I had prepared nano particles, I checked that using SEM, it shows spherical nanoparticles .... but when measuring its surface area it has no surface area ( the curve showed negative values)...
I know that by going to smaller size, surface area must increase..
What is happening??? Kindly could anybody help?
Thanks in advance.
I can only suggest that the nanoparticles degraded in course of sample preparation for surface area measurements. This could be due to heating, degassing etc.
Dear Farah:
Specific characteristics of dispersion is the curvature of the surface H. It is determined by the derivative of the surface area by volume of H = 1 / 2dS / dV. The curvature can be positive or negative. If the surface is convex, the positive curvature. If the surface is concave, the negative curvature. Your particles has more concave surface that convexs.
Regards.
thanks all ^^
Prof. Mirgorod , your answer was so helpful, thanks again .
prof. Mirgorod could you please explain more about negative surface area....
Is it a solid compound?
are you using Nitrogen adsorption (BET method) or another method for determination of surface area?
yes, I am using BET method and my sample is solid.
Sample Mass: 0.0247g
Analysis Adsorptive: N2
Analysis Temp.: -197.075 °C
Degassing temp. 300 under negative pressure.
Can someone explain why I am having negative adsorption isotherms , and the final surface area is 3.00 m2/g or 0.00 m2/g ,, ?
there are two possibilities from my point of view either the opinion of Dr.Vladimir K Ivanov that your nanoparticles broke down releasing gases under the conditions of degassing so the net uptake of Nitrogen gas is negative
or
the apparatus itself has technical problems (i have seen it before) that the technician did not perform maintenance periodically, so you can send a reference sample (a sample you know its surface area) to check the accuracy of the apparatus.
Thank you Dr. Mahmoud Hazem, I will check the accuracy of the apparatus.
I think it is experimental error
you can check this from the y plotting the Relative Pressure (P/Po) against 1/[Q(Po/P - 1)] and it should give a straight line with R2 (correlation coefficient), 1
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