The Density of an Iron Nucleus

Q. Given the mass of iron nucleus as 55.85 u and A = 56, and taking R₀ as 1.2 fm, what is the nuclear density, in kg/m³? expects 2.29 × 10¹⁷ kg/m³

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Q. Given the mass of iron nucleus as 55.85 u $m = 55.85$ u and A = 56 $A = 56$ , and taking R₀ as 1.2 fm $R_{0} = 1.2$ fm, what is the nuclear density, in kg/m³?

The playground

Inputs

mass

m

55.85 u

nucleons

A

size

R_{0}

1.2 fm

Outputs

density

ρ

The solution

Solved with NuclearSize

Given

$m = 55.85$ u
$A = 56$
$R_{0} = 1.2$ fm

Find $ρ$ .

Working

$m = 55.85$ u $= 9.27411069 \times 10^{−26}$ kg
$R_{0} = 1.2$ fm $= 1.2 \times 10^{−15}$ m
radius grows as the cube root(iron) $R = R_{0} \cdot A^{0.333333333}$
$R = 1.2 \times 10^{−15} \cdot 56^{0.333333333}$
$R = 4.59103484 \times 10^{−15}$ m
density of a nucleus(iron) $ρ = \frac{m}{\frac{4 \cdot π \cdot R^{3}}{3}}$
$ρ = \frac{9.27411069 \times 10^{−26}}{\frac{4 \cdot 3.14159265 \cdot {(4.59103484 \times 10^{−15})}^{3}}{3}}$
$ρ = 2.288 \times 10^{17}$ kg/m³
$ρ \approx 2.29 \times 10^{17}$ kg/m³to 3 significant figures

Conclusion

A nucleus's volume grows with the number of nucleons, so every nucleus has about the same enormous density, whatever the element.

The Ruby

1require_relative "../physics"23question :the_density_of_an_iron_nucleus do4  kind :numerical5  describes "Given the mass of iron nucleus as %{mass} and A = %{nucleons}, and taking R₀ as %{size}, %{density}, in " \6            "kg/m³?"7  given mass: "55.85 u"8  given nucleons: "56"9  given size: "1.2 fm"10  asks density: "what is the nuclear density"11end

1require_relative "../questions/the_density_of_an_iron_nucleus"2require_relative "../science/nuclei/nuclear_size"34solve :the_density_of_an_iron_nucleus, using: NuclearSize do5  objects :iron67  take mass: iron.mass8  take nucleons: iron.mass_number9  take size: R_01011  density = iron.nuclear_density12  answer density:, in: "kg/m3", figures: 31314  conclusion "A nucleus's volume grows with the number of nucleons, so every nucleus has about the same " \15             "enormous density, whatever the element."16end

1require_relative "../solutions/the_density_of_an_iron_nucleus"23play_with :the_density_of_an_iron_nucleus do4  title "The Density of an Iron Nucleus"5end

The laws

quantity	written	units	within	from
pi	$π$	—	0 … ∞	NuclearSize
coulomb constant	$k$	N m² C⁻²	—	NuclearSize
elementary charge	$e$	C	—	NuclearSize
size of a nucleon s share	$R_{0}$	m	—	NuclearSize
mass number	$A$	—	0 … ∞	NuclearSize
atomic number	$Z$	—	0 … ∞	NuclearSize
mass	$m$	kg	0 … ∞	NuclearSize
nuclear radius	$R$	m	—	NuclearSize
nuclear density	$ρ$	kg/m³	—	NuclearSize
coulomb barrier	$U_{b}$	J	—	NuclearSize
nuclear radius of ground	$R_{g}$	m	—
mass number of ground	$A_{g}$	—	0 … ∞
nuclear radius of iron	$R$	m	—
mass number of iron	$A$	—	0 … ∞
nuclear density of ground	$ρ_{g}$	kg/m³	—
mass of ground	$m_{g}$	kg	0 … ∞
nuclear density of iron	$ρ$	kg/m³	—
mass of iron	$m$	kg	0 … ∞
coulomb barrier of ground	$U_{b,g}$	J	—
atomic number of ground	$Z_{g}$	—	0 … ∞
coulomb barrier of iron	$U_{b}$	J	—
atomic number of iron	$Z$	—	0 … ∞

R_{n} = R_{0} \cdot {A_{n}}^{0.333333333}

radius grows as the cube root

ρ_{n} = \frac{m_{n}}{\frac{4 \cdot π \cdot {R_{n}}^{3}}{3}}

density of a nucleus

U_{b,n} = \frac{k \cdot {(Z_{n} \cdot e)}^{2}}{2 \cdot R_{n}}

two alike nuclei just touching

R_{g} = R_{0} \cdot {A_{g}}^{0.333333333}

radius grows as the cube root(ground)

R = R_{0} \cdot A^{0.333333333}

radius grows as the cube root(iron)

ρ_{g} = \frac{m_{g}}{\frac{4 \cdot π \cdot {R_{g}}^{3}}{3}}

density of a nucleus(ground)

ρ = \frac{m}{\frac{4 \cdot π \cdot R^{3}}{3}}

density of a nucleus(iron)

U_{b,g} = \frac{k \cdot {(Z_{g} \cdot e)}^{2}}{2 \cdot R_{g}}

two alike nuclei just touching(ground)

U_{b} = \frac{k \cdot {(Z \cdot e)}^{2}}{2 \cdot R}

two alike nuclei just touching(iron)