Splitting an Iron Nucleus

solved MassEnergy

Q. Suppose, we think of fission of an iron nucleus, ⁵⁶₂₆Fe, into two equal fragments ${dm}_{f} = m_{i} - 2 \cdot m_{a}$ , ²⁸₁₃Al. Given m(⁵⁶₂₆Fe) = 55.93494 u $m_{i} = 55.93494$ u and m(²⁸₁₃Al) = 27.98191 u $m_{a} = 27.98191$ u, what is the Q-value of the process, in MeV?

Inputs

iron

m_{i}

55.93 u

aluminium

m_{a}

27.98 u

Outputs

Q_{f}

How it's solved

Solved with MassEnergy

Given

$m_{i} = 55.93494$ u
$m_{a} = 27.98191$ u
${dm}_{f} = m_{i} - 2 \cdot m_{a}$ from “into two equal fragments”

Find $Q_{f}$ .

Working

$m_{i} = 55.93494$ u $= 9.28821531 \times 10^{−26}$ kg
$m_{a} = 27.98191$ u $= 4.64650547 \times 10^{−26}$ kg
into two equal fragments ${dm}_{f} = m_{i} - 2 \cdot m_{a}$
${dm}_{f} = 9.28821531 \times 10^{−26} - 2 \cdot 4.64650547 \times 10^{−26}$
${dm}_{f} = −4.79563682 \times 10^{−29}$ kg
energy a reaction gives out(fission) $Q_{f} = {dm}_{f} \cdot c^{2}$
$Q_{f} = (−4.79563682 \times 10^{−29}) \cdot {(2.99792458 \times 10^{8})}^{2}$
$Q_{f} = −26.9$ MeV
$Q_{f} = −4.31010343 \times 10^{−12}$ J $= −26.9015497$ MeV
$Q_{f} \approx −26.9$ MeVto 3 significant figures

Conclusion

The two fragments together weigh more than the iron nucleus, so splitting it would need energy rather than give it: the fission is not energetically possible.

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 :splitting_an_iron_nucleus do4  kind :numerical5  describes "Suppose, we think of fission of an iron nucleus, ⁵⁶₂₆Fe, into two equal fragments, ²⁸₁₃Al. Given " \6            "m(⁵⁶₂₆Fe) = %{iron} and m(²⁸₁₃Al) = %{aluminium}, %{q} of the process, in MeV?"7  given iron: "55.93494 u"8  given aluminium: "27.98191 u"9  asks q: "what is the Q-value"10end

solution.rb

1require_relative "../questions/splitting_an_iron_nucleus"2require_relative "../science/nuclei/mass_energy"34solve :splitting_an_iron_nucleus, using: MassEnergy do5  objects :fission, :iron, :aluminium67  take iron: iron.mass8  take aluminium: aluminium.mass910  deduce_from "into two equal fragments", fission.mass_defect => iron.mass - 2 * aluminium.mass1112  q = fission.q_value13  answer q:, in: "MeV", figures: 31415  conclusion "The two fragments together weigh more than the iron nucleus, so splitting it would need energy " \16             "rather than give it: the fission is not energetically possible."17end

playground.rb

1require_relative "../solutions/splitting_an_iron_nucleus"23play_with :splitting_an_iron_nucleus do4  title "Splitting an Iron Nucleus"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 fission	$E_{f}$	J	—
mass of fission	$m_{f}$	kg	0 … ∞
energy of iron	$E_{i}$	J	—
mass of iron	$m_{i}$	kg	0 … ∞
energy of aluminium	$E_{a}$	J	—
mass of aluminium	$m_{a}$	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 fission	${dm}_{f}$	kg	—
atomic number of fission	$Z_{f}$	—	—
mass number of fission	$A_{f}$	—	0 … ∞
mass defect of iron	${dm}_{i}$	kg	—
atomic number of iron	$Z_{i}$	—	—
mass number of iron	$A_{i}$	—	0 … ∞
mass defect of aluminium	${dm}_{a}$	kg	—
atomic number of aluminium	$Z_{a}$	—	—
mass number of aluminium	$A_{a}$	—	0 … ∞
binding energy of ground	$E_{b,g}$	J	—
binding energy of fission	$E_{b,f}$	J	—
binding energy of iron	$E_{b,i}$	J	—
binding energy of aluminium	$E_{b,a}$	J	—
q value of ground	$Q_{g}$	J	—
q value of fission	$Q_{f}$	J	—
q value of iron	$Q_{i}$	J	—
q value of aluminium	$Q_{a}$	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_{f} = m_{f} \cdot c^{2}

einsteins mass energy(fission)

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

einsteins mass energy(iron)

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

einsteins mass energy(aluminium)

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

mass lost in binding(ground)

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

mass lost in binding(fission)

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

mass lost in binding(iron)

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

mass lost in binding(aluminium)

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

energy of the lost mass(ground)

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

energy of the lost mass(fission)

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

energy of the lost mass(iron)

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

energy of the lost mass(aluminium)

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

energy a reaction gives out(ground)

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

energy a reaction gives out(fission)

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

energy a reaction gives out(iron)

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

energy a reaction gives out(aluminium)

{dm}_{f} = m_{i} - 2 \cdot m_{a}

into two equal fragments