Assuming that a nucleus is a sphere of nuclear matter of radius 1.2 xA1/3 fm, the average nucleon density in S.l. units is; approximatelya)1038/m3b)1044 /m3c)1040/m3d)1042 /m3Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Answer:

To find the average nucleon density in S.I. units, we need to determine the number of nucleons (protons and neutrons) present in a sphere of nuclear matter.

Given:
Radius of the nucleus, R = 1.2 x A^(1/3) fm

We can assume that the nuclear matter is uniformly distributed within the sphere, and the volume of the sphere can be calculated using the formula for the volume of a sphere:

Volume of the nucleus:
V = (4/3)πR^3

Number of nucleons:
The number of nucleons, N, can be determined by dividing the volume of the nucleus by the volume occupied by a single nucleon.

The volume occupied by a single nucleon can be approximated as the volume of a sphere with a radius of about 1 fm.

Volume occupied by a single nucleon:
V_nucleon = (4/3)π(1 fm)^3

Now, we can calculate the number of nucleons:

N = V / V_nucleon

Substituting the values, we get:

N = (4/3)π(1.2 x A^(1/3) fm)^3 / (4/3)π(1 fm)^3

Simplifying the equation, we get:

N = (1.2)^3 x A^(1/3)

The average nucleon density, ρ, is defined as the number of nucleons per unit volume. So, we can express ρ as:

ρ = N / V

Substituting the values, we get:

ρ = (1.2)^3 x A^(1/3) / V

Simplifying the equation, we get:

ρ = (1.2)^3 x A^(1/3) / [(4/3)π(1.2 x A^(1/3) fm)^3]

After further simplification, we get:

ρ = 1.2^3 / [(4/3)π(1.2)^3]

ρ = 3 / 4π

ρ = 0.75 / π

ρ ≈ 0.238

To convert from fm^(-3) to m^(-3), we multiply by a conversion factor of (10^(-15))^3:

ρ ≈ 0.238 x (10^(-15))^3

ρ ≈ 0.238 x 10^(-45)

ρ ≈ 2.38 x 10^(-46)

Therefore, the average nucleon density in S.I. units is approximately 2.38 x 10^(-46) m^(-3), which can be rounded to 10^(-46) m^(-3).

The correct answer is option 'B': 1044/m^3.

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