Two Reacting Hydrogen Isotopes

Q. The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb – mC – md]c², where the masses refer to the respective nuclei. Given m(¹₁H) = 1.007825 u, m(²₁H) = 2.014102 u and m(³₁H) = 3.016049 u, what is the Q-value of the reaction ¹₁H + ³₁H → ²₁H + ²₁H, in MeV? expects -4.03 MeV

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Q. The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb – mC – md]c², where the masses refer to the respective nuclei. Given m(¹₁H) = 1.007825 u $m_{h} = 1.007825$ u, m(²₁H) = 2.014102 u $m_{d} = 2.014102$ u and m(³₁H) = 3.016049 u $m_{t} = 3.016049$ u, what is the Q-value of the reaction ¹₁H + ³₁H → ²₁H + ²₁H ${dm}_{r} = m_{h} + m_{t} - 2 \cdot m_{d}$ , in MeV?

The playground

Inputs

hydrogen

m_{h}

1.008 u

deuterium

m_{d}

2.014 u

tritium

m_{t}

3.016 u

Outputs

Q_{r}

The solution

Solved with MassEnergy

Given

$m_{h} = 1.007825$ u
$m_{d} = 2.014102$ u
$m_{t} = 3.016049$ u
${dm}_{r} = m_{h} + m_{t} - 2 \cdot m_{d}$ from “¹₁H + ³₁H → ²₁H + ²₁H”

Find $Q_{r}$ .

Working

$m_{h} = 1.007825$ u $= 1.67353278 \times 10^{−27}$ kg
$m_{d} = 2.014102$ u $= 3.34449506 \times 10^{−27}$ kg
$m_{t} = 3.016049$ u $= 5.00826719 \times 10^{−27}$ kg
¹₁H + ³₁H → ²₁H + ²₁H ${dm}_{r} = m_{h} + m_{t} - 2 \cdot m_{d}$
${dm}_{r} = 1.67353278 \times 10^{−27} + 5.00826719 \times 10^{−27} - 2 \cdot 3.34449506 \times 10^{−27}$
${dm}_{r} = −7.19013416 \times 10^{−30}$ kg
energy a reaction gives out(reaction) $Q_{r} = {dm}_{r} \cdot c^{2}$
$Q_{r} = (−7.19013416 \times 10^{−30}) \cdot {(2.99792458 \times 10^{8})}^{2}$
$Q_{r} = −4.033$ MeV
$Q_{r} = −6.46217031 \times 10^{−13}$ J $= −4.03336946$ MeV
$Q_{r} \approx −4.03$ MeVto 3 significant figures

Conclusion

The two deuterium nuclei weigh more than the nuclei that made them, so the reaction must be given energy to happen: it is endothermic.

The Ruby

1require_relative "../physics"23question :two_reacting_hydrogen_isotopes do4  kind :numerical5  describes "The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb – mC – md]c², where the " \6            "masses refer to the respective nuclei. Given m(¹₁H) = %{hydrogen}, m(²₁H) = %{deuterium} and " \7            "m(³₁H) = %{tritium}, %{q} of the reaction ¹₁H + ³₁H → ²₁H + ²₁H, in MeV?"8  given hydrogen: "1.007825 u"9  given deuterium: "2.014102 u"10  given tritium: "3.016049 u"11  asks q: "what is the Q-value"12end

1require_relative "../questions/two_reacting_hydrogen_isotopes"2require_relative "../science/nuclei/mass_energy"34solve :two_reacting_hydrogen_isotopes, using: MassEnergy do5  objects :reaction, :hydrogen, :deuterium, :tritium67  take hydrogen: hydrogen.mass8  take deuterium: deuterium.mass9  take tritium: tritium.mass1011  deduce_from "¹₁H + ³₁H → ²₁H + ²₁H",12              reaction.mass_defect => hydrogen.mass + tritium.mass - 2 * deuterium.mass1314  q = reaction.q_value15  answer q:, in: "MeV", figures: 31617  conclusion "The two deuterium nuclei weigh more than the nuclei that made them, so the reaction must be given " \18             "energy to happen: it is endothermic."19end

1require_relative "../solutions/two_reacting_hydrogen_isotopes"23play_with :two_reacting_hydrogen_isotopes do4  title "Two Reacting Hydrogen Isotopes"5end

