For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.
n(Cd) / n(Fe)=1
Generally, the equation for the density is mathematically given as
d=\frac{A}{4/3}\piR^3
Therefore
n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3
n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3
n(Cd) / n(Fe)=1
In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same
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Compared to the nucleus ⁵⁶₂₆Fe, the density of the nucleus ¹¹²₄₈Cd is same.
Density of any nucleus will be calculated by using the below equation as:
ρ = m/V, where
m = mass of the atom
V = volume = 4/3πR³
For the given question calculation will be proceed as:
n(Cd) / n(Fe) = [m(Cd)/m(Fe)].[R(Fe)/R(Cd)]³
From the given formulas of iron and cadmuim we put values on the above equation then we get,
n(Cd) / n(Fe) = (112 / 56 ) × (1/1.26)3
n(Cd) / n(Fe) = 1
Hence the density of nucleus ¹¹²₄₈Cd is same as of nucleus ⁵⁶₂₆Fe nucleus.
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