Splitting an Iron Nucleus

Q. Suppose, we think of fission of an iron nucleus, ⁵⁶₂₆Fe, into two equal fragments, ²⁸₁₃Al. Given m(⁵⁶₂₆Fe) = 55.93494 u and m(²⁸₁₃Al) = 27.98191 u, what is the Q-value of the process, in MeV? expects -26.9 MeV

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The web of concepts

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The equations it used

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Check an answer

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?

The playground

Inputs

iron

m_{i}

55.93 u

aluminium

m_{a}

27.98 u

Outputs

Q_{f}

The solution

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 Ruby

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

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

1require_relative "../solutions/splitting_an_iron_nucleus"23play_with :splitting_an_iron_nucleus do4  title "Splitting an Iron Nucleus"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 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