Solve this following


In the ground state of atomic $\mathrm{Fe}(\mathrm{Z}=26)$, the spin-only magnetic moment is

$\times 10^{-1} \mathrm{BM}$. (Round off to the Nearest Integer).

$[$ Given $: \sqrt{3}=1.73, \sqrt{2}=1.41]$


Number of unpaired $\mathrm{e}^{-}=4$

$\mu=\sqrt{4(4+2)}$ B.M.

$\mu=\sqrt{24}$ B.M.

$\mu=4.89$ B.M.

$\mu=48.9 \times 10^{-1}$ B.M.

Nearest integer value will be 49 .


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now