Question

Assuming no exit aperture, what is the theoretical resolving power of a magnetic sector field with a mass dispersion of ∂M = 0.950m, an object width of OS = 250µm and magnification d=1? Give resolving power as R = m/∆m. Is R sufficient to resolve the isobars 87Rb and 87Sr?Hint: 87Rb = 86.909187u, 87Sr = 86.908884u

Answer 1

Resolving power in mass spectrometry is defined asthe ability of an instrument to distinguish the two peaks at m/z valuesdiffering by small amounts. Basically mass resolving power is the ability how fat the peak, the bigger the number, more skinner the peak and it is ability of one peak to take another peak beneath it. Mass resolving power and resolution are two related terms, more the resolving power, the more likely you have resolved the two peaks.

Two cases given resolving power can be calculated by:

R = m/∆m

m = mass of the ion

∆m = peak width at a height which is a specified fraction of the maximum peak height(50% of full peak height) in cm

For Rb

R_Rb = 86.909187/0.025= 3476.37

R_Sr = 86.908884/0.025 = 3476.35

As we can see value of resolving power for Rb and Sr are too close.

Resolving power alone can not help in finding out the peaks we also need m/z ratio

Charge on Rb is +1 and on Sr is +2

So m/z ratio of Rb = 86.909187/1 = 86.909187

m/z ratio of Sr = 86.908884 /2 = 43.4544

Only R is not sufficient to resolve the isobars 87Rb and 87Sr, to get the two peaks seperately we need resoving power and resolution both. Resolution tells us how apart our peak from each other. And resolving power tells us the shape of our peak big or small.