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How to convert particles/mL to microgram/mL?
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Krister Sundelin
How to convert particles/mL to microgram/mL?
Yes Kåre surely it depends on their average size. The calculation formula is: Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm) If you take the gold nanoparticles true density: 19.32 g/cm3 Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3 Best regards
Yes Kåre surely it depends on their average size. The calculation formula is: Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm) If you take the gold nanoparticles true density: 19.32 g/cm3 Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3 Best regards
Dear Alan F Rawle , you are right: a) Pi is not a ratio, and b) 355/113 is closer to Pi than 22/7.
But, nearly everywhere there is a 'but'. The term '22/7' is well known to me as a simple ratio, which is very close to Pi. I remember, I have learned that '22/7' as a numerical substitute*) for Pi in a very early course of theoretical physics (last millenium).
One might also use your '355/113', but dividing by 7 is much easier done by head (mental arithmetic) or by pencil, at least for me, than by 113. You know, we had times without any electronic calculator. Incredible nowadays...
Dear Alan F Rawle , you are right: a) Pi is not a ratio, and b) 355/113 is closer to Pi than 22/7.
But, nearly everywhere there is a 'but'. The term '22/7' is well known to me as a simple ratio, which is very close to Pi. I remember, I have learned that '22/7' as a numerical substitute*) for Pi in a very early course of theoretical physics (last millenium).
One might also use your '355/113', but dividing by 7 is much easier done by head (mental arithmetic) or by pencil, at least for me, than by 113. You know, we had times without any electronic calculator. Incredible nowadays...
Gerhard Martens Yes, I think that all scientists will know that 22/7 is an approximation for pi. I was being a little facetious in suggesting 355/113 (I could have used the even better 52163/16604). I collect old text books (among other things) and a couple of the more interesting ones are:
The 'proof' that pi is 3 1/6 or 19/6 (The Measure of the Circle Perfected in January 1845 John Davis Providence 1854). There's no way that Mr Davis is convinced that he is wrong in the 156 pages of the book. Laborious calculations. Sad... I attach a Google Books download
Squaring the circle. Numerous examples
Yes, many times approximations are useful (we still teach Dalton's Atomic Theory in schools - all atoms are indivisible - even though it's patently untrue. It has uses.). Reminds me of Rothchild's Law: 'For every phenomenon, however complex, someone will eventually come up with a simple and elegant theory. This theory will be wrong'. Maybe in your field of XRD then the Scherrer equation for crystallite size by line broadening could be an example - sure, small size increases the broadening but there's also strain and instrumental broadening to consider. We can then work with Williamson-Hall or Rietveld...
Gerhard Martens Yes, I think that all scientists will know that 22/7 is an approximation for pi. I was being a little facetious in suggesting 355/113 (I could have used the even better 52163/16604). I collect old text books (among other things) and a couple of the more interesting ones are:
The 'proof' that pi is 3 1/6 or 19/6 (The Measure of the Circle Perfected in January 1845 John Davis Providence 1854). There's no way that Mr Davis is convinced that he is wrong in the 156 pages of the book. Laborious calculations. Sad... I attach a Google Books download
Squaring the circle. Numerous examples
Yes, many times approximations are useful (we still teach Dalton's Atomic Theory in schools - all atoms are indivisible - even though it's patently untrue. It has uses.). Reminds me of Rothchild's Law: 'For every phenomenon, however complex, someone will eventually come up with a simple and elegant theory. This theory will be wrong'. Maybe in your field of XRD then the Scherrer equation for crystallite size by line broadening could be an example - sure, small size increases the broadening but there's also strain and instrumental broadening to consider. We can then work with Williamson-Hall or Rietveld...
Mohammed Hussein j. H. Al'Atia You state 'Where π is the constant ratio = 22/7'. Pi is not a ratio - it is a completely irrational number that cannot be expressed as a fraction. If so, why not use 355/113 which is closer?
Mohammed Hussein j. H. Al'Atia You state 'Where π is the constant ratio = 22/7'. Pi is not a ratio - it is a completely irrational number that cannot be expressed as a fraction. If so, why not use 355/113 which is closer?
Dear Mayra Beltran Thank you for your feedback 10^-21 is the conversion factors from cm3 to nm3 10^6 is the conversion factors from ugm to gm Best regards
Dear Mayra Beltran Thank you for your feedback 10^-21 is the conversion factors from cm3 to nm3 10^6 is the conversion factors from ugm to gm Best regards
Yes Kåre surely it depends on their average size. The calculation formula is: Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm) If you take the gold nanoparticles true density: 19.32 g/cm3 Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3.
Is there any formula like the above for calculating microplastic particle conversion from particles/mL to microgram/mL?
