Question

Given the mass of iron nucleus as 55.85u and a=56. Find the nuclear density

Answer 1

Answer:

According to rutherford's relations radius of nucleus = ro×A13 ro = constant = 1.2×10−15 m

therefore

r = 1.2×10−15 ×56−−√3

r = 4.59 ×10−15 m

also,

1u = 1.67×10−27 kg

55.85 u = 9.327 ×10−26 kg = mass of nucleus of iron atom

nuclear density = mass of nucleus of iron atomvolume of nucleus

nuclear density = 9.327 ×10−26 kg43π×r3 =9.327 ×10−26 kg43π×(4.59 ×10−15 )3

nuclear density = 4.851×1018 kg/m^3

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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.