- #1
alfredbester
- 40
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Hi, I appear to have lost some mass/density.
Q: Calculate the atom density, n, of aluminium at 600k.
Data given (some of it not relevant to this part of the ? but i'll post it all).
The debye temperature [tex]T_D = 428 K[/tex] . The velocity of sound at room temp, [tex] v = 5100m s^{-1}[/tex]. The interatomic spacing a = 405pm, and its relative atomic mass is 27.
I found the atom density n to be:
[tex] n = (KT_D / \hbar v )^3.(1 / 6\pi^2)[/tex]
I used [tex]{\omega_m}^3 = 6\pi^2 v^3 n[/tex], and the fact [tex]{\omega_m} = K T_D / \hbar[/tex] to get my equations for n.
I assumed the atom density was the same at 600k as at room temperature (the way the question was worded I couldn't see any other method). Plugging the numbers in I found [tex]n = 2.24x10^{28} m^{-3}[/tex].
Then I'm asked to compare the density found with aluminiums true density of [tex]2700 kg m^{-3}[/tex] and explain any difference.
My density is just the atom density multiplied by the atomic mass (assuming the mass is just contained within the aluminium).
Therefore
[tex]/rho = n m(amu) = n = 2.24x10^{28} * (27 / 6.022x10^{26})) = 1000 kg m^{-3}[/tex].
There in lies my problem. I'd expect my approximations to overestimate the density if anything.
Q: Calculate the atom density, n, of aluminium at 600k.
Data given (some of it not relevant to this part of the ? but i'll post it all).
The debye temperature [tex]T_D = 428 K[/tex] . The velocity of sound at room temp, [tex] v = 5100m s^{-1}[/tex]. The interatomic spacing a = 405pm, and its relative atomic mass is 27.
I found the atom density n to be:
[tex] n = (KT_D / \hbar v )^3.(1 / 6\pi^2)[/tex]
I used [tex]{\omega_m}^3 = 6\pi^2 v^3 n[/tex], and the fact [tex]{\omega_m} = K T_D / \hbar[/tex] to get my equations for n.
I assumed the atom density was the same at 600k as at room temperature (the way the question was worded I couldn't see any other method). Plugging the numbers in I found [tex]n = 2.24x10^{28} m^{-3}[/tex].
Then I'm asked to compare the density found with aluminiums true density of [tex]2700 kg m^{-3}[/tex] and explain any difference.
My density is just the atom density multiplied by the atomic mass (assuming the mass is just contained within the aluminium).
Therefore
[tex]/rho = n m(amu) = n = 2.24x10^{28} * (27 / 6.022x10^{26})) = 1000 kg m^{-3}[/tex].
There in lies my problem. I'd expect my approximations to overestimate the density if anything.
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