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The frequency, $\\nu$, of an electron spin transition of an unknown inorganic radical is measured to be {Frequency_MHz} MHz in a magnetic field, B, of {magfield} T.

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(i) First, note that;

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\$\\frac{\\Delta E}{h}=\\frac{g\\mu_BB}{h} =\\nu\$

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in an electron spin resonance experiment. Therefore;

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\$\\frac{h\\nu}{\\mu_BB}=g\$

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so;

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\$\\frac{6.62607~\\times10^{-34}~{\\rm~J~s}~\\times~\\times~\\var{Frequency_mantissa}~\\times~10^\\var{Frequency_log}~{\\rm~Hz}}{9.274 \\times 10^{-24}{\\rm ~J~T^{-1}}\\times \\var{magfield}{\\rm~T}}=\\var{Lande_g}\$

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(ii) Note that;

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\$E=h\\nu\$

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so

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\$6.62607~\\times~10^{-34}~{\\rm J~s}~\\times\\var{Frequency_mantissa}~\\times~10^\\var{Frequency_log}~{\\rm~Hz}=\\var{photon_energy_mantissa}\\times10^{\\var{photon_energy_log}} {\\rm J}\$

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What is the Lande g-factor for the unpaired electron in this radical?

[[0]]$\\times$10[[1]]