PSA powerpoint slide.

If the grounded cube has a side length of 10 micrometres then it would have a surface area of 60 micrometres^2. Which is equivalent to 60 x10^-12 metres^2.
To calculate the amount of particles I did 60 x10^-12/ 6x10^-4 (the original size of the MgO cube.) which came out to be 10^6 particles.
For the overall surface area I did the S.A of one particle x by the overall amount of particles so 60x10^-12 x 10^6 = 0.00006 metres^2.

This is the first time I have done this calculation so I don't feel confident in challenging my lecturers note calculations. But, they are different and I cannot see why?

Any suggestions?


ps. if you can't read the photo. it was said that 10^9 crystals or particles would be present in the grounded MgO. 1 particle having a surface area of 6x10^-10 m^2

Edited October 24, 201411 yr by shaneo

If the grounded cube has a side length of 10 micrometres then it would have a surface area of 60 micrometres^2. Which is equivalent to 60 x10^-12 metres^2.

 

No.

10 um = 10*10^-6 m

( 10*10^-6)^2 = 10^-10 m^2

6*10^-10 m^2 = 6*10^-10 m^2

six sides box

Oh it makes sense now. but, you missed out a decimal point "= 10^-10 m^2" on the second line should be "= 1.0^-10"

Thanks for your input!

Shaneo

Oh it makes sense now. but, you missed out a decimal point "= 10^-10 m^2" on the second line should be "= 1.0^-10"

 

Thanks for your input!

 

No, I didn't miss it..

 

1.0 to any power is still 1.0

To evaluate the number of grains, you'd like to compare volumes.