A Copper Coin Pulled Apart

Q. A given coin has a mass of 3.0 g. For simplicity assume that the coin is entirely made of copper atoms (A = 63, Z = 29) of mass 62.92960 u. Taking mH = 1.007825 u and mn = 1.008665 u, what is the nuclear energy that would be required to separate all the neutrons and protons from each other, in MeV? expects 1.58 × 10²⁵ MeV

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Q. A given coin has a mass of 3.0 g $m_{c} = 3.0$ g. For simplicity assume that the coin is entirely made of copper atoms $E_{b,c} = \frac{m_{c}}{m_{Cu}} \cdot E_{b,Cu}$ (A = 63 $A_{Cu} = 63$ , Z = 29 $Z_{Cu} = 29$ ) of mass 62.92960 u $m_{Cu} = 62.92960$ u. Taking mH = 1.007825 u $m_{H} = 1.007825$ u and mn = 1.008665 u $m_{n} = 1.008665$ u, what is the nuclear energy that would be required to separate all the neutrons and protons from each other, in MeV?

The playground

Inputs

coin

m_{c}

3 g

nucleons

A_{Cu}

protons

Z_{Cu}

atom

m_{Cu}

62.93 u

hydrogen

m_{H}

1.008 u

neutron

m_{n}

1.009 u

Outputs

energy

E_{b,c}

The solution

Solved with MassEnergy

Given

$m_{c} = 3.0$ g
$A_{Cu} = 63$
$Z_{Cu} = 29$
$m_{Cu} = 62.92960$ u
$m_{H} = 1.007825$ u
$m_{n} = 1.008665$ u
$E_{b,c} = \frac{m_{c}}{m_{Cu}} \cdot E_{b,Cu}$ from “entirely made of copper atoms”

Find $E_{b,c}$ .

Working

$m_{c} = 3.0$ g $= 0.003$ kg
$m_{Cu} = 62.92960$ u $= 1.04497059 \times 10^{−25}$ 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(copper) ${dm}_{Cu} = Z_{Cu} \cdot m_{H} + (A_{Cu} - Z_{Cu}) \cdot m_{n} - m_{Cu}$
${dm}_{Cu} = 29 \cdot 1.67353278 \times 10^{−27} + (63 - 29) \cdot 1.67492764 \times 10^{−27} - 1.04497059 \times 10^{−25}$
${dm}_{Cu} = 9.82931192 \times 10^{−28}$ kg
energy of the lost mass(copper) $E_{b,Cu} = {dm}_{Cu} \cdot c^{2}$
$E_{b,Cu} = 9.82931192 \times 10^{−28} \cdot {(2.99792458 \times 10^{8})}^{2}$
$E_{b,Cu} = 8.834145 \times 10^{−11}$ J
entirely made of copper atoms $E_{b,c} = \frac{m_{c}}{m_{Cu}} \cdot E_{b,Cu}$
$E_{b,c} = \frac{0.003}{1.04497059 \times 10^{−25}} \cdot 8.834145 \times 10^{−11}$
$E_{b,c} = 1.583 \times 10^{25}$ MeV
$E_{b,c} = 2.53618955 \times 10^{12}$ J $= 1.58296501 \times 10^{25}$ MeV
$E_{b,c} \approx 1.58 \times 10^{25}$ MeVto 3 significant figures

Conclusion

Every one of the coin's countless copper nuclei must be pulled apart, so even a small coin holds an enormous nuclear binding energy.

The Ruby

1require_relative "../physics"23question :a_copper_coin_pulled_apart do4  kind :numerical5  describes "A given coin has a mass of %{coin}. For simplicity assume that the coin is entirely made of copper " \6            "atoms (A = %{nucleons}, Z = %{protons}) of mass %{atom}. Taking mH = %{hydrogen} and mn = %{neutron}, " \7            "%{energy}, in MeV?"8  given coin: "3.0 g"9  given nucleons: "63"10  given protons: "29"11  given atom: "62.92960 u"12  given hydrogen: "1.007825 u"13  given neutron: "1.008665 u"14  asks energy: "what is the nuclear energy that would be required to separate all the neutrons and protons from " \15               "each other"16end

1require_relative "../questions/a_copper_coin_pulled_apart"2require_relative "../science/nuclei/mass_energy"34solve :a_copper_coin_pulled_apart, using: MassEnergy do5  objects :coin, copper: :Cu67  take coin: coin.mass8  take nucleons: copper.mass_number9  take protons: copper.atomic_number10  take atom: copper.mass11  take hydrogen: m_H12  take neutron: m_n1314  deduce_from "entirely made of copper atoms", coin.binding_energy => coin.mass / copper.mass * copper.binding_energy1516  energy = coin.binding_energy17  answer energy:, in: "MeV", figures: 31819  conclusion "Every one of the coin's countless copper nuclei must be pulled apart, so even a small coin holds " \20             "an enormous nuclear binding energy."21end

1require_relative "../solutions/a_copper_coin_pulled_apart"23play_with :a_copper_coin_pulled_apart do4  title "A Copper Coin Pulled Apart"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 coin	$E_{c}$	J	—
mass of coin	$m_{c}$	kg	0 … ∞
energy of copper	$E_{Cu}$	J	—
mass of copper	$m_{Cu}$	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 coin	${dm}_{c}$	kg	—
atomic number of coin	$Z_{c}$	—	—
mass number of coin	$A_{c}$	—	0 … ∞
mass defect of copper	${dm}_{Cu}$	kg	—
atomic number of copper	$Z_{Cu}$	—	—
mass number of copper	$A_{Cu}$	—	0 … ∞
binding energy of ground	$E_{b,g}$	J	—
binding energy of coin	$E_{b,c}$	J	—
binding energy of copper	$E_{b,Cu}$	J	—
q value of ground	$Q_{g}$	J	—
q value of coin	$Q_{c}$	J	—
q value of copper	$Q_{Cu}$	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_{c} = m_{c} \cdot c^{2}

einsteins mass energy(coin)

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

einsteins mass energy(copper)

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

mass lost in binding(ground)

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

mass lost in binding(coin)

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

mass lost in binding(copper)

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

energy of the lost mass(ground)

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

energy of the lost mass(coin)

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

energy of the lost mass(copper)

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

energy a reaction gives out(ground)

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

energy a reaction gives out(coin)

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

energy a reaction gives out(copper)

E_{b,c} = \frac{m_{c}}{m_{Cu}} \cdot E_{b,Cu}

entirely made of copper atoms