Yes Kåre surely it depends on their average size. The calculation formula is: Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm) If you take the gold nanoparticles true density: 19.32 g/cm3 Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3.
Is there any formula like the above for calculating microplastic particle conversion from particles/mL to microgram/mL?
I am working with 3 sizes of gold nano particles 5nm, 20nm and 40nm with 4.92E+13 - 6.01E+13, 5.89E+11-7.19E+11 and 6.44E+10-7.87E+10 concentrations particles/ml respectively. Could you please help me to covert these concentrations to ug/ml?
I am working with 3 sizes of gold nano particles 5nm, 20nm and 40nm with 4.92E+13 - 6.01E+13, 5.89E+11-7.19E+11 and 6.44E+10-7.87E+10 concentrations particles/ml respectively. Could you please help me to covert these concentrations to ug/ml?
Ridhwan Haliq You need to open a separate question (as you have done) rather than confusing this thread. It is likely that you'll not get a reply here as people will not look at old questions generally.
Ridhwan Haliq You need to open a separate question (as you have done) rather than confusing this thread. It is likely that you'll not get a reply here as people will not look at old questions generally.
Yes Kåre surely it depends on their average size. The calculation formula is:
Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm)
If you take the gold nanoparticles true density: 19.32 g/cm3
Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3
Best regards
Yes Kåre surely it depends on their average size. The calculation formula is:
Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm)
If you take the gold nanoparticles true density: 19.32 g/cm3
Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3
Best regards
More
VOTE
Dear Alan F Rawle ,
you are right:
a) Pi is not a ratio, and
b) 355/113 is closer to Pi than 22/7.
But, nearly everywhere there is a 'but'.
The term '22/7' is well known to me as a simple ratio, which is very close to Pi.
I remember, I have learned that '22/7' as a numerical substitute*) for Pi in a very early course of theoretical physics (last millenium).
One might also use your '355/113', but dividing by 7 is much easier done by head (mental arithmetic) or by pencil, at least for me, than by 113. You know, we had times without any electronic calculator. Incredible nowadays...
*) According to the 'Background' section of
https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80
such a substitute by a ratio is called 'Diophantine approximation'
(I must admit, that I have never heard that term up to now)
https://en.wikipedia.org/wiki/Diophantine_approximation
Best regards
G.M.
Dear Alan F Rawle ,
you are right:
a) Pi is not a ratio, and
b) 355/113 is closer to Pi than 22/7.
But, nearly everywhere there is a 'but'.
The term '22/7' is well known to me as a simple ratio, which is very close to Pi.
I remember, I have learned that '22/7' as a numerical substitute*) for Pi in a very early course of theoretical physics (last millenium).
One might also use your '355/113', but dividing by 7 is much easier done by head (mental arithmetic) or by pencil, at least for me, than by 113. You know, we had times without any electronic calculator. Incredible nowadays...
*) According to the 'Background' section of
https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80
such a substitute by a ratio is called 'Diophantine approximation'
(I must admit, that I have never heard that term up to now)
https://en.wikipedia.org/wiki/Diophantine_approximation
Best regards
G.M.
More
VOTE
Gerhard Martens
Yes, I think that all scientists will know that 22/7 is an approximation for pi. I was being a little facetious in suggesting 355/113 (I could have used the even better 52163/16604). I collect old text books (among other things) and a couple of the more interesting ones are:
- The 'proof' that pi is 3 1/6 or 19/6 (The Measure of the Circle Perfected in January 1845 John Davis Providence 1854). There's no way that Mr Davis is convinced that he is wrong in the 156 pages of the book. Laborious calculations. Sad... I attach a Google Books download
- Squaring the circle. Numerous examples
Yes, many times approximations are useful (we still teach Dalton's Atomic Theory in schools - all atoms are indivisible - even though it's patently untrue. It has uses.). Reminds me of Rothchild's Law: 'For every phenomenon, however complex, someone will eventually come up with a simple and elegant theory. This theory will be wrong'. Maybe in your field of XRD then the Scherrer equation for crystallite size by line broadening could be an example - sure, small size increases the broadening but there's also strain and instrumental broadening to consider. We can then work with Williamson-Hall or Rietveld...Gerhard Martens
Yes, I think that all scientists will know that 22/7 is an approximation for pi. I was being a little facetious in suggesting 355/113 (I could have used the even better 52163/16604). I collect old text books (among other things) and a couple of the more interesting ones are:
- The 'proof' that pi is 3 1/6 or 19/6 (The Measure of the Circle Perfected in January 1845 John Davis Providence 1854). There's no way that Mr Davis is convinced that he is wrong in the 156 pages of the book. Laborious calculations. Sad... I attach a Google Books download
- Squaring the circle. Numerous examples
Yes, many times approximations are useful (we still teach Dalton's Atomic Theory in schools - all atoms are indivisible - even though it's patently untrue. It has uses.). Reminds me of Rothchild's Law: 'For every phenomenon, however complex, someone will eventually come up with a simple and elegant theory. This theory will be wrong'. Maybe in your field of XRD then the Scherrer equation for crystallite size by line broadening could be an example - sure, small size increases the broadening but there's also strain and instrumental broadening to consider. We can then work with Williamson-Hall or Rietveld...More
VOTE
Mehwish Fida
Your question has been answered separately: https://www.researchgate.net/post/How_to_calculate_the_concentration_of_gold_Nanoparticles
Mehwish Fida
Your question has been answered separately: https://www.researchgate.net/post/How_to_calculate_the_concentration_of_gold_Nanoparticles
More
VOTE
Thanks Dr Mohammed just a last question the numbers 10E-21 y 10E6 in the equation what that it's mean?. Thanks again.
