A Copper Coin Pulled Apart

solved MassEnergy

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?

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}

How it's solved

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 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 :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

solution.rb

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

playground.rb

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 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 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