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The atomic fraction (d) of tin in bronze...

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`

Text Solution

Verified by Experts

The correct Answer is:
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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