A Nitrogen Nucleus's Binding Energy

Q. What is the binding energy of a nitrogen nucleus (A = 14, Z = 7), given its mass 14.00307 u, with mH = 1.007825 u and mn = 1.008665 u, in MeV? expects 104.7 MeV

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Q. What is the binding energy of a nitrogen nucleus (A = 14 $A = 14$ , Z = 7 $Z = 7$ ), given its mass 14.00307 u $m = 14.00307$ u, with mH = 1.007825 u $m_{H} = 1.007825$ u and mn = 1.008665 u $m_{n} = 1.008665$ u, in MeV?

The playground

Inputs

nucleons

A

protons

Z

mass

m

14 u

hydrogen

m_{H}

1.008 u

neutron

m_{n}

1.009 u

Outputs

energy

E_{b}

The solution

Solved with MassEnergy

Given

$A = 14$
$Z = 7$
$m = 14.00307$ u
$m_{H} = 1.007825$ u
$m_{n} = 1.008665$ u

Find $E_{b}$ .

Working

$m = 14.00307$ u $= 2.32526448 \times 10^{−26}$ kg
$m_{H} = 1.007825$ u $= 1.67353278 \times 10^{−27}$ kg
$m_{n} = 1.008665$ u $= 1.67492764 \times 10^{−27}$ kg
mass lost in binding(nitrogen) $dm = Z \cdot m_{H} + (A - Z) \cdot m_{n} - m$
$dm = 7 \cdot 1.67353278 \times 10^{−27} + (14 - 7) \cdot 1.67492764 \times 10^{−27} - 2.32526448 \times 10^{−26}$
$dm = 1.8657817 \times 10^{−28}$ kg
energy of the lost mass(nitrogen) $E_{b} = dm \cdot c^{2}$
$E_{b} = 1.8657817 \times 10^{−28} \cdot {(2.99792458 \times 10^{8})}^{2}$
$E_{b} = 104.7$ MeV
$E_{b} = 1.67688096 \times 10^{−11}$ J $= 104.662677$ MeV
$E_{b} \approx 104.7$ MeVto 4 significant figures

Conclusion

The nucleus weighs less than the protons and neutrons it is made of; the missing mass is the energy that binds them, and the energy it would take to pull them apart.

The Ruby

1require_relative "../physics"23question :a_nitrogen_nucleus_s_binding_energy do4  kind :numerical5  describes "%{energy} of a nitrogen nucleus (A = %{nucleons}, Z = %{protons}), given its mass %{mass}, with " \6            "mH = %{hydrogen} and mn = %{neutron}, in MeV?"7  given nucleons: "14"8  given protons: "7"9  given mass: "14.00307 u"10  given hydrogen: "1.007825 u"11  given neutron: "1.008665 u"12  asks energy: "What is the binding energy"13end

1require_relative "../questions/a_nitrogen_nucleus_s_binding_energy"2require_relative "../science/nuclei/mass_energy"34solve :a_nitrogen_nucleus_s_binding_energy, using: MassEnergy do5  objects :nitrogen67  take nucleons: nitrogen.mass_number8  take protons: nitrogen.atomic_number9  take mass: nitrogen.mass10  take hydrogen: m_H11  take neutron: m_n1213  energy = nitrogen.binding_energy14  answer energy:, in: "MeV", figures: 41516  conclusion "The nucleus weighs less than the protons and neutrons it is made of; the missing mass is the energy " \17             "that binds them, and the energy it would take to pull them apart."18end

1require_relative "../solutions/a_nitrogen_nucleus_s_binding_energy"23play_with :a_nitrogen_nucleus_s_binding_energy do4  title "A Nitrogen Nucleus's Binding Energy"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 nitrogen	$E$	J	—
mass of nitrogen	$m$	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 nitrogen	$dm$	kg	—
atomic number of nitrogen	$Z$	—	—
mass number of nitrogen	$A$	—	0 … ∞
binding energy of ground	$E_{b,g}$	J	—
binding energy of nitrogen	$E_{b}$	J	—
q value of ground	$Q_{g}$	J	—
q value of nitrogen	$Q$	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 = m \cdot c^{2}

einsteins mass energy(nitrogen)

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

mass lost in binding(ground)

dm = Z \cdot m_{H} + (A - Z) \cdot m_{n} - m

mass lost in binding(nitrogen)

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

energy of the lost mass(ground)

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

energy of the lost mass(nitrogen)

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

energy a reaction gives out(ground)

Q = dm \cdot c^{2}

energy a reaction gives out(nitrogen)