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Given data:

The density of mercury (D) is given as $13.6g/ml$. The diameter of a mercury atom is$a=2.9004\times {{10}^{-x}}cm$which is also equal to the edge length of the cube inside which this atom is placed. The shape is somewhat like this:

We also know the mass of a mercury atom. It is $200.59u$ where “u” is the atomic mass unit. Converting it into grams it becomes

\[200.59u\times 1.66\times {{10}^{-24}}=3.329\times {{10}^{-22}}g\]

The volume (V) of atom is $\dfrac{4}{3}\pi {{r}^{3}}$

\[\begin{align}

& \Rightarrow V=\dfrac{4}{3}\times 3.14\times {{(1.4502\times {{10}^{-X}})}^{3}} \\

& \Rightarrow V=12.768\times {{10}^{-3X}}c{{m}^{3}} \\

\end{align}\]

Now we have the values of all the variables required for this question. Below we put them together systematically and try to solve them. We start with the formula for density:

\[\begin{align}

& Density(D)=\dfrac{Mass(M)}{Volume(V)}=\dfrac{3.329\times {{10}^{-22}}g}{12.768\times {{10}^{-3x}}c{{m}^{3}}} \\

& D=0.26\times {{10}^{-22+3x}}g/c{{m}^{3}} \\

& \text{As we know density is given as D=13}\text{.6g/ml or 13}\text{.6c}{{\text{m}}^{3}} \\

& \text{So the above equation becomes:} \\

& \Rightarrow 13.6=0.26\times {{10}^{-22+3x}} \\

& \Rightarrow {{10}^{-22+3x}}=\dfrac{13.6}{0.26}=52.3 \\

& \text{Taking the log of both sides we get:} \\

& \Rightarrow (3x-22)log(10)=log(52.3) \\

& \Rightarrow 3x-22=1.718 \\

& \Rightarrow 3x=23.718 \\

& \Rightarrow x=7.906\approx 8 \\

\end{align}\]

The value of the unknown x = 8

So, the diameter of the atom or the edge of the cube $a=2.9004\times {{10}^{-8}}cm$.

As you can see in the picture represented at the start, the volume of the cube with the given value of edge length is more than that of the sphere.

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