JEE Class main Answered

. Niobium crystallizes in bcc structure. If its density is 8.55 g/cm3. Calculate its edge length. Atomic mass of Niobium = 93amu

Asked by remanikasohal24 | 14 Dec, 2020, 09:52: AM

Expert Answer

Given:
No. of particles in b.c.c. type unit cell (Z) = 2
Atomic mass of the element (M) = 93 g mol−193 g mol-1

Mass space of space the space unit space cell space equals space fraction numerator straight Z cross times straight M over denominator straight N subscript 0 end fraction space equals space fraction numerator 2 cross times 93 over denominator 6.022 cross times 10 to the power of 23 space end fraction space equals space 30.89 space cross times space 10 to the power of negative 23 end exponent straight g

Density of unit cell (ρ)=8.55g cm⁻³(ρ)=8.55g cm^-3

Volume space of space unit space cell space left parenthesis straight a cubed right parenthesis equals fraction numerator Mass space of space unit space cell over denominator Density space of space unit space cell end fraction equals space fraction numerator 30.89 cross times 10 to the power of negative 23 end exponent over denominator 8.55 end fraction space equals space 3.613 cross times 10 to the power of negative 23 end exponent cm cubed space equals space 36.13 cross times 10 to the power of negative 24 end exponent cm cubed

Edge length of unit cell (a) = (36.13×10⁻²⁴cm³)^1/3= 3.31×10⁻⁸cm = 331 pm

Step II. Calculation of radius of unit cell

For space straight b. straight c. straight c. space structure comma space straight r space equals fraction numerator square root of 3 over denominator 4 end fraction space equals space fraction numerator square root of 3 space cross times 3.31 cross times 10 to the power of negative 8 end exponent over denominator 4 end fraction space equals space 1.43 cross times 10 to the power of negative 8 end exponent cm space equals space 143 space pm.

Answered by Ramandeep | 14 Dec, 2020, 12:37: PM

Concept Videos