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Compared to the nucleus 5626fe, what is the density of the nucleus 112 48cd?

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okpalawalter8

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

What is the density of the nucleus 112 48cd?

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.

How do we calculate the density of nucleus?

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.

To know more about nucleus density, visit the below link:
https://brainly.com/question/5722114

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