Question

Niobium (Nb) has an atomic radius of 0.1430 nm and a density of
8.57 g/cm^3. Determine whether it has an FCC or a BCC crystal
structure.

Answer #1

density = 8.57 g/cm3

atomic radius = 0.1430 nm

if Nb is BCC.

4r = sqrt(3)a

a = 4 r /sqrt(3) = 4 x 0.1430 / 1.732

= 0.330 nm

= 0.33 x 10^-7 cm

Atomic mass = 92.906 g/mol

density = Z M / N a^3

= 2 x 92.906 / 6.023 x 10^23 x (0.33 x 10^-7)^3

= 8.57 g/cm3

here the density is 8.57 g/cm3

this matches with the given density value. so this is
**confirmed to be BCC.**

Niobium (Nb) has an atomic radios of 0.1430 nm and a density of
8.57 g/cm^3. Determine whether it has an FCC or a BCC crystal
structure.

1. Aluminum and iron are both important structural metals.
Suppose they are processed into the FCC and BCC structures. Of
these four possible structures (FCC Al, BCC Al, FCC Fe, BCC Fe),
which one has the a) lowest unit cell volume, and b) lowest
theoretical density? The radius of Al is 0.1431 nm and the radius
of Fe is 0.1241 nm.
2. An unknown metal has a BCC structure, density of 7.25 g/cm3,
and atomic weight of 50.99 g/mol. Determine...

The concentration of vacancies in niobium (Nb)
at 765˚C is 9.836 x 1018 cm-3. If niobium has
a BCC crystal structure and a lattice parameter of 0.3302 nm,
calculate the energy for vacancy formation in
niobium.

Show if lead is an FCC or a BCC crystal structure.
Lead (Pb) has an atomic radius of 0.175 nm and a density
of 11.35 g/cm3 . Use this information to determine whether it has
an FCC or a BCC crystal structure.
Please also provide the correct units.

Below are listed the atomic weight, density, and atomic radius
for three hypothetical alloys at room temperature. For each
determine whether its crystal structure is FCC, BCC, or simple
cubic and then justify your answer.
Alloy
Atomic weight
Density
Atomic radius
(g/mol)
(g/cm3)
(nm)
A
195.08
21.45
0.139
B
209
9.32
0.335
C
55.85
7.87
0.124

Niobium has a density of 8.57 g/cm3 and crystallizes
with the body-centered cubic unit cell. Calculate the radius of a
niobium atom.

Calculate the radius of an silver atom in cm, given that Ag has
an FCC crystal structure, a density of 10.5 g/cm3, and
an atomic weight of 107.87 g/mol.

1) For a metal that has the face-centered cubic (FCC) crystal
structure, calculate the atomic radius if the metal has a density
of (8.000x10^0) g/cm3 and an atomic weight of
(5.80x10^1) g/mol. Express your answer in
nm.
2) Consider a copper-aluminum solid solution containing
(7.82x10^1) at% Al. How many atoms per cubic centimeter
(atoms/cm^3) of copper are there in this solution?
Take the density of copper to be 8.94 g/cm3 and the
density of aluminum to be 2.71 g/cm3.

compute the atomic weight of an atom whose theoretical density
is 133g/mol .the atomic radius of an atom is 0.1445 nano meters.the
unit cell of the atom has FCC crystal structure

--Given Values--
Atomic Radius (nm) = 0.116
FCC Metal = Gold
BCC Metal: = Sodium
Temperature ( C ) = 1017
Metal A = Tin
Equilibrium Number of Vacancies (m-3) = 6.02E+23
Temperature for Metal A = 369
Metal B = Gallium
1) If the atomic radius of a metal is the value shown above and it
has the face-centered cubic crystal structure, calculate the volume
of its unit cell in nm3? Write your answers in
Engineering Notation.
...

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