Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 132

Look up the diameter of a silicon atom, in pm. The latest semiconductor chips have fabricated lines as small as \(14 \mathrm{nm}\). How many silicon atoms does this correspond to?

Short Answer

Expert verified
The diameter of a silicon atom is approximately 220 pm, which is 0.22 nm. In a 14 nm fabrication line, there are approximately 64 silicon atoms.

Step by step solution

01

Find the diameter of a silicon atom in pm

Look up the diameter of a silicon atom in a reliable source like a textbook, scientific article, or reputable online source. The diameter of a silicon atom is approximately 220 pm.
02

Convert the diameter of the silicon atom to nanometers

Now, we need to convert the diameter of the silicon atom from picometers to nanometers. There are 1000 pm in 1 nm. To convert the diameter of a silicon atom to nm, divide its diameter (in pm) by 1000. \( Diameter_{nm} = \frac{Diameter_{pm}}{1000} \) \( Diameter_{nm} = \frac{220}{1000} \) \( Diameter_{nm} = 0.22 \, nm \) The diameter of a silicon atom is 0.22 nm.
03

Calculate the number of silicon atoms in a 14 nm fabrication line

To calculate the number of silicon atoms in a 14 nm fabrication line, divide the diameter of the fabrication line (14 nm) by the diameter of a silicon atom (0.22 nm). \( Number \, of \, atoms = \frac{14 \, nm}{0.22 \, nm} \) \( Number \, of \, atoms \approx 63.64 \) Since we can't have a fraction of an atom, we round to the nearest whole number: 64 atoms.
04

State the result

In a 14 nm fabrication line, there are approximately 64 silicon atoms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Gold adopts a face-centered cubic structure with a unit cell edge of 408 pm (Figure 12.11). How many gold atoms are there in a sphere that is $20 \mathrm{nm}\( in diameter? Recall that the volume of a sphere is \)\frac{4}{3} \pi r^{3}$.

An ester is a compound formed by a condensation reaction between a carboxylic acid and an alcohol that eliminates a water molecule. Read the discussion of esters in Section 24.4 and then give an example of a reaction forming an ester. How might this kind of reaction be extended to form a polymer (a polyester)?

At room temperature and pressure RbCl crystallizes with the NaCl-type structure. (a) Use ionic radii to predict the length of the cubic unit cell edge. (b) Use this value to estimate the density. (c) At high temperature and pressure, the structure transforms to one with a CsCl-type structure. Use ionic radii to predict the length of the cubic unit cell edge for the high- pressure form of RbCl. (d) Use this value to estimate the density. How does this density compare with the density you calculated in part (b)?

Indicate whether each statement is true or false: (a) Substitutional alloys tend to be more ductile than interstitial alloys. (b) Interstitial alloys tend to form between elements with similar ionic radii. (c) Nonmetallic elements are never found in alloys.

Calculate the volume in \(\AA^{3}\) of each of the following types of cubic unit cells if it is composed of atoms with an atomic radius of \(182 \mathrm{pm}\). (a) primitive (b) face-centered cubic.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free