Question

Suppose that the saturation magnetization Mmax of the ferromagnetic metal nickel is 4.41 × 105 A/m. Calculate the magnetic moment of a single nickel atom. (The density of nickel is 8.90 g/cm3 and its molar mass is 58.71 g/mol.)

Answer 1

Answer : The magnetic moment of a single nickel atom is $4.82\times 10^{-24}Am^2$

Explanation :

Formula used :

$M_{max}=(\frac{N}{V})\times \mu$

where,

$M_{max}$ = saturation magnetization = $4.41\times 10^5A/m$

N = Avogadro's number = $6.022\times 10^{23}atoms/mol$

V = volume of nickel atom

$\mu$ = magnetic moment

First we have to calculate the volume of nickel atom.

$Density=\frac{Mass}{Volume}$

$8.90g/cm^3=\frac{58.71g/mol}{Volume}$

$Volume=\frac{58.71g}{8.90g/cm^3}$

$Volume=6.59cm^3=6.59\times (100)^3m^3$

Now we have to calculate the magnetic moment of a single nickel atom.

$M_{max}=(\frac{N}{V})\times \mu$

$4.41\times 10^5=(\frac{(6.022\times 10^{23})}{6.59})\times (100^3)\times \mu$

$\mu=4.82\times 10^{-24}Am^2$

Therefore, the magnetic moment of a single nickel atom is $4.82\times 10^{-24}Am^2$