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Assuming that each nucleus is roughlyspherical and that its mass is roughly equal to A (in atomic mass units {\rm u}), what is the density rho of a nucleus with nucleon number A?
Express your answer in terms ofA,R_0, and pi.
I know rho= F/Area... but whatabout in this case where related with the nucleon number??! IS itthe same A tho?!

Respuesta :

Answer:

ρ/ρ2 = 3 / R₀       the two densities are different

Explanation:

Density is defined as

       ρ = M / V

As the nucleus is spherical

       V = 4/3 π r³

Let's replace

      ρ = A / (4/3 π R₀³)

      ρ = ¾ A / π R₀³

b)

      ρ2 = F / area

The area of ​​a sphere is

     A = 4π R₀²

     ρ2 = F / 4π R₀²

     ρ2 = F / 4π R₀²

Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.

Let's look for the relationship of the two densities

     ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)

     ρ /ρ2 = 3 (A / F) (1 / R₀)

In this case it does not say that the nucleon number is A (F = A), the relationship is

     ρ/ρ2 = 3 / R₀

I see that the two densities are different

okpalawalter8

The density and relationship with the nucleon number is mathematically given as

d= 3/4 A / π R₀^3

d/d2 = 3 / R₀

What is the density and relationship with the nucleon number?

Question Parameter(s):

Generally, the equation for the Density   is mathematically given as

 d = M / V

Where

V = 4/3 π r^3

Hence

d = A / (4/3 π R₀^3)

d= 3/4 A / π R₀^3

In conclusion,

d2 = F / area

Hence

d2 = F / 4π R₀^2

d /d2 = 3 (A / F) (1 / R₀)

d/d2 = 3 / R₀

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