 eV | 1.6021773 × 10^{−19} J | ||
| 1 u | 931.4943 MeV/c² | ||
| 1.6605402 × 10^{−27} kg | |||
| 1 atomic unit | 27.2114 eV | ||
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