A Nucleus Three Times the Unit Radius

solved NuclearSize

Q. The radius of a nucleus is found to be 3.6 fm $R = 3.6$ fm. Taking R₀ = 1.2 fm $R_{0} = 1.2$ fm, how many nucleons does it hold?

Inputs

radius

R

3.6 fm

unit radius

R_{0}

1.2 fm

Outputs

nucleons

A

How it's solved

Solved with NuclearSize

Given

$R = 3.6$ fm
$R_{0} = 1.2$ fm

Find $A$ .

Working

$R = 3.6$ fm $= 3.6 \times 10^{−15}$ m
$R_{0} = 1.2$ fm $= 1.2 \times 10^{−15}$ m
radius grows as the cube root(nucleus) $R = R_{0} \cdot A^{0.333333333}$
$3.6 \times 10^{−15} = 1.2 \times 10^{−15} \cdot A^{0.333333333}$
$A = 27$
$A \approx 27$ to 2 significant figures

Conclusion

The radius goes as the cube root of the number of nucleons, so a nucleus three times as wide as R₀ holds three cubed, 27, of them: an aluminium nucleus.

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_nucleus_three_times_the_unit_radius do4  kind :numerical5  describes "The radius of a nucleus is found to be %{radius}. Taking R₀ = %{unit_radius}, %{nucleons}?"6  given radius: "3.6 fm"7  given unit_radius: "1.2 fm"8  asks nucleons: "how many nucleons does it hold"9end

solution.rb

1require_relative "../questions/a_nucleus_three_times_the_unit_radius"2require_relative "../science/nuclei/nuclear_size"34solve :a_nucleus_three_times_the_unit_radius, using: NuclearSize do5  objects :nucleus67  take radius: nucleus.nuclear_radius8  take unit_radius: R_0910  nucleons = nucleus.mass_number11  answer nucleons:, figures: 21213  conclusion "The radius goes as the cube root of the number of nucleons, so a nucleus three times as wide " \14             "as R₀ holds three cubed, 27, of them: an aluminium nucleus."15end

playground.rb

1require_relative "../solutions/a_nucleus_three_times_the_unit_radius"23play_with :a_nucleus_three_times_the_unit_radius do4  title "A Nucleus Three Times the Unit Radius"5end

The laws it drew on

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 nucleus	$R$	m	—
mass number of nucleus	$A$	—	0 … ∞
nuclear density of ground	$ρ_{g}$	kg/m³	—
mass of ground	$m_{g}$	kg	0 … ∞
nuclear density of nucleus	$ρ$	kg/m³	—
mass of nucleus	$m$	kg	0 … ∞
coulomb barrier of ground	$U_{b,g}$	J	—
atomic number of ground	$Z_{g}$	—	0 … ∞
coulomb barrier of nucleus	$U_{b}$	J	—
atomic number of nucleus	$Z$	—	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(nucleus)

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(nucleus)

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(nucleus)