Answer to Question #163791 in Mechanics | Relativity for Sage Topaz

Question

ONE GRAM OF GODCAN BE BEATEN OUT INTO A FOIL 1.O M2 IN AREA. HOW MANY ATOMS THICK IS SUCH A FOIL? THE MASS OF GOLD ATOM IS 3.27 x 10 –25 KG , DENSITY OF GOLD 1.9 X 104 KG.

Answer 1

Explanations & Calculations

• If the thickness of the sheet is $\small h(m)$, then the mass of it can be expressed as (by $\small m=V\rho$ )

$\qquad\qquad \begin{aligned} \small (h\times1)m^3\times1.9\times10^4\,kgm^{-3}&=\small 0.001kg\\ \small &hence\\ \small h&= \small 5.263158\times10^{-8}\,m \end{aligned}$

• Assuming a complete gold particle to be a solid sphere of radius $\small r(A^o)$,

$\qquad\qquad \begin{aligned} \small \frac{4}{3}\pi (r)^3\times(1.9\times10^{4}\,kgm^{-3})&= \small 3.27\times10^{-25}\,kg\\ \small r&= \small 1.601563\times 10^{-10}\,m\\ &= \small 1.601563A^0 \end{aligned}$

• Assuming that the height of the sheet consists of some $\small n$ number of piled gold atoms,

$\qquad\qquad \begin{aligned} \small n\times r& \small h\\ \small n &= \small \frac{5.263158\times10^{-8}\,m}{1.601563\times10^{-10}\,m}\\ &= \small 328.63\,atoms\\ &\approx\small \bold{328 \,atoms} \end{aligned}$