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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` |
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Answer» Correct Answer - 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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