EP PHYSICS F/SCI.+ENGR.W/MOD..-MOD MAST
4th Edition
ISBN: 9780133899634
Author: GIANCOLI
Publisher: PEARSON CO
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Question
Chapter 40, Problem 8Q
To determine
Why the sodium chloride is a good insulator?
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Students have asked these similar questions
The ionic crystal of NaCl has an fcc structure as
shown below with the position of the Cl atom
located at : (0 0 0); (½ ½ 0); ( ½ 0 %) and (0
½) and Na atom is located at : (½ ½½) + fcc
translation.
The following are the physical data of NaCl:
- Ionic radius Na = 0.98 A, Cl ion radius
1.81 A
Atomic mass of Na = 22.99 amu, Atomic
mass of Cl = 35.45 amu
NaCl bulk modulus = 2.40 x 1010 N/m
- Madelung's constant = 1.75
Young's modulus in the direction [100] = 5
x 1010 Nm-2
Debye temperature = 281 K
-
State the structural factors of NaCl, F in
fNa and fCl (f = atomic scattering factor).
Also determine the condition (h k 1) so that
the value of F = 0.
The elements A and B crystallize in bcc and fcc structures, respectively. Suppose that (i) the crystals of A
and B have the same density, and (ii) the nearest-neighbor distances in these crystals are equal. Calculate the
ratio MA/MB, where MA and MB denote the masses of the A and B atoms, respectively.
Consider a face centered cubic (fcc) crystal in which one atom is placed at each lattice point. The lattice constant is a = 0.7 \: nma=0.7nm. Calculate the volume density of atoms (i.e. number of atoms per unit volume) in units of cm^{-3}cm−3. Values within 5% error will be considered correct.
Chapter 40 Solutions
EP PHYSICS F/SCI.+ENGR.W/MOD..-MOD MAST
Ch. 40.4 - Determine the three lowest rotational energy...Ch. 40.6 - Prob. 1BECh. 40.6 - Prob. 1CECh. 40.8 - Prob. 1DECh. 40 - What type of bond would you expect for (a) the N2...Ch. 40 - Describe how the molecule CaCl2 could be formed.Ch. 40 - Does the H2 molecule have a permanent dipole...Ch. 40 - Although the molecule H3 is not stable, the ion...Ch. 40 - The energy of a molecule can be divided into four...Ch. 40 - Would you expect the molecule H2+ to be stable? If...
Ch. 40 - Explain why the carbon atom (Z = 6) usually forms...Ch. 40 - Prob. 8QCh. 40 - Prob. 9QCh. 40 - Prob. 10QCh. 40 - Prob. 11QCh. 40 - Prob. 12QCh. 40 - Prob. 13QCh. 40 - Prob. 14QCh. 40 - Prob. 15QCh. 40 - Prob. 16QCh. 40 - Prob. 17QCh. 40 - Prob. 18QCh. 40 - Prob. 19QCh. 40 - Prob. 20QCh. 40 - Prob. 21QCh. 40 - Prob. 22QCh. 40 - Prob. 23QCh. 40 - Prob. 1PCh. 40 - (II) The measured binding energy of KCl is 4.43eV....Ch. 40 - (II) Estimate the binding energy of the H2...Ch. 40 - (II) The equilibrium distance r0 between two atoms...Ch. 40 - Prob. 5PCh. 40 - Prob. 6PCh. 40 - (III) (a) Apply reasoning similar to that in the...Ch. 40 - (I) Show that the quantity 2/I has units of...Ch. 40 - Prob. 9PCh. 40 - Prob. 10PCh. 40 - Prob. 11PCh. 40 - Prob. 12PCh. 40 - Prob. 13PCh. 40 - Prob. 14PCh. 40 - Prob. 15PCh. 40 - Prob. 16PCh. 40 - (II) Calculate the bond length for the NaCl...Ch. 40 - Prob. 18PCh. 40 - Prob. 19PCh. 40 - Prob. 20PCh. 40 - Prob. 21PCh. 40 - Prob. 22PCh. 40 - Prob. 23PCh. 40 - Prob. 24PCh. 40 - Prob. 25PCh. 40 - Prob. 26PCh. 40 - Prob. 27PCh. 40 - Prob. 28PCh. 40 - Prob. 29PCh. 40 - Prob. 30PCh. 40 - Prob. 31PCh. 40 - Prob. 32PCh. 40 - Prob. 33PCh. 40 - Prob. 34PCh. 40 - Prob. 35PCh. 40 - Prob. 36PCh. 40 - Prob. 37PCh. 40 - Prob. 38PCh. 40 - Prob. 39PCh. 40 - Prob. 40PCh. 40 - Prob. 41PCh. 40 - Prob. 42PCh. 40 - Prob. 43PCh. 40 - Prob. 44PCh. 40 - Prob. 45PCh. 40 - Prob. 46PCh. 40 - Prob. 47PCh. 40 - Prob. 48PCh. 40 - Prob. 49PCh. 40 - Prob. 50PCh. 40 - Prob. 51PCh. 40 - Prob. 52PCh. 40 - Prob. 53PCh. 40 - Prob. 54PCh. 