As computer processor speeds increase, it is necessary for engineers to increase the number of circuit elements packed into a given area. Individual circuit elements are often connected using very small copper "wires" deposited directly onto the surface of the chip. In some current generation processors, these copper interconnects are about 45 nm wide. What mass of copper would be in a 1-mm length of such an interconnect, assuming a square cross-section. (The density of copper is 8.96 g/cm3.)

Question

contestada

Respuesta :

ACCESS MORE ACCESS MORE ACCESS MORE ACCESS MORE

MrRoyal MrRoyal · Answer 1 · 2020-02-21T04:11:08+01:00

Answer:

1.72E11 atomCu

Explanation:

Assumption: A square intersection

Since the cross section wire is assumed to be a square intersection, the height and the width of the wire are the same (equal)

and height = h = 45nm

Convert 45nm to centimetres (cm), we have

45nm * 10^-9m * 100cm / (1nm * 1m)

= 4.5 * 10^-6cm

Length. = 1mm

Convert 1mm to centimetre (cm) as well, we have

1mm * 1m * 100cm/(1000mm * 1m)

= 0.1cm

Now, we'll calculate the volume of the wire.

Using

Volume = length * height * width ---_--_ width = height

So,

Volume = 0.1 * 4.5 * 10^-6 * 4.5 * 10^-6

Volume = 2.025E−12 cm³

Lastly, we'll calculate the numbers of copper atoms using the following steps

1. We'll begin with the volume of the wire

2. We'll use the density of copper as a conversion factor to convert cubic centimetres (cm³) of copper to grams of copper.

3. We'll use the molar mass of copper as a conversion factor to grams of copper to moles of copper.

4. We'll use the Avogadro's number as a conversion factor to convert moles of copper to atoms of copper.

So, we have

2.025E−12 cm³Cu * 8.96g/cm³ * 1molCu * 6.022E23atomCu / (1cm³Cu * 63.546gCu * 1 molCu)

= 1.72E11 atomCu