Two Reacting Hydrogen Isotopes

solved MassEnergy

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?

Inputs

hydrogen

m_{h}

1.008 u

deuterium

m_{d}

2.014 u

tritium

m_{t}

3.016 u

Outputs

Q_{r}

How it's solved

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 web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

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

solution.rb

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

playground.rb

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

The laws it drew on

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