Question:
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 $: \sqrt{3}=1.73, \sqrt{2}=1.41]$
Solution:
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 .
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