Question

Gold has a density of 19 700 kg m-3. 193 g of gold

contains 6.02 × 1023 atoms. Use this information

to estimate the volume of a gold atom, and hence

its radius. State any assumptions you make.

Answer 1

$6.02 \times 10^{23}$ atoms of gold $= 193 \times 10^{-3} \;kg$

mass of one atom of gold $= \frac{193 \times 10^{-3}}{6.02 \times 10^{23}}$

$= 32.06 \times 10^{-26} \;kg$

Volume of one atom of gold :

$V= \frac{Mass \;of\; one \;atom \;of \;gold}{Density\; of\; gold} \\ V= \frac{32.06 \times 10^{-26}}{19700} = 1.63 \times 10^{-29} \;m^3$

Assumption: gold atom is spherical.

Volume of one atom of gold $= \frac{4}{3} \pi R^3$

R= radius of atom of gold

$V= \frac{4}{3} \pi R^3 \\ R =( \frac{3V}{4 \pi})^{\frac{1}{3}} \\ R =( \frac{3 \times 1.63 \times 10^{-29}}{4 \times 3.14})^{\frac{1}{3}} \\ R= (0.3889 \times 10^{-29})^{\frac{1}{3}} \\ R= 1.57 \times 10^{-10} \;m$