Respuesta :
Answer:
Density of 127 I = [tex]\rm 1.79\times 10^{14}\ g/cm^3.[/tex]
Also, [tex]\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]
Explanation:
Given, the radius of a nucleus is given as
[tex]\rm r=kA^{1/3}[/tex].
where,
- [tex]\rm k = 1.3\times 10^{-13} cm.[/tex]
- A is the mass number of the nucleus.
The density of the nucleus is defined as the mass of the nucleus M per unit volume V.
[tex]\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.[/tex]
For the nucleus 127 I,
Mass, M = [tex]\rm 2.1\times 10^{-22}\ g.[/tex]
Mass number, A = 127.
Therefore, the density of the 127 I nucleus is given by
[tex]\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.[/tex]
On comparing with the density of the solid iodine,
[tex]\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]
