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Homework Statement
A column of water contains fine metal particles of radius 20nm, which are in thermal
equilibrium at 25°C.
The density of the metal is [tex]2\times10^{4} kg m^{-3}[/tex].
If there are 1000 particles per unit volume at given height, what will the particle density per unit volume be at a position of 1 mm higher?
Homework Equations
Stated within the question / solution attempt.
The Attempt at a Solution
For each particle:
Radius = [tex]20\times10^{-9}m[/tex]
Hence Volume = [tex]\frac{4}{3} \pi (20\times10^{-9}m^{3} = 3.351\times^{-23}m^{3}[/tex]
Therefore since density is mass divided by volume, [tex]\rho = \frac{m}{V}[/tex]:
[tex]m_{total} = \rho m = (2\times10^{4} kg m^{-3})(3.351\times^{-23}m^{3}) = 6.702\times^{-19}kg [/tex] which is the total contained mass.
It is given that at any height there are 1000 particles, therefore:
[tex]m_{particle} = \frac{6.702\times^{-19}kg }{1000} = 6.702\times^{-22} kg[/tex]
I thought this would probably help somehow, not sure how anymore though.
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