# Electron Spin Resonance (ESR) Experiment: the two peaks

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I am unable to figure out how we get two peaks in the Electron Spin Resonance (ESR) experiment, in which we use Helmholtz coils as B source with modulation, and a basic unit as a source of radio waves.

This post imported from StackExchange Physics at 2014-05-04 11:25 (UCT), posted by SE-user kalkanistovinko
retagged May 4, 2014
ESR is electron spin rotation?

This post imported from StackExchange Physics at 2014-05-04 11:25 (UCT), posted by SE-user rob
electron spin resonance

This post imported from StackExchange Physics at 2014-05-04 11:25 (UCT), posted by SE-user kalkanistovinko

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You really need to provide more detailed information on your experimental if you hope to get a response. In what follows, I will guess the details of your setup (since you did not provide much), based on a similar setup I once used.

Assuming that the word "modulation" refers to Zeeman modulation created by superimposing a DC and AC voltage on the Helmholtz coil, and that a fixed-frequency radio wave source is used, the amplitude of the Zeeman-modulation-induced fluctuations in radio-wave intensity after passing through the sample is given by $$\Delta I\approx\frac{dI}{dH}\Delta H$$ where $I(H)$ is the absorption intensity as a function of field-strength $H$ and $\Delta H$ is the depth of the Zeeman modulation. Assuming that $\Delta H\ll\sigma$, where $\sigma$ is the width of the peaks, this means that the fluctuation signal measured is actually proportional to $dI/dH$, rather than $I(h)$.

For a Lorentz profile, this gives the following lineshape:

expr = D[1/(x^2 + 1), x];
Plot[expr, {x, -5, 5}]


This is basically the origin of the double-peak shape associated with Zeeman-modulated detection. In essence, you're detecting the derivative of the spectrum, rather than the spectrum itself.

This post imported from StackExchange Physics at 2014-05-04 11:25 (UCT), posted by SE-user DumpsterDoofus
answered Apr 29, 2014 by (50 points)

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