Thanks Dr Mohammed just a last question the numbers 10E-21 y 10E6 in the equation what that it's mean?. Thanks again.
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VOTE
Mohammed Hussein j. H. Al'Atia
You state 'Where π is the constant ratio = 22/7'.
Pi is not a ratio - it is a completely irrational number that cannot be expressed as a fraction. If so, why not use 355/113 which is closer?
Mohammed Hussein j. H. Al'Atia
You state 'Where π is the constant ratio = 22/7'.
Pi is not a ratio - it is a completely irrational number that cannot be expressed as a fraction. If so, why not use 355/113 which is closer?
More
VOTE
Dear Mayra Beltran
Thank you for your feedback
10^-21 is the conversion factors from cm3 to nm3
10^6 is the conversion factors from ugm to gm
Best regards
Dear Mayra Beltran
Thank you for your feedback
10^-21 is the conversion factors from cm3 to nm3
10^6 is the conversion factors from ugm to gm
Best regards
More
VOTE
Yes Kåre surely it depends on their average size. The calculation formula is:
Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm)
If you take the gold nanoparticles true density: 19.32 g/cm3
Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3.
Is there any formula like the above for calculating microplastic particle conversion from particles/mL to microgram/mL?
Yes Kåre surely it depends on their average size. The calculation formula is:
Microgram/ml =2 x 10^11 (particles/ml) * gold density (gm/cm3) * (1/6) (22/7) * (average diameter of particles nm)3 *10^-21 (cm3/nm3) * 10^6 (ugm/gm)
If you take the gold nanoparticles true density: 19.32 g/cm3
Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3.
Is there any formula like the above for calculating microplastic particle conversion from particles/mL to microgram/mL?
More
VOTE
Yes Monique
Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3
Microgram/ml = 20.24^ -4 *(30nm)3
= 54.64
Best regards
Yes Monique
Microgram/ml = 20.24^ -4 *(average diameter of particles in nanometer)3
Microgram/ml = 20.24^ -4 *(30nm)3
= 54.64
Best regards
More
VOTE
Dear Dr Muhammad,
I am working with 3 sizes of gold nano particles 5nm, 20nm and 40nm with 4.92E+13 - 6.01E+13, 5.89E+11-7.19E+11 and 6.44E+10-7.87E+10 concentrations particles/ml respectively. Could you please help me to covert these concentrations to ug/ml?
Best regards
Mehwish
Dear Dr Muhammad,
I am working with 3 sizes of gold nano particles 5nm, 20nm and 40nm with 4.92E+13 - 6.01E+13, 5.89E+11-7.19E+11 and 6.44E+10-7.87E+10 concentrations particles/ml respectively. Could you please help me to covert these concentrations to ug/ml?
Best regards
Mehwish
More
VOTE
may I know the reference or Paper that uses the formula (Concentration Particle / ml into mg/ml) like your discussion in this link (https://www.researchgate.net/post/How-to-convert-particles-mL-to- microgram-mL) Prof Mohammed Hussein j. H. Al'Atia and Monique Engelbrecht
may I know the reference or Paper that uses the formula (Concentration Particle / ml into mg/ml) like your discussion in this link (https://www.researchgate.net/post/How-to-convert-particles-mL-to- microgram-mL) Prof Mohammed Hussein j. H. Al'Atia and Monique Engelbrecht
More
VOTE
Ridhwan Haliq You need to open a separate question (as you have done) rather than confusing this thread. It is likely that you'll not get a reply here as people will not look at old questions generally.
Ridhwan Haliq You need to open a separate question (as you have done) rather than confusing this thread. It is likely that you'll not get a reply here as people will not look at old questions generally.
More
VOTE