A Nucleus Three Times the Unit Radius

Q. The radius of a nucleus is found to be 3.6 fm. Taking R₀ = 1.2 fm, how many nucleons does it hold? expects 27

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

radius

R

3.6 fm

unit radius

R_{0}

1.2 fm

Outputs

nucleons

A

The solution

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 Ruby

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

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

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

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)