The unit of length convenient on the nuclear scale is a fermi: 1 f = 10^{–15} m. Nuclear sizes obey roughly the following empirical relation :**r = r _{0} **

Asked by Abhisek | 1 year ago | 294

Radius of the nucleus

r = r_{0} A^{1/3}

r_{o} = 1.2 f = 1.2 x 10^{-15} m

Considering the nucleus is spherical. Volume of nucleus

= \( \dfrac{4}{3}\) πr^{3} = \( \dfrac{4}{3}\) π [r_{0} A^{1/3}]^{3} = \( \dfrac{4}{3}\) πr_{0}^{3}A

Mass of nucleus = mA

m is the average mass of the nucleon

A is the number of nucleons

Nuclear mass density = Mass of nucleus/Volume of nucleus

= \(( \dfrac{mA}{\dfrac{4}{3πr^3}} )\) = \(
\dfrac{ 3mA}{4πr^3}\)^{ }= \(
\dfrac{ 3mA}{4πr_0^3A}\)

= \( \dfrac{ 3m}{4πr_0^3}\)

Using m = 1.66 x 10^{-27} kg and r_{o} = 1.2 f

= 1.2 x 10^{-15} m in the above equation

= \( \dfrac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times ( 1.2 \times 10^{-15})^3}\)

= \( \dfrac{4.98 \times 10^{-27}}{21. 703 \times 10^{-45}}= 2.29 \times 10^{17} kg/m^3\)

So, the nuclear mass density is much larger than atomic mass density for a sodium atom we got in 2.27.

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