Answer

https://www.researchgate.net/post/Are_the_areas_of_EDX_peaks_of_different_elements_in_a_sample_correlated_to_their_percentage_abundance

There are a number of reasons for this:

(1) First and foremost - the intensity measured is related to the number of atoms, or the atomic fraction. Gold is about 16 times heavier than carbon. So if wt% of carbon is 28%, it isn't enough to have 28 C atoms per 100 atoms; you need many more C atoms to make up for the weight of (fewer, but heavier) Au atoms. The worksheet provided by Tran Thanh-Nhan

This should answer your specific question. However, there are additional problems if you want to make these measurements with any accuracy:

(2) EDS is generally called a semi-quantitative method, because accurate measurement of light elements is difficult. Carbon is one such element. You should take the numbers for elements like B,C,N with a pinch of salt, definitely not for quantitation/compositional analysis. The discussion shared by Peter Farrington explains this quite well.

(3) From the pattern, this looks like an SEM-EDS spectrum. If so, the measurement of C is all the more suspect, because patterns typically have a high background from the window, and this background has an energy very close to the carbon peak. You may like to subtract that from a reference pattern and verify. If this is from a TEM, you may ignore the last point.

(4) Concerns also arise from use of carbon tape, carbon coating (may not have been used here) and other carbonaceous contamination, so specimen preparation and mounting requires some care.

I am completely agreed with T. Thanh-Nhan's Answer.

The table shows the higher value of Gold (wt.%) in comparison to carbon, but the EDS peak of Gold is smaller. You will find the answer by converting from at.% to wt.%.

Dear H. Y. Chem, EDS spectrum displayed the peaks of all elements in atomic percentage (at.%). However, the wt.% of the elements is calculated with the help of at.%, it does not observe by EDX spectra. It depends on the atomic weight of the element.

Generally, EDX spectra are plotted as counts vs photon energy. I believe that is the case here. A point that has not been mentioned yet, which is important when trying to understand peak heights/areas, is that different elements have different cross sections (which is further possibly dependent on the electron microscope accelerating voltage). Simply put, different elements do not have the same likelyhood of going through the processes that ultimately leads to emission of an X-ray photon. There is also a bunch of instrumental factors that might influence the size of a given peak, such as collection efficiencies, angular dependence of emission etc. All of those things are normally calibrated to produce some sensitivity factors that convert counts to at%/wt%. If you can trust the calibration, you don't need to worry much about the peak size but can rely on the quantification.

Dear Hooi-Ying,

The ionization efficiency is maximum for a value of the accelerating voltage approximately between two and three times the energy of the peak considered. If the accelerating voltage of the microscope is too low it could be difficult to ionize the gold peak which can then be little intense (even non-existent if the accelerating voltage is too low). The very low energy peak of the carbon needs only few KVs to appears (even 500eV) and may be is quite intense because the accelerating voltage is quite low.