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Charandeep Matthai
Asked: 2022-11-06T15:50:54+05:30 In:

The atomic fraction `(d)` of tin in bronze (fcc) with a density of `7717 kg m^(-3)` and a lattice parameter of `3.903 Å` is `(Aw Cu = 63.54, Sn = 118.7, 1 amu = 1.66 xx 10^(-27 kg))`
A. `0.01`
B. `0.05`
C. `0.10`
D. `3.8`

The atomic fraction `(d)` of tin in bronze (fcc) with a density of `7717 kg m^(-3)` and a lattice parameter of `3.903 Å` is `(Aw Cu = 63.54, Sn = 118.7, 1 amu = 1.66 xx 10^(-27 kg))`
A. `0.01`
B. `0.05`
C. `0.10`
D. `3.8`
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  1. Correct Answer – B
    (Refer Section `1.13` alternative method)
    `rho=[{:[(sum(“Number of atom of each kind”)),(xx(“Mw of each kind”) xx 1.66 xx 10^(-27)kg)]:}]/a^(3)`
    `7717kgm^(-3) = [[{:((“Number of Sn atoms”)xx),((118.7xx1.66xx10^(-27))+),((“Number of Cu atoms”)xx),((63.54 xx 1.66 xx 10^(-27))):}]]/((3.903 xx 10^(-10))^(2)m^(3))`
    `276.4 = n_(Sn)(118.7) + n_(Cu)(63.54)`
    `4.35 = 1.86n_(Sn) + n_(Cu)`
    `n_(Cu) = 4 rArr n_(Sn) = 0.188`
    Atomic fraction `= (n_(Sn))/(n_(Sn) + n_(Cu)) = 0.05`

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