/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q3P Copper, a monovalent metal, has ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Copper, a monovalent metal, has molar mass 63.54 g/mol and density 8.96g/cm3. What is the number density nof conduction electrons in copper?

Short Answer

Expert verified

Conduction electrons, in copper, have a number density n=8.49×1028m-3.

Step by step solution

01

The given data

  1. Molar mass of copper,A=63.54g/mol
  2. Density of copper,d=8.96g/cm3
02

Understanding the concept of density and molar mass

The mass of one mole of a substance is called molar mass. To get the mass of the copper atom, we divide the molar mass of copper by the Avogadro number. Now, the number of atoms per unit volume can be calculated by the formula for the density of the substance.

Formulae:

The mass of an atom, M=A/NA...............................1

where NA=6.022×1023mol-1 and A is the molar mass.

The number density of conduction electrons,n=dM..................................2

Where d= density of the atom,M = mass of a single atom

03

Calculation of the number density of conductions electrons of copper

Since each atom contributes one conduction electron, the number of atoms per unit volume is equal to the number of conduction electrons per unit volume.

Now, the mass of the copper atom can be given using the data of molar mass in equation (1) as follows:

M=63.54g/mol6.022×1023mol-1=1.055×10-22g

Thus, the value of the number density of the conductions electrons of the copper atom can be calculated using the given data in equation (2) as follows:

n=8.96g/cm31.055×10-22g=8.49×1028m-3

Hence, the value of the number density of free electrons is 8.49×1028m-3.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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

(a) Show that the density of states at the Fermi energy is given by

N(EF)=4(31/3)(π2/3)(mn1/3)h2=(4.11×1018m-2eV-1)n1/3

in which nis the number density of conduction electrons.

(b) Calculate N(EF)for copper, which is a monovalent metal with molar mass 63.54g/mol and density 8.96g/cm3.

Verify your calculation with the curve of Fig. 41-6, recalling that EF=7.0eV=for copper.

Figure 41-1ashows 14 atoms that represent the unit cell of copper. However, because each of these atoms is shared with one or more adjoining unit cells, only a fraction of each atom belongs to the unit cell shown. What is the number of atoms per unit cell for copper? (To answer, count up the fractional atoms belonging to a single unit cell.)

A sample of a certain metal has a volume of 4.0×10-5m3. The metal has a density of 9.0 g/cm3and a molar mass of 60 g/mol. The atoms are bivalent. How many conduction electrons (or valence electrons) are in the sample?

Calculate N0(E)the density of occupied states, for copper at 10000K for energy Eof (a)4.00eV , (b) 6.75eV, (c) 7.00eV, (d) 7.25eV, and (e) 9.00eV. Compare your results with the graph of Fig. 41-8b.The Fermi energy for copper is 7.00eV.

A certain computer chip that is about the size of a postage stamp 2.54cm×2.22cmcontains about 3.5 million transistors. If the transistors are square, what must be their maximumdimension? (Note:Devices other than transistors are also on the chip, and there must be room for the interconnections among the circuit elements. Transistors smaller than 0.7μmare now commonly and inexpensively fabricated.)

See all solutions

Recommended explanations on Physics Textbooks

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.