The laws

quantity	written	units	within	from
speed of light	$c$	m/s	—	MassEnergy
hydrogen atom mass	$m_{H}$	kg	—	MassEnergy
neutron mass	$m_{n}$	kg	—	MassEnergy
mass	$m$	kg	0 … ∞	MassEnergy
energy	$E$	J	—	MassEnergy
atomic number	$Z$	—	—	MassEnergy
mass number	$A$	—	0 … ∞	MassEnergy
mass defect	$dm$	kg	—	MassEnergy
binding energy	$E_{b}$	J	—	MassEnergy
q value	$Q$	J	—	MassEnergy
energy of ground	$E_{g}$	J	—
mass of ground	$m_{g}$	kg	0 … ∞
energy of reaction	$E_{r}$	J	—
mass of reaction	$m_{r}$	kg	0 … ∞
energy of hydrogen	$E_{h}$	J	—
mass of hydrogen	$m_{h}$	kg	0 … ∞
energy of deuterium	$E_{d}$	J	—
mass of deuterium	$m_{d}$	kg	0 … ∞
energy of tritium	$E_{t}$	J	—
mass of tritium	$m_{t}$	kg	0 … ∞
mass defect of ground	${dm}_{g}$	kg	—
atomic number of ground	$Z_{g}$	—	—
mass number of ground	$A_{g}$	—	0 … ∞
mass defect of reaction	${dm}_{r}$	kg	—
atomic number of reaction	$Z_{r}$	—	—
mass number of reaction	$A_{r}$	—	0 … ∞
mass defect of hydrogen	${dm}_{h}$	kg	—
atomic number of hydrogen	$Z_{h}$	—	—
mass number of hydrogen	$A_{h}$	—	0 … ∞
mass defect of deuterium	${dm}_{d}$	kg	—
atomic number of deuterium	$Z_{d}$	—	—
mass number of deuterium	$A_{d}$	—	0 … ∞
mass defect of tritium	${dm}_{t}$	kg	—
atomic number of tritium	$Z_{t}$	—	—
mass number of tritium	$A_{t}$	—	0 … ∞
binding energy of ground	$E_{b,g}$	J	—
binding energy of reaction	$E_{b,r}$	J	—
binding energy of hydrogen	$E_{b,h}$	J	—
binding energy of deuterium	$E_{b,d}$	J	—
binding energy of tritium	$E_{b,t}$	J	—
q value of ground	$Q_{g}$	J	—
q value of reaction	$Q_{r}$	J	—
q value of hydrogen	$Q_{h}$	J	—
q value of deuterium	$Q_{d}$	J	—
q value of tritium	$Q_{t}$	J	—

E_{b} = m_{b} \cdot c^{2}

einsteins mass energy

{dm}_{n} = Z_{n} \cdot m_{H} + (A_{n} - Z_{n}) \cdot m_{n} - m_{n}

mass lost in binding

E_{b,n} = {dm}_{n} \cdot c^{2}

energy of the lost mass

Q_{r} = {dm}_{r} \cdot c^{2}

energy a reaction gives out

E_{g} = m_{g} \cdot c^{2}

einsteins mass energy(ground)

E_{r} = m_{r} \cdot c^{2}

einsteins mass energy(reaction)

E_{h} = m_{h} \cdot c^{2}

einsteins mass energy(hydrogen)

E_{d} = m_{d} \cdot c^{2}

einsteins mass energy(deuterium)

E_{t} = m_{t} \cdot c^{2}

einsteins mass energy(tritium)

{dm}_{g} = Z_{g} \cdot m_{H} + (A_{g} - Z_{g}) \cdot m_{n} - m_{g}

mass lost in binding(ground)

{dm}_{r} = Z_{r} \cdot m_{H} + (A_{r} - Z_{r}) \cdot m_{n} - m_{r}

mass lost in binding(reaction)

{dm}_{h} = Z_{h} \cdot m_{H} + (A_{h} - Z_{h}) \cdot m_{n} - m_{h}

mass lost in binding(hydrogen)

{dm}_{d} = Z_{d} \cdot m_{H} + (A_{d} - Z_{d}) \cdot m_{n} - m_{d}

mass lost in binding(deuterium)

{dm}_{t} = Z_{t} \cdot m_{H} + (A_{t} - Z_{t}) \cdot m_{n} - m_{t}

mass lost in binding(tritium)

E_{b,g} = {dm}_{g} \cdot c^{2}

energy of the lost mass(ground)

E_{b,r} = {dm}_{r} \cdot c^{2}

energy of the lost mass(reaction)

E_{b,h} = {dm}_{h} \cdot c^{2}

energy of the lost mass(hydrogen)

E_{b,d} = {dm}_{d} \cdot c^{2}

energy of the lost mass(deuterium)

E_{b,t} = {dm}_{t} \cdot c^{2}

energy of the lost mass(tritium)

Q_{g} = {dm}_{g} \cdot c^{2}

energy a reaction gives out(ground)

Q_{r} = {dm}_{r} \cdot c^{2}

energy a reaction gives out(reaction)

Q_{h} = {dm}_{h} \cdot c^{2}

energy a reaction gives out(hydrogen)

Q_{d} = {dm}_{d} \cdot c^{2}

energy a reaction gives out(deuterium)

Q_{t} = {dm}_{t} \cdot c^{2}

energy a reaction gives out(tritium)

{dm}_{r} = m_{h} + m_{t} - 2 \cdot m_{d}

¹₁H + ³₁H → ²₁H + ²₁H