This spectroscopic technique can be used only to study atomic nuclei with an odd number of protons or neutrons (or both). This situation occurs in the atoms of 1H, 13C, 19F and 31P. Such cores are magnetically active, ie possess spin, like electrons, since the nuclei possess a positive charge and have a rotational movement about an axis which makes them behave like tiny magnets.

Thanks.

Dear Anjaly

Only complemented the response of Genaro, so that a given nucleus is visible by NMR are some very important factors, the first is that the nuclei being studied must possess spin magnetic moment nonzero if the nucleo does not possess it is invisivle by NMR. Another important factor is the natural abundance of the isotope to be studied. For example, let's think about the chemical element silicon (Si), this element has three natural isotopes: 28Si, 29Si and 30Si, with natural abundances (approximate values​​) of 92%, 5% and 2% respectively. The only isotope of Si that has non-zero nuclear spin is the 29 Si, that is, only 5% of all nuclei of Si present in a material contribute to the NMR signal.

Thank you for your information. Just one question.

How can we detect the presence of Si in compounds if we are having 28Si?

Hi,

As like all other NMR, 29Si-NMR (1/2 spin) will also tell you the chemical environment that surronded 'Si'..

you can also look at this below site which will give ou more information,

http://www.pascal-man.com/periodic-table/29Si.pdf

Si compounds naturally have 28Si, 29Si and 30Si is Mother Nature did well, so when the experiments are made of only the 29 Si NMR can be observed and the information you get is this neighborhood nucleus or better, like the neighbors affects the 29Si nucleus.