A domain in ferromagnetic iron is in the form of a ube of side length `2mum` then the number of iron atoms in the domain are `("Molecular mass of iron"= 55gmol^(-1) and "density"=7.92 g cm^(-3))`

Options

A. `6.92xx10^(12)"atoms"`
B. `6.92xx10^(11)"atoms"`
C. `6.92xx10^(10)"atoms"`
D. `6.92xx10^(13)"atoms"`

Answer

B

Explanation

The volume of the cubic domain
`V=(2mu m)^(3)=(2xx10^(-6)m)^(3)`
`=8xx10^(-18)m^(3)=8xx10^(-12)cm^(3)`
and mass = volume `xx` density
`= 8 xx 10^(-12) cm^(3) xx 7.9 g cm^(-3)`
`=63.2xx10^(-12)g`
Now the Avagadro number `(6.023xx10^(23))` of iron atoms have a mass of 55g. Hence the number of atoms in the domain are
`N=(63.2xx10^(-12)xx6.023xx10^(23))/(55)=6.92xx10^(11)"atoms"`