Respuesta :
Answer : The magnetic moment of a single nickel atom is [tex]4.82\times 10^{-24}Am^2[/tex]
Explanation :
Formula used :
[tex]M_{max}=(\frac{N}{V})\times \mu[/tex]
where,
[tex]M_{max}[/tex] = saturation magnetization = [tex]4.41\times 10^5A/m[/tex]
N = Avogadro's number = [tex]6.022\times 10^{23}atoms/mol[/tex]
V = volume of nickel atom
[tex]\mu[/tex] = magnetic moment
First we have to calculate the volume of nickel atom.
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]8.90g/cm^3=\frac{58.71g/mol}{Volume}[/tex]
[tex]Volume=\frac{58.71g}{8.90g/cm^3}[/tex]
[tex]Volume=6.59cm^3=6.59\times (100)^3m^3[/tex]
Now we have to calculate the magnetic moment of a single nickel atom.
[tex]M_{max}=(\frac{N}{V})\times \mu[/tex]
[tex]4.41\times 10^5=(\frac{(6.022\times 10^{23})}{6.59})\times (100^3)\times \mu[/tex]
[tex]\mu=4.82\times 10^{-24}Am^2[/tex]
Therefore, the magnetic moment of a single nickel atom is [tex]4.82\times 10^{-24}Am^2[/tex]
