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In the lattice parameter of Si = 5.43 Å ...

In the lattice parameter of `Si = 5.43 Å` and the mass of `Si` atom is `28.08 xx 1.66 xx10^(-27) kg`, the density of silicon in `kg m^(-3)` is (Given: Silicon has diamondcubic structure)

A

`2330`

B

`1115`

C

`3445`

D

`1673`

Text Solution

Verified by Experts

The correct Answer is:
A
`Si` has fcc sturcture `(Z_(eff) = 4` and `Si` is also present in alternate `TVs (= 4)`. So tha total number of atoms
`= 4 + 4 = 8//"unit cell"`.
`:. rho =m (Z_(eff))/(N_(A) xx a^(3))` [Mas of `Si` atom `= (Aw)(N_(A))`]
`= (8 xx "Mass of Si atom")/(a^(3))`
`= (8 xx (28.08 xx 1.66 xx 10^(-27))kg)/((5.43 xx 10^(-10))^(3)m^(3)) [1Å = 10^(-10)m]`
`= 2330 kg m^(-3)`.
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