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Answer 1 · 2018-02-17T09:12:44

$1.310976 \times 10^{- 23} J/T$ $1.310976xx 10 ^-23"J/T"$

The magnetic orbital moment is given by

$μ_{orb} = - g_{L} (\frac{e}{2 m_{e}} L)$ $mu_"orb" = -g_"L"(e/(2m_e)"L")$

where $L$ $"L"$ is orbital angular momentum

$| L | = \sqrt{l (l + 1)} \frac{h}{2 π}$ $|"L"| = sqrt(l(l+1))h/(2pi)$

hyperphysics.phy-astr.gsu.edu

$g_{L}$ $g_L$ is the electron orbital g- factor which is equal to 1

$l$ $l$ for a the ground state orbital or 1s orbital is 0

so the magnetic orbital moment is also 0

sites.google.com

$l$ $l$ for the 4p orbital is 1

$μ_{orb} = - g_{L} (\frac{e}{2 m_{e}} \sqrt{l (l + 1)} \frac{h}{2 π})$ $mu_"orb" = -g_"L"(e/(2m_e)sqrt(l(l+1))h/(2pi))$

$μ_{orb} = - g_{L} (\frac{e}{2 m_{e}} \sqrt{1 (1 + 1)} \frac{h}{2 π})$ $mu_"orb" = -g_"L"(e/(2m_e)sqrt(1(1+1))h/(2pi))$

A unit of magnetic moment called the "Bohr magneton" is introduced here.

$μ_{B} = \frac{e ¯ h}{2 m_{e}} \approx 9.27 \times 10^{24} J/T$ $mu_"B" = (ebarh)/(2m_e) ~~ 9.27xx10^24"J/T"$

$¯ h = \frac{h}{2 π}$ $barh = h/(2pi)$

$\frac{e}{2 m_{e}} = \frac{μ_{B}}{¯ h}$ $e/(2m_e) = mu_"B"/barh$

$μ_{orb} = - g_{L} (\frac{e}{2 m_{e}} \sqrt{1 (1 + 1)} \frac{h}{2 π})$ $mu_"orb" = -g_"L"(e/(2m_e)sqrt(1(1+1))h/(2pi))$

$= - μ_{B} \sqrt{2}$ $=-mu_"B" sqrt(2)$

$- 9.27 \times 10^{- 24} J/T \times \sqrt{2}$ $-9.27xx10^-24"J/T" xx sqrt(2)$

$\approx - 9.27 \times 10^{- 24} J/T \times 1.41421356237$ $~~-9.27xx10^-24"J/T" xx 1.41421356237$

$- 1.310976 \times 10^{- 23} J/T$ $-1.310976xx 10 ^-23"J/T"$

So the difference between them is
$1.310976 \times 10^{- 23} J/T$ $1.310976xx 10 ^-23"J/T"$