Answer:
Using
Radius = RoA^ 1/3
And we are given
To= 1.42 fm A= 208
So R = 1.4E-15 x 208^1/3
Radius= 8.4fm
But we know that density =mass/volume
So D = 3/4 x A/ πR³
The approximate radius R is [tex]7\times10^{-15}[/tex] and the density of a nucleus with nucleon number A is [tex]\rm 3M/4\pi R^3_0A[/tex] and this can be determined by using the given data.
Given :
Lead (Pb) atomic mass number = 208
[tex]\rm R_0[/tex] = 1.25 fm
The following steps can be used in order to determine the approximate radius R and the density of a nucleus with nucleon number A:
Step 1 - The approximate radius R can be determined by using the below formula:
[tex]\rm R = R_0A^{1/3}[/tex]
Step 2 - Substitute the value of the known terms in the above formula.
[tex]\rm R = \left((1.2) \times\left(\dfrac{1\times 10^{-15}}{1}\right) \right)\times (208)^{1/3}[/tex]
Step 3 - Simplify the above expression.
[tex]\rm R = 7\times 10^{-15}[/tex]
Step 4 - The formula of density is used in order to determine the density of a nucleus with nucleon number A.
[tex]\rm Density = \dfrac{Mass}{Volume}[/tex] --- (1)
Step 5 - The volume of a sphere is given below:
[tex]\rm V = \dfrac{4}{3}\pi R^3[/tex]
Step 6 - Now, substitute the value of known terms in the expression (1).
[tex]\rm Density = \dfrac{M}{\dfrac{4}{3}\pi (R_0A^{1/3})^3}[/tex]
Step 7 - Simplify the above expression.
[tex]\rm Density = \dfrac{3M}{4\pi R^3_0A}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/4206478
