Silver (atomic weight $=108 \mathrm{~g} \mathrm{~mol}^{-1}$ ) has a density of $10.5 \mathrm{~g}…

Silver (atomic weight $=108 \mathrm{~g} \mathrm{~mol}^{-1}$ ) has a density of $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12} \mathrm{~m}^2$ can be expressed in scientific notation as $y \times 10^x$. The value of $x$ is

Solution

Volume of one mole of silver atoms $=\frac{108}{10.5} \mathrm{~cm}^3 /$ mole Volume of one silver atom $=\frac{108}{10.5} \times \frac{1}{6.022 \times 10^{23}} \mathrm{~cm}^3$ So, $ \begin{aligned} \frac{4}{3} \pi R^3 & =\frac{108}{10.5} \times \frac{1}{6.022 \times 10^{23}}=1.708 \times 10^{-23} \\ R^3 & =0.407 \times 10^{-23} \mathrm{~cm}^3 \\ & =0.407 \times 10^{-29} \mathrm{~m}^3 \end{aligned} $ Area of each silver atom $ \pi R^2=\pi \times\left(0.407 \times 10^{-29} \mathrm{~m}^3\right)^{2 / 3} $ So number of silver atoms in given area $ =\frac{10^{-12}}{\left(0.407 \times 10^{-29} \mathrm{~m}^3\right)^{2 / 3}}=\frac{10^8}{(\pi \times 2)}=1.6 \times 10^7=y \times 10^x \quad \text { so, } \quad x=7 $ Solid state Mathematical understanding III

Asked in: JEE Advanced 2010 (Paper 2)

Practice more Solid State questions on Aicharya