An elemental crystal has density of `8570 kg m^(-3)`. The packing efficiency is `0.67`. If the closest distance between neighbouring atoms is `2.86 Å`. The mass of one atom is `(1 amu = 1.66 xx 10^(-27))kg)`

Options

A. `186` amu
B. `93` amu
C. `46.5` amu
D. `43` amu

Answer

B

Explanation

Let the volume of unit cell `= V`
Volume occupied by atoms `= 0.68 V`
Volume occupied by atoms `= 0.68V`
Thus,
`Z_(eff) (4/3pir^(3)) = 0.68V`
Also, `2r = 2.86 Å rArr 1.43 Å`
`rho (Z_(eff) xx Aw)/(a^(3) xx N_(A)) (Aw/(N_(A))=` Mass of an atom in amu)
Applying another fourmula of `rho`.
(Refer Section 1.13, alternative method)
`rho = (Z_(eff) xx "Mass in amu" xx 1.66 xx 10^(-27)kg)/(V)`
`8570 kg m^(-3) = (0.68 xx m xx 1.66 xx 10^(-27)kg)/(4/3pi (1.43)^(3) xx 10^(-30))`
`:. m rArr 93 amu`