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Question
Niobium crystallises as body centered cube (BCC) and has density of 8.55kg/dm3. Calculate the atomic radius of niobium. (Given: Atomic mass of niobium=93).
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Solution
Density of Niobium(Nb) crystal is =8.55kg/dm3.
Crystalline structure is bcc.
Atomic mass no. of Nb=93g/mol.
Avogadro numberNA=6.022×1023mol−1.
Atomic radius of Niobium=?
In bcc unit cells, there are 8 atoms at 8-corners and 1 atom at the body centre.
∴ Number of Nb atoms=18×8+1=1+1=2
Mass of one Nb atom=a26.022×1023
=1.544×10−22kg
∴ Mass of 2Nb atoms=2×1.544×10−22
=3.088×10−22kg
Mass of unit cell=Mass of 2Nb atoms=3.088×10−22g
If a is edge length of bcc unit cell.
Volume of unit cell=a3.
Density=MassofunitcellVolumeofunitcell
d=3.088×10−22a3
∴a3=3.088×10−22d
=3.088×10−228.55=0.361×10−22dm3
=36.1×10−24dm3
∴a=(36.1×10−24)1/3
=3.33×10−8dm.
If r is the radius of 1Nb atom, then in bcc structure
r=√34a=√34×3.33×10−8
=1.43×10−8dm
or =14.3nm(where, 1dm=109nm)
Hence, the atomic radius of niobium is 14.3nm.
2) The first law of thermodynamics states that energy can neither be created nor destroyed. The total energy of an isolated system (cannot exchange energy with its surroundings) remains the same.
Mathematical expression:
In a system that exchanges both heat and mechanical work with its surroundings, the total energy or the energy change is described by taking both-heat and work in it's account.
U2−U1=ΔU=q+W
q=ΔU−W
ΔU=q+W
This is the mathematical expression for first law of thermodynamics, from the point of view of conservation of energy. This expression justifies first law of thermodynamics.
where, U2−U1=Energies of the system in the final and initial states, respectively
ΔU=Change in internal energy
q= heat absorbed or evolved by the system
W= Work done by the system or on the system.
Reactions involved in the zone of reduction:
3Fe2O3+Co→2Fe3O4+CO2o
Fe3O4+Co→3FeO+CO2
FeO+Co→Fe+CO2.
Crystalline structure is bcc.
Atomic mass no. of Nb=93g/mol.
Avogadro numberNA=6.022×1023mol−1.
Atomic radius of Niobium=?
In bcc unit cells, there are 8 atoms at 8-corners and 1 atom at the body centre.
∴ Number of Nb atoms=18×8+1=1+1=2
Mass of one Nb atom=a26.022×1023
=1.544×10−22kg
∴ Mass of 2Nb atoms=2×1.544×10−22
=3.088×10−22kg
Mass of unit cell=Mass of 2Nb atoms=3.088×10−22g
If a is edge length of bcc unit cell.
Volume of unit cell=a3.
Density=MassofunitcellVolumeofunitcell
d=3.088×10−22a3
∴a3=3.088×10−22d
=3.088×10−228.55=0.361×10−22dm3
=36.1×10−24dm3
∴a=(36.1×10−24)1/3
=3.33×10−8dm.
If r is the radius of 1Nb atom, then in bcc structure
r=√34a=√34×3.33×10−8
=1.43×10−8dm
or =14.3nm(where, 1dm=109nm)
Hence, the atomic radius of niobium is 14.3nm.
2) The first law of thermodynamics states that energy can neither be created nor destroyed. The total energy of an isolated system (cannot exchange energy with its surroundings) remains the same.
Mathematical expression:
In a system that exchanges both heat and mechanical work with its surroundings, the total energy or the energy change is described by taking both-heat and work in it's account.
U2−U1=ΔU=q+W
q=ΔU−W
ΔU=q+W
This is the mathematical expression for first law of thermodynamics, from the point of view of conservation of energy. This expression justifies first law of thermodynamics.
where, U2−U1=Energies of the system in the final and initial states, respectively
ΔU=Change in internal energy
q= heat absorbed or evolved by the system
W= Work done by the system or on the system.
Reactions involved in the zone of reduction:
3Fe2O3+Co→2Fe3O4+CO2o
Fe3O4+Co→3FeO+CO2
FeO+Co→Fe+CO2.
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