40 - Prob. 55PCh. 40 - Prob. 56PCh. 40 - Prob. 57PCh. 40 - Prob. 58PCh. 40 - Prob. 59PCh. 40 - Prob. 60PCh. 40 - Prob. 61PCh. 40 - Prob. 62GPCh. 40 - Prob. 63GPCh. 40 - Prob. 64GPCh. 40 - Prob. 65GPCh. 40 - Prob. 66GPCh. 40 - Prob. 67GPCh. 40 - Prob. 68GPCh. 40 - Prob. 69GPCh. 40 - Prob. 70GPCh. 40 - Prob. 71GPCh. 40 - Prob. 72GPCh. 40 - Prob. 73GPCh. 40 - Prob. 74GPCh. 40 - Prob. 75GPCh. 40 - Prob. 76GPCh. 40 - Prob. 77GPCh. 40 - Prob. 78GPCh. 40 - Prob. 79GPCh. 40 - Prob. 80GPCh. 40 - Prob. 81GPCh. 40 - Prob. 82GPCh. 40 - Prob. 83GPCh. 40 - Prob. 84GPCh. 40 - Prob. 85GPCh. 40 - Prob. 86GPCh. 40 - Prob. 87GPCh. 40 - Prob. 88GPCh. 40 - Prob. 89GP
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Similar questions
- In solid KCI the smallest distance between the centers of a. potassium ion and a chloride ion is 314 pm. Calculate the length of the edge of the unit cell and the density of KCI, assuming it has the same structure as sodium chloride.arrow_forwardCalculate the radius of a nickel atom in cm, given that Ni has an FCC crystal structure, a density of 7.982 g/cm³, and an atomic weight of 58.69 g/mol.arrow_forwardBody-centered cubic structure. (a) Show that for BCC the lattice length a in terms of the atomic radius is 4R/√3. (b) Calculate the volume of a BCC unit cell in terms of the atomic radius R. (c) Show that the atomic packing factor for the BCC crystal structure is 0.68.arrow_forward
- Evaluate the Packing fractions of the sc, fcc, bcc and hcp crystals.arrow_forward3. (a) Determine the percentage of ionic polarizability in the sodium chloride crystal which hasthe optical index of refraction and the static dielectric constant as 6.5 and 6.6respectively.arrow_forwardSuppose a pure Si crystal has 5 × 1028 atoms m-3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that ni =1.5 × 1016 m-3.arrow_forward
- The crystal structure of sodium chloride (NaCI) is face centered cubic. If the minimum separation between the Na* and Cl ion is 'r', then assuming the lattice only upto the two nearest neighbouring ions, the Madelung constant will be -0.8665 0.8665 -1.7476. O 1.7476arrow_forwardThe interplanar spacing of (220) planes of a FCC structure is 1.7458 Å. Calculate the lattice constant. a) 4.983 Å b) 2.458 Å c) 0 d) 5.125 Åarrow_forwardAt 23° K, the crystal structure transforms from BCC to HCP keeping density same. The cubic face in BCC is a 3.42 A and in HCP is-1.53. The lattice constant in HCP structure is, (n) 2.64 A (b) 3.16 A () 2.16 A (4) 2.42 Aarrow_forward
- (e) Intrinsic silicon has effective densities of states in the conduction band and the valence band of 3.2 × 10¹⁹ cm−³ and 1.8 × 10¹⁹ cm-³, respectively. If the band gap is 1.12 eV, what is the concentration of intrinsic charge carriers in silicon at 300 K? A. 9.46 x 10⁹ m-³ 9.46 x 10⁹ cm-³ 0 m-3 2.40 x 1019 cm-3 B. C. D.arrow_forwardWhen the ions at their equilibrium interionic separation, the force of attraction between a divalent (valency of 2) cation and a monovalent (valency of 1) anion is 7.32 × 10−9 N. If the ionic radius of the cation is 0.05 nm, what is the anion radius (nm)? Round your result to 2 decimal place.arrow_forwardCu Assume that the crystal structure of metallic copper (Cu) results in a density of atoms p = 8.46 × 10²m 3. Each Cu atom in the crystal donates one electron to the conduction band, which leads, for the 3-D Fermi gas, to a densityu of states g(ɛ) = 2 x = ( 2 m ² ) ² 1/2 where m is the effective mass of the conduction electrons. In the low temperature limit (i.c. T = 0 K), find the Fermi energy E, in units of eV. You may assume m* to be equal to the free electron mass marrow_forward
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