Answer:
- The radius of a rhodium atom, [tex]r = 134.55pm[/tex]
Explanation:
From,
Density, [tex]d = \frac{Z*atomic mass}{N_o*a^3*10^{-30}}[/tex]
where
Z = non of atoms in a unit cell = 4
[tex]N_o[/tex] = Avagrado number = [tex]6.023*10^{23}[/tex]
a = edge length
Atomic mass of Rhodium = [tex]103 g/mol[/tex]
density [tex]d = 12410 kg/m^3 = 12.41 g/cm^3[/tex]
Therefore,
[tex]a^3 = \frac{Z*atomic mass}{N_o*10^{-30}*d}\\\\a^3 = \frac{4*103}{6.023*10^{23}*10^{-30}*12.41}\\\\a^3 = 55120426.76\\\\a = 380.57 pm[/tex]
For fcc arrangement,
[tex]r = \frac{\sqrt2}{4}*a\\\\r = \frac{\sqrt2}{4}*380.57\\\\r = 134.55pm[/tex]
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