Using the atomic masses of the 1 1H and 4 2H e isotopes given in Table 15.1 and the mass of the neutron given in Appendix A, calculate the binding energy of the 4 2H e nucleus.

Question

Using the atomic masses of the [latex]^{1}_{ 1}H [/latex] and [latex]^{4}_{ 2}H e[/latex] isotopes given in Table 15.1 and the mass of the neutron given in Appendix A, calculate the binding energy of the [latex] ^{4}_{ 2}H e[/latex] 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%
H	1	2	2.014102	1	0.011%
He	2	3	3.016029	1/2	[latex]1.37 × 10^{−4}[/latex]%
He	2	4	4.002603	0	99.99986%
Li	3	6	6.015122	1	7.59%
Li	3	7	7.016004	3/2	92.41%
Be	4	9	9.012182	3/2	100%
B	5	10	10.012937	3	19.9%
B	5	11	11.009306	3/2	80.1%
C	6	12	12.000000	0	98.93%
C	6	13	13.003355	1/2	1.07%
N	7	14	14.003074	1	99.632%
N	7	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%

Accepted Answer

The reaction in which two hydrogen atoms combine with two neutrons to form the [latex] ^{4}_{ 2}H e[/latex] atom is

[latex]2 ^{1}_{1}H+2n ightarrow ^{4}_{2}He. [/latex]

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

[latex]\begin{matrix}\begin{matrix}2 ^{1}_{1}H & +\ 2(1.007825) u \end{matrix} &\begin{matrix} 2n & ightarrow \2(1.008665) u \end{matrix} & \begin{matrix} ^{4}_{2}He\ 4.002602u \end{matrix} \end{matrix} [/latex]

The mass lost when two [latex]^{1}_{ 1}H [/latex] atoms combine with two neutrons to form the [latex] ^{4}_{ 2}H e[/latex] isotope is

[latex]\Delta m = 2(1.007825) u + 2(1.008665) u − 4.002602 u = 0.030378 u.[/latex]

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	[latex]2.99792458 × 10^{8} m/s[/latex]
Charge of electron	e	[latex]1.6021773 × 10^{−19}[/latex] C
Plank’s constant	h	[latex]6.626076 × 10^{−34} [/latex] J s
		[latex]4.135670 × 10^{−15}[/latex] eV s
	[latex] \hbar=h/2π[/latex]	[latex]1.054573 × 10^{−34}[/latex] J s
		[latex]6.582122 × 10^{−16}[/latex] eV s
	hc	1239.8424 eV nm
		1239.8424 MeV fm
Hydrogen ionization energy		13.605698 eV
Rydberg constant		[latex]1.0972 × 10^{5} cm^{−1}[/latex]
Bohr radius	[latex]a_{0} = (4π \epsilon _{0})/(me²) [/latex]	[latex]5.2917725 × 10^{−11} [/latex] m
Bohr magneton	[latex]μ_{B}[/latex]	[latex]9.2740154 × 10^{−24}[/latex] J/T
		[latex]5.7883826 × 10^{−5}[/latex] eV/T
Nuclear magneton	[latex]μ_{N}[/latex]	[latex]5.0507865 × 10^{−27}[/latex] J/T
		[latex]3.1524517 × 10^{−8}[/latex] eV/T
Fine structure constant	[latex]α = e^{2}/(4π\epsilon _{0} c \hbar)[/latex]	1/137.035989
	[latex]e^{2}/4π\epsilon _{0}[/latex]	1.439965 eV nm
Boltzmann constant	k	[latex]1.38066 × 10^{−23}[/latex] J/K
		[latex]8.6174 × 10^{−5}[/latex] eV/K
Avogadro’s constant	[latex]N_{A}[/latex]	[latex]6.022137 × 10^{23}[/latex] mole
Stefan-Boltzmann constant	σ	[latex]5.6705 × 10^{−8} W/m² K^{4}[/latex]
Particle masses
	kg	u	MeV/c²
Electron	[latex]9.1093897 × 10^{−31} [/latex]	[latex]5.485798 × 10^{−4} [/latex]	0.5109991
Proton	[latex]1.6726231 × 10^{−27} [/latex]	1.00727647	938.2723
Neutron	[latex]1.674955 × 10^{−27} [/latex]	1.008664924	939.5656
Deuteron	[latex]3.343586 × 10^{−27} [/latex]	2.013553	1875.6134
Conversion factors
	1 eV	[latex]1.6021773 × 10^{−19} [/latex] J
	1 u	931.4943 MeV/c²
		[latex]1.6605402 × 10^{−27} [/latex] kg
	1 atomic unit	27.2114 eV

Question 15.2: Using the atomic masses of the 1 1H and 4 2H e isotopes give...

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