Question

given the mass of iron nucleus as 55.85 amu and A=56 find nuclear density

with explanation

Answer 1

Hi friend,

The formula is

Density (n)= Mass of nuclei/nuclear volume

Nuclear volume= 4/3 π r³

To find radius:-

r= r° (a)^¹/³

R= 1.3×10^-15(56)^1/3

R=3.825×1.3×10^-15

R=4.97×10^-15

R³==122.76×10^-45

4/3πr³= 513.96×10^-45

Density = mass/volume

Density = 55×1.6×10^-27

Density= 88×10^18/513.96

Density = 0.1812×10^18

Density =1.812×10^17 kg /m³.

Hope this helped you a little!!!

Answer 2

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\purple{R}}}}$.

• ${mFe}$=${55.85}$ ${u}$=${9.27×10^-26 kg}$
• ${Nuclear density}$=$\frac{Mass}{Volume}$ =$\frac{9.27×10^-26}{(4\pi/3)(1.2×10^-15)^3}$×$\frac{1}{56}$
• ${2.29×10^17kgm^-3}$
• The density of matter in neutron starts (an astrophysical object) is comparable to this density.
• This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.