Question

A monolayer (single layer of atoms) of Cuon the (100) crystal face shown in the margin has$\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}$atoms$\mathbf{/}{\mathbf{cm}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{54}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{mol}\mathbf{/}{\mathbf{cm}}^{\mathbf{2}}$. What current can deposit one layer of Cuatoms on$\mathbf{1}{\mathbf{cm}}^{\mathbf{2}}$in 1 s?

Short answer

The current required to form a monolayer of copper atoms is 0.4.90mA.

Step-by-step solution

Define electric circuit

At any point in an electric circuit the charge is the product of current I and time t.

q = I.t

The unit of q is Coulombs, I is Ampere and t is seconds

$\mathrm{q}=\mathrm{molesofelectrons}\times \mathrm{Faraday}\text{'}\mathrm{sconstant}$

Determine the current required to form a monolayer of copper atoms.

Number of electrons required to deposit copper is two. Then number of moles of copper is

Moles of copper$=2\times 2.54\times {10}^{-9}$

$=5.08\times {10}^{-9}\mathrm{mol}$

$$\begin{gathered} \mathrm{q}=\mathrm{molesofelectrons}\times \mathrm{Faraday}\mathrm{constant} \\ \mathrm{It}=\mathrm{molesofelectrons}\times \mathrm{Faraday}\text{'}\mathrm{sconstant} \\ \mathrm{I}=\frac{\mathrm{moles}\mathrm{of}\mathrm{electrons}\times \mathrm{Faraday}\mathrm{constant}}{\mathrm{t}} \\ \mathrm{I}=\frac{(5.08\times {10}^{-8}\mathrm{mol}\times 96485\mathrm{C}/\mathrm{mol})}{1\mathrm{s}} \end{gathered}$$

$\mathrm{I}=0.490\mathrm{